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This topic has expert replies
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Geometry
by [email protected] » Mon Jun 02, 2014 9:04 pm
Hello, Pls help with this question
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by [email protected] » Tue Jun 03, 2014 12:47 am
Hi shibsriz,
I'm going to give you a couple of hints and let you attempt this question again:
1) To calculate the perimeter of triangle ABE, you'll NEED some actual numbers to work with.
2) Draw a picture with a radius running from the center of the circle to point A and another running to point B. With the knowledge that arc AB = 120 degrees, what do you know about arc AE and arc BE (hint: how many degrees are in a circle)?
GMAT assassins aren't born, they're made,
Rich
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by theCodeToGMAT » Tue Jun 03, 2014 12:55 am
{C}?
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R A H U L
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by sukriti2hats » Tue Jun 03, 2014 1:54 am
Let the centre of the circle be O.
AO=BO=EO because they represent the radius.
Statement 1- measure of arc AB is 120 degrees.
The angle AOB is 120. The angles AOE and BOE will also be equal to 120. So the measure of arc AE = BE =120.
In triangles AOB , AOE and BOE, sides AO=BO=EO because they are radii and angle AOB=BOE=AOE=120 degrees. So by side-angle -side condition these three triangles are congruent. Thus the sides of the triangle, AB =AE=BE. It is an equilateral triangle.
But as Rich said, in order to find out the perimeter we need actual numbers to work with, statement 1 is Insufficient.
Statement 2- AB is 2 units
Knowing the length of AB alone will not help in finding the perimeter of the triangle.
So statement 2 is also Insufficient.
Togther using 1 and 2, we can find out by 1 that triangle is equilateral and then by using the information that AB=2 we can find out the perimeter i.e. AB+BE+AE= 2+2+2=6
So the correct answer is C.
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Search a number
593 is a prime number
BaseRepresentation
bin1001010001
3210222
421101
54333
62425
71505
oct1121
9728
10593
1149a
12415
13368
14305
15298
hex251
593 has 2 divisors, whose sum is σ = 594. Its totient is φ = 592.
The previous prime is 587. The next prime is 599. The reversal of 593 is 395.
It is a balanced prime because it is at equal distance from previous prime (587) and next prime (599).
It can be written as a sum of positive squares in only one way, i.e., 529 + 64 = 23^2 + 8^2 .
593 is a truncatable prime.
It is a cyclic number.
It is not a de Polignac number, because 593 - 24 = 577 is a prime.
It is a Sophie Germain prime.
It is a Leyland number of the form 92 + 29.
It is a Curzon number.
It is a plaindrome in base 11 and base 13.
It is a nialpdrome in base 5.
It is not a weakly prime, because it can be changed into another prime (599) by changing a digit.
It is a good prime.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 296 + 297.
It is an arithmetic number, because the mean of its divisors is an integer number (297).
It is an amenable number.
593 is a deficient number, since it is larger than the sum of its proper divisors (1).
593 is an equidigital number, since it uses as much as digits as its factorization.
593 is an evil number, because the sum of its binary digits is even.
The product of its digits is 135, while the sum is 17.
The square root of 593 is about 24.3515913238. The cubic root of 593 is about 8.4013981044.
Subtracting from 593 its sum of digits (17), we obtain a square (576 = 242).
It can be divided in two parts, 5 and 93, that multiplied together give a triangular number (465 = T30).
The spelling of 593 in words is "five hundred ninety-three", and thus it is an aban number and an oban number.
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# Normal Distribution in Python
Normal distribution is also called as Gaussian distribution or Laplace-Gauss distribution.
## Normal Distribution with Python Example
Normal distribution represents a symmetric distribution where most of the observations cluster around the central peak called as mean of the distribution. The parameter used to measure the variability of observations around the mean is called as standard deviation. The probabilities for values occurring near mean are higher than the values far away from the mean. Normal distribution is a probability distribution plot. The parameters of the normal distribution plot defining the shape and the probabilities are mean and standard deviation. The area of the plot between two different points in the normal distribution plot represents the probability of the value occurring between those two points.
Here are some of the properties of normal distribution of the population:
• The points in the normal distribution are symmetric. Normal distribution can not be used to model skewed distributions.
• The mean, median and mode of normal distribution are equal.
• Half of the population is less than the mean and half is greater than the mean.
• The empirical rule of the normal distribution goes like the following: 68% of the observations fall within +/- 1 standard deviation from the mean, 95% of the observations fall within +/- 2 standard deviation from the mean and 99.7% of the observations fall within +/- 3 standard deviation from the mean.
Here is the probability density function for normal distribution:
Fig 1. Normal distribution probability density function
In above function, μμ represents the mean and σσ represents the standard deviation. Given different values of random variable (x), one could calculate the probability using the above probability density function.
Here is a sample probability distribution plot representing normal distribution with a mean of 5 and standard deviation of 10. The plot is created for random variable taking values between -100 and 100.
Fig 2. Normal Distribution Plot
The following code can be used to generate above normal distribution plot.
```#
# Create a normal distribution with mean as 5 and standard deviation as 10
#
mu = 5
std = 10
snd = stats.norm(mu, std)
#
# Generate 1000 random values between -100, 100
#
x = np.linspace(-100, 100, 1000)
#
# Plot the standard normal distribution for different values of random variable
# falling in the range -100, 100
#
plt.figure(figsize=(7.5,7.5))
plt.plot(x, snd.pdf(x))
plt.xlim(-60, 60)
plt.title('Normal Distribution (Mean = 5, STD = 10)', fontsize='15')
plt.xlabel('Values of Random Variable X', fontsize='15')
plt.ylabel('Probability', fontsize='15')
plt.show()```
Here is the code representing multiple normal distribution plots which looks like the following:
The following code can be used to create above shown multiple normal distribution plots having different means and standard deviation.
```#
# Values of random variable
#
x = np.linspace(-10, 10, 100)
#
plt.figure(figsize=(7.5,7.5))
#
# Normal distribution with mean 0 and std as 1
#
plt.plot(x, stats.norm(0, 1).pdf(x))
#
# Normal distribution with mean 1 and std as 0.75
#
plt.plot(x, stats.norm(1, 0.75).pdf(x))
#
# Normal distribution with mean 2 and std as 1.5
#
plt.plot(x, stats.norm(2, 1.5).pdf(x))
plt.xlim(-10, 10)
plt.title('Normal Distribution', fontsize='15')
plt.xlabel('Values of Random Variable X', fontsize='15')
plt.ylabel('Probability', fontsize='15')
plt.show()```
## Standard Normal Distribution with Python Example
Standard Normal Distribution is normal distribution with mean as 0 and standard deviation as 1.
Here is the Python code and plot for standard normal distribution. Note that the standard normal distribution has a mean of 0 and standard deviation of 1. Pay attention to some of the following in the code below:
The following is the Python code used to generate the above standard normal distribution plot. Pay attention to some of the following in the code given below:
• Scipy Stats module is used to create an instance of standard normal distribution with mean as 0 and standard deviation as 1 (stats.norm)
• Probability density function pdf() is invoked on the instance of stats.norm to generate probability estimates of different values of random variable given the standard normal distribution
```import numpy as np
import matplotlib.pyplot as plt
from scipy import stats
#
# Create a standard normal distribution with mean as 0 and standard deviation as 1
#
mu = 0
std = 1
snd = stats.norm(mu, std)
#
# Generate 100 random values between -5, 5
#
x = np.linspace(-5, 5, 100)
#
# Plot the standard normal distribution for different values of random variable
# falling in the range -5, 5
#
plt.figure(figsize=(7.5,7.5))
plt.plot(x, snd.pdf(x))
plt.xlim(-5, 5)
plt.title('Normal Distribution', fontsize='15')
plt.xlabel('Values of Random Variable X', fontsize='15')
plt.ylabel('Probability', fontsize='15')
plt.show()```
## Conclusions
Here is the summary of what you learned in this post in relation to Normal distribution:
• Normal distribution is a symmetric probability distribution with equal number of observations on either half of the mean.
• The parameters representing the shape and probabilities of the normal distribution are mean and standard deviation
• Python Scipy stats module can be used to create a normal distribution with meand and standard deviation parameters using method norm.
• Standard normal distribution is normal distribution with mean as 0 and standard deviation as 1.
• In normal distribution, 68% of observations lie within 1 standard deviation, 95% of observations lie within 2 standard deviations and 99.7% observations lie within 3 standard deviations from the mean.
Even if you are not in the field of statistics, you must have come across the term “Normal Distribution”.
A probability distribution is a statistical function that describes the likelihood of obtaining the possible values that a random variable can take. By this, we mean the range of values that a parameter can take when we randomly pick up values from it.
A probability distribution can be discrete or continuous.https://imasdk.googleapis.com/js/core/bridge3.476.0_en.html#goog_1402046895
Suppose in a city we have heights of adults between the age group of 20-30 years ranging from 4.5 ft. to 7 ft.
If we were asked to pick up 1 adult randomly and asked what his/her (assuming gender does not affect height) height would be? There’s no way to know what the height will be. But if we have the distribution of heights of adults in the city, we can bet on the most probable outcome.
## What is Normal Distribution?
Normal Distribution is also known as a Gaussian distribution or famously Bell Curve. People use both words interchangeably, but it means the same thing. It is a continuous probability distribution.
The probability density function (pdf) for Normal Distribution:
where, μ = Mean , σ = Standard deviation , x = input value.
Terminology:
• Mean – The mean is the usual average. The sum of total points divided by the total number of points.
• Standard Deviation – Standard deviation tells us how “spread out” the data is. It is a measure of how far each observed value is from the mean.
Looks daunting, isn’t it? But it is very simple.
### 1. Example Implementation of Normal Distribution
Let’s have a look at the code below. We’ll use numpy and matplotlib for this demonstration:
```# Importing required libraries
import numpy as np
import matplotlib.pyplot as plt
# Creating a series of data of in range of 1-50.
x = np.linspace(1,50,200)
#Creating a Function.
def normal_dist(x , mean , sd):
prob_density = (np.pi*sd) * np.exp(-0.5*((x-mean)/sd)**2)
return prob_density
#Calculate mean and Standard deviation.
mean = np.mean(x)
sd = np.std(x)
#Apply function to the data.
pdf = normal_dist(x,mean,sd)
#Plotting the Results
plt.plot(x,pdf , color = 'red')
plt.xlabel('Data points')
plt.ylabel('Probability Density')```
### 2. Properties of Normal Distribution
The normal distribution density function simply accepts a data point along with a mean value and a standard deviation and throws a value which we call probability density.
We can alter the shape of the bell curve by changing the mean and standard deviation.
Changing the mean will shift the curve towards that mean value, this means we can change the position of the curve by altering the mean value while the shape of the curve remains intact.
The shape of the curve can be controlled by the value of Standard deviation. A smaller standard deviation will result in a closely bounded curve while a high value will result in a more spread out curve.
Some excellent properties of a normal distribution:
• The mean, mode, and median are all equal.
• The total area under the curve is equal to 1.
• The curve is symmetric around the mean.
Empirical rule tells us that:
• 68% of the data falls within one standard deviation of the mean.
• 95% of the data falls within two standard deviations of the mean.
• 99.7% of the data falls within three standard deviations of the mean.
It is by far one of the most important distributions in all of the Statistics. The normal distribution is magical because most of the naturally occurring phenomenon follows a normal distribution. For example, blood pressure, IQ scores, heights follow the normal distribution.
## Calculating Probabilities with Normal Distribution
To find the probability of a value occurring within a range in a normal distribution, we just need to find the area under the curve in that range. i.e. we need to integrate the density function.
Since the normal distribution is a continuous distribution, the area under the curve represents the probabilities.
Before getting into details first let’s just know what a Standard Normal Distribution is.
A standard normal distribution is just similar to a normal distribution with mean = 0 and standard deviation = 1.
`Z = (x-μ)/ σ`
The z value above is also known as a z-score. A z-score gives you an idea of how far from the mean a data point is.
If we intend to calculate the probabilities manually we will need to lookup our z-value in a z-table to see the cumulative percentage value. Python provides us with modules to do this work for us. Let’s get into it.
### 1. Creating the Normal Curve
We’ll use `scipy.norm` class function to calculate probabilities from the normal distribution.
Suppose we have data of the heights of adults in a town and the data follows a normal distribution, we have a sufficient sample size with mean equals 5.3 and the standard deviation is 1.
This information is sufficient to make a normal curve.
```# import required libraries
from scipy.stats import norm
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sb
# Creating the distribution
data = np.arange(1,10,0.01)
pdf = norm.pdf(data , loc = 5.3 , scale = 1 )
#Visualizing the distribution
sb.set_style('whitegrid')
sb.lineplot(data, pdf , color = 'black')
plt.xlabel('Heights')
plt.ylabel('Probability Density')```
The `norm.pdf( )` class method requires `loc` and `scale` along with the data as an input argument and gives the probability density value. `loc` is nothing but the mean and the `scale` is the standard deviation of data. the code is similar to what we created in the prior section but much shorter.
### 2. Calculating Probability of Specific Data Occurance
Now, if we were asked to pick one person randomly from this distribution, then what is the probability that the height of the person will be smaller than 4.5 ft. ?
The area under the curve as shown in the figure above will be the probability that the height of the person will be smaller than 4.5 ft if chosen randomly from the distribution. Let’s see how we can calculate this in python.
The area under the curve is nothing but just the Integration of the density function with limits equals -∞ to 4.5.
`0.211855 or 21.185 %`
The single line of code above finds the probability that there is a 21.18% chance that if a person is chosen randomly from the normal distribution with a mean of 5.3 and a standard deviation of 1, then the height of the person will be below 4.5 ft.
We initialize the object of class `norm` with mean and standard deviation, then using `.cdf( )` method passing a value up to which we need to find the cumulative probability value. The cumulative distribution function (CDF) calculates the cumulative probability for a given x-value.
Cumulative probability value from -∞ to ∞ will be equal to 1.
Now, again we were asked to pick one person randomly from this distribution, then what is the probability that the height of the person will be between 6.5 and 4.5 ft. ?
```cdf_upper_limit = norm(loc = 5.3 , scale = 1).cdf(6.5)
cdf_lower_limit = norm(loc = 5.3 , scale = 1).cdf(4.5)
prob = cdf_upper_limit - cdf_lower_limit
print(prob)```
`0.673074 or 67.30 %`
The above code first calculated the cumulative probability value from -∞ to 6.5 and then the cumulative probability value from -∞ to 4.5. if we subtract cdf of 4.5 from cdf of 6.5 the result we get is the area under the curve between the limits 6.5 and 4.5.
Now, what if we were asked about the probability that the height of a person chosen randomly will be above 6.5ft?
It’s simple, as we know the total area under the curve equals 1, and if we calculate the cumulative probability value from -∞ to 6.5 and subtract it from 1, the result will be the probability that the height of a person chosen randomly will be above 6.5ft.
`0.115069 or 11.50 %.`
That’s a lot to sink in, but I encourage all to keep practicing this essential concept along with the implementation using python.
The complete code from above implementation:
```# import required libraries
from scipy.stats import norm
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sb
# Creating the distribution
data = np.arange(1,10,0.01)
pdf = norm.pdf(data , loc = 5.3 , scale = 1 )
#Probability of height to be under 4.5 ft.
prob_1 = norm(loc = 5.3 , scale = 1).cdf(4.5)
print(prob_1)
#probability that the height of the person will be between 6.5 and 4.5 ft.
cdf_upper_limit = norm(loc = 5.3 , scale = 1).cdf(6.5)
cdf_lower_limit = norm(loc = 5.3 , scale = 1).cdf(4.5)
prob_2 = cdf_upper_limit - cdf_lower_limit
print(prob_2)
#probability that the height of a person chosen randomly will be above 6.5ft
cdf_value = norm(loc = 5.3 , scale = 1).cdf(6.5)
prob_3 = 1- cdf_value
print(prob_3)```
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Enter the total number of grams of a substance, and the moles per gram of that substance to calculate the total number of moles
## Grams to moles formula
The following formula is used to convert the total number of grams of a substance to the total number of moles.
m = M * mpg
• Where m is the total number of moles
• M is the total mass in grams
• mpg is the moles per gram of the substance
For this formula, the mass and moles per unit pass could be switched to any unit you have. For example, if you have a mass of 2 lbs, you must calculate or determine the number of moles per pound, and then the total number of moles can be calculated.
## Grams to Moles Definition
Converting grams to moles is as simple as multiplying the total mass by the moles per unit of mass.
## How to convert grams to moles
The following is a step-by-step guide on how one might convert the total amount of grams of a substance to the total number of moles. For this example, we will assume we have a solution with two different substances, at 65% and 35% concentration by weight.
1. The first step is to calculate the total number of grams for the substance we want to convert to moles. It’s assumed that the total weight is 100 grams and we are trying to find the weight of the 65% concentration by weight substance. Therefore, we must multiply 100*65% = grams.
2. Next, the total number of moles per gram of the substance must be either calculated or measured. This is solely determined by the elements in the substance. The learn more about how this is calculated, visit this blog post. For this example, we are going to assume that we know this to be 1000 moles per gram. Note that this number is very very low, and only used to make this simpler.
3. The final step is the plug the information we gathered above into the formula, or calculator if you’re feeling lazy. We find m = 65 * 1000 = 65,000 moles.
4. Analyze the results to see if the answer makes logical sense. if this was a real problem, in this step you would find that the answer does not make logical sense because the total moles in a gram of substance should be orders of magnitude greater than what it was in this problem.
## FAQ
What is grams to moles?
Converting grams to moles is as simple as multiplying the total mass by the moles per unit of mass.
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# Christopher Dougherty EC220 - Introduction to econometrics (chapter 10) Slideshow: introduction to maximum likelihood estimation Original citation: Dougherty,
## Presentation on theme: "Christopher Dougherty EC220 - Introduction to econometrics (chapter 10) Slideshow: introduction to maximum likelihood estimation Original citation: Dougherty,"— Presentation transcript:
1 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION This sequence introduces the principle of maximum likelihood estimation and illustrates it with some simple examples. L p
2 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Suppose that you have a normally-distributed random variable X with unknown population mean and standard deviation , and that you have a sample of two observations, 4 and 6. For the time being, we will assume that is equal to 1. L p
3 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Suppose initially you consider the hypothesis = 3.5. Under this hypothesis the probability density at 4 would be 0.3521 and that at 6 would be 0.0175. L p p(4) p(6) 3.50.35210.0175 0.3521 0.0175
4 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION The joint probability density, shown in the bottom chart, is the product of these, 0.0062. p(4) p(6) L 3.50.35210.01750.0062 L p 0.3521 0.0175
5 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Next consider the hypothesis = 4.0. Under this hypothesis the probability densities associated with the two observations are 0.3989 and 0.0540, and the joint probability density is 0.0215. p(4) p(6) L 3.50.35210.01750.0062 4.00.39890.05400.0215 L p 0.3989 0.0540
6 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Under the hypothesis = 4.5, the probability densities are 0.3521 and 0.1295, and the joint probability density is 0.0456. p(4) p(6) L 3.50.35210.01750.0062 4.00.39890.05400.0215 4.50.35210.12950.0456 L p 0.3521 0.1295
7 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Under the hypothesis = 5.0, the probability densities are both 0.2420 and the joint probability density is 0.0585. p(4) p(6) L 3.50.35210.01750.0062 4.00.39890.05400.0215 4.50.35210.12950.0456 5.00.24200.24200.0585 L p 0.2420
8 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Under the hypothesis = 5.5, the probability densities are 0.1295 and 0.3521 and the joint probability density is 0.0456. p(4) p(6) L 3.50.35210.01750.0062 4.00.39890.05400.0215 4.50.35210.12950.0456 5.00.24200.24200.0585 5.50.12950.35210.0456 L p 0.3521 0.1295
9 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION The complete joint density function for all values of has now been plotted in the lower diagram. We see that it peaks at = 5. p(4) p(6) L 3.50.35210.01750.0062 4.00.39890.05400.0215 4.50.35210.12950.0456 5.00.24200.24200.0585 5.50.12950.35210.0456 p L 0.1295 0.3521
10 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Now we will look at the mathematics of the example. If X is normally distributed with mean and standard deviation , its density function is as shown.
11 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION For the time being, we are assuming is equal to 1, so the density function simplifies to the second expression.
12 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Hence we obtain the probability densities for the observations where X = 4 and X = 6.
13 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION The joint probability density for the two observations in the sample is just the product of their individual densities. joint density
14 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION In maximum likelihood estimation we choose as our estimate of the value that gives us the greatest joint density for the observations in our sample. This value is associated with the greatest probability, or maximum likelihood, of obtaining the observations in the sample. joint density
15 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION In the graphical treatment we saw that this occurs when is equal to 5. We will prove this must be the case mathematically. p(4) p(6) L 3.50.35210.01750.0062 4.00.39890.05400.0215 4.50.35210.12950.0456 5.00.24200.24200.0585 5.50.12950.35210.0456 p L 0.1295 0.3521
16 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION To do this, we treat the sample values X = 4 and X = 6 as given and we use the calculus to determine the value of that maximizes the expression.
17 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION When it is regarded in this way, the expression is called the likelihood function for , given the sample observations 4 and 6. This is the meaning of L( | 4,6).
18 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION To maximize the expression, we could differentiate with respect to and set the result equal to 0. This would be a little laborious. Fortunately, we can simplify the problem with a trick.
19 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION log L is a monotonically increasing function of L (meaning that log L increases if L increases and decreases if L decreases).
20 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION It follows that the value of which maximizes log L is the same as the one that maximizes L. As it so happens, it is easier to maximize log L with respect to than it is to maximize L.
21 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION The logarithm of the product of the density functions can be decomposed as the sum of their logarithms.
22 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Using the product rule a second time, we can decompose each term as shown.
23 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Now one of the basic rules for manipulating logarithms allows us to rewrite the second term as shown.
24 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION log e is equal to 1, another basic logarithm result. (Remember, as always, we are using natural logarithms, that is, logarithms to base e.)
25 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Hence the second term reduces to a simple quadratic in X. And so does the fourth.
26 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION We will now choose so as to maximize this expression.
27 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Quadratic terms of the type in the expression can be expanded as shown.
28 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Thus we obtain the differential of the quadratic term.
29 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Applying this result, we obtain the differential of log L with respect to . (The first term in the expression for log L disappears completely since it is not a function of .)
30 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Thus from the first order condition we confirm that 5 is the value of that maximizes the log-likelihood function, and hence the likelihood function.
31 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Note that a caret mark has been placed over , because we are now talking about an estimate of , not its true value.
32 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Note also that the second differential of log L with respect to is -2. Since this is negative, we have found a maximum, not a minimum.
33 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION We will generalize this result to a sample of n observations X 1,...,X n. The probability density for X i is given by the first line.
34 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION The joint density function for a sample of n observations is the product of their individual densities.
35 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Now treating the sample values as fixed, we can re-interpret the joint density function as the likelihood function for , given this sample. We will find the value of that maximizes it.
36 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION We will do this indirectly, as before, by maximizing log L with respect to . The logarithm decomposes as shown.
37 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION We differentiate log L with respect to .
38 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION The first order condition for a minimum is that the differential be equal to zero.
39 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Thus we have demonstrated that the maximum likelihood estimator of is the sample mean. The second differential, -n, is negative, confirming that we have maximized log L.
40 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION So far we have assumed that , the standard deviation of the distribution of X, is equal to 1. We will now relax this assumption and find the maximum likelihood estimator of it.
41 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION We will illustrate the process graphically with the two-observation example, keeping fixed at 5. We will start with equal to 2. L p
42 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION With equal to 2, the probability density is 0.1760 for both X = 4 and X = 6, and the joint density is 0.0310. L p p(4) p(6) L 2.00.17600.17600.0310
43 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Now try equal to 1. The individual densities are 0.2420 and so the joint density, 0.0586, has increased. L p p(4) p(6) L 2.00.17600.17600.0310 1.00.24200.24200.0586
44 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Now try putting equal to 0.5. The individual densities have fallen and the joint density is only 0.0117. L p p(4) p(6) L 2.00.17600.17600.0310 1.00.24200.24200.0586 0.50.10800.10800.0117
45 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION The joint density has now been plotted as a function of in the lower diagram. You can see that in this example it is greatest for equal to 1. p(4) p(6) L 2.00.17600.17600.0310 1.00.24200.24200.0586 0.50.10800.10800.0117 L p
46 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION We will now look at this mathematically, starting with the probability density function for X given and .
47 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION The joint density function for the sample of n observations is given by the second line.
48 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION As before, we can re-interpret this function as the likelihood function for and , given the sample of observations.
49 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION We will find the values of and that maximize this function. We will do this indirectly by maximizing log L.
50 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION We can decompose the logarithm as shown. To maximize it, we will set the partial derivatives with respect to and equal to zero.
51 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION When differentiating with respect to , the first two terms disappear. We have already seen how to differentiate the other terms.
52 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Setting the first differential equal to 0, the maximum likelihood estimate of is the sample mean, as before.
53 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Next, we take the partial differential of the log-likelihood function with respect to .
54 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Before doing so, it is convenient to rewrite the equation.
55 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION The derivative of log with respect to is 1/ . The derivative of --2 is –2 --3.
56 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Setting the first derivative of log L to zero gives us a condition that must be satisfied by the maximum likelihood estimator.
57 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION We have already demonstrated that the maximum likelihood estimator of is the sample mean.
58 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Hence the maximum likelihood estimator of the population variance is the mean square deviation of X.
59 INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION Note that it is biased. The unbiased estimator is obtained by dividing by n – 1, not n.
INTRODUCTION TO MAXIMUM LIKELIHOOD ESTIMATION However it can be shown that the maximum likelihood estimator is asymptotically efficient, in the sense of having a smaller mean square error than the unbiased estimator in large samples. 60
Copyright Christopher Dougherty 2011. These slideshows may be downloaded by anyone, anywhere for personal use. Subject to respect for copyright and, where appropriate, attribution, they may be used as a resource for teaching an econometrics course. There is no need to refer to the author. The content of this slideshow comes from Section 10.6 of C. Dougherty, Introduction to Econometrics, fourth edition 2011, Oxford University Press. Additional (free) resources for both students and instructors may be downloaded from the OUP Online Resource Centre http://www.oup.com/uk/orc/bin/9780199567089/http://www.oup.com/uk/orc/bin/9780199567089/. Individuals studying econometrics on their own and who feel that they might benefit from participation in a formal course should consider the London School of Economics summer school course EC212 Introduction to Econometrics http://www2.lse.ac.uk/study/summerSchools/summerSchool/Home.aspx http://www2.lse.ac.uk/study/summerSchools/summerSchool/Home.aspx or the University of London International Programmes distance learning course 20 Elements of Econometrics www.londoninternational.ac.uk/lsewww.londoninternational.ac.uk/lse. 11.07.25
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DENSITY
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# DENSITY - PowerPoint PPT Presentation
DENSITY. Density = amount of matter per unit volume. D = m/v (g/cm 3 ) Mass usually expressed in grams Volume usually expressed in cm 3 or ml etc. M ÷ ÷ D X V. Mass =. Density =. Volume =. M V. M D. The “DMV” Triangle for Volume, Mass, and Density. D x V.
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DENSITY
• Density = amount of matter per unit volume
D = m/v (g/cm3)
• Mass usually expressed in grams
• Volume usually expressed in cm3 or ml etc.
M
÷ ÷
DXV
Mass=
Density =
Volume=
M
V
M
D
The “DMV” Triangle forVolume, Mass, and Density
DxV
What would take up more space??? A kilogram of feathers…..or a kilogram of steel??
OR
How close the atoms or molecules are to each other
More than “heaviness” - density includes how much space an object takes up!!
All substances have density including liquids, solids, and gases
Gases
• How much kinetic energy do the molecules have??
• The greater the kinetic energy
• ……the greater the volume
• …… and the less dense that gas is!!
• Therefore, cold air is more dense than warm air
Low pressure weather system means warmer air tends to rise,
High pressure systems indicate a colder more dense air mass that will……. SINK!!!
LIQUIDS
• The more dissolved solids in a solution, the more dense (such as ocean water)
• Cold water in lakes tend to sink (this creates a constant mixing of water, nutrients, and other substances)
• Kinetic energy again!!
Denser layers to less dense layers…..
What would happen????
• Mercury density = 13600kg/m3
Solids
Ice vs. water…..
SOLIDS
• Ice is less dense than water (which is why lakes and ponds have a thin layer of ice covering in winter, with water underneath)
• Various rocks, woods, metals have a characteristic density specific to that substance
Wouldn’t you like to have a bunch of THIS dense material?
Factors affecting Density
• Temperature
• Pressure
Factors affecting Density
• Dissolved solids – in liquids
• Concentration and kind of substances
DETERMINING DENSITY
• Regular Shapes – mass, then determine the volume by formula
EX: cubes, cylinders, spheres, cones, etc.
• Irregular shapes – mass, then measure displacement of a liquid (usually water) by that irregularly shaped object
Add water to a predetermined level - record.
Gently drop in the irregularly shaped object.
Subtract the first water level from the second – this is the volume
Density Table
SINK or FLOAT
In Water (D = 1.0 g/mL)
Float
Float
Float
Sink
Sink
Sink
Float
(alcohol)
Float
(fuel)
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# How to solve a problem on river speed
In tasks on addition of speeds the motion of bodies happens, as a rule, uniform and rectilinear and is described by the simple equations. Nevertheless, these tasks can be carried to the most difficult problems of mechanics. At the solution of such tasks use the rule of addition of classical speeds. To understand the principle of the decision, it is better to consider it on concrete examples of tasks.
## Instruction
1. Example on the rule of addition of speeds. Let speed currents of the river v0, and speed of the boat crossing this river concerning water be equal to v1 and is directed perpendicular to the coast (cm figure 1). The boat at the same time participates in two independent movements: it for some time of t crosses the river width of N with v1 speed concerning water and for the same time bears her downstream the river on l distance. As a result the boat floats a way of S with v speed concerning the coast, equal on the module: v equally in a root in v1, kvardratny from expression, in a square + v0 in a square for same is a high time for t. Therefore it is possible to write down the equations which solve similar problems: H=v1t, l = v0t? S = root square of expression: v1 in a square + v0 in a square increased by t.
2. Other type of such tasks asks questions: under what corner to the coast to the dolena do to row the oarsman in the boat to appear on the opposite coast, having passed the minimum way during a crossing? For what time will this way be passed? With what speed will the boat pass this way? To answer these questions it is necessary to make the drawing (cm rice 2). It is obvious that the minimum way which there can pass the boat, crossing the river, is equal to width of the river of N. To float this way, the oarsman has to direct the boat under such corner and to a breg, at kotorm the vector of absolute speed of boat v will be directed perpendicular to the coast. Then from a rectangular triangle it is possible to find: cos a=v0/v1. From here it is possible to take a corner and. To determine speed from the same triangle by Pythagorean theorem: v = root square of expression: v1 in a square - v0 in a square. And at last time of t for which the boat will cross the river width of N, moving with v speed, there will be t=H/v.
Author: «MirrorInfo» Dream Team
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# math
explain how to solve r/-3 = 12?
1. 👍 0
2. 👎 0
3. 👁 73
1. r/-3 = 12
Multiply both sides by -3
r = -36
1. 👍 0
2. 👎 0
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PDA
View Full Version : Tricky Simulation
WeeG
06-19-2009, 03:45 PM
Hi again, I got a tricky simulation to do in R...
In a kindergarden there are 12 kids. Every day at lunchtime, they sit randomly next to 3 tables, 4 kids next to each. What is the probabilty that in 10 days, each kid will sit at least once in the same table with each one of the other kids ?
hint: I need to use the function sample(), like: sample(1:10), .....
cheers
Tart
06-22-2009, 09:05 PM
Hey WeeG,
I always hate those type of problems, combinatorics always gave me a headache, so, please verify everything I write below, I'm not entirely certain that what I do is right. Also, I think my way of solving this problem is not very good one, and maybe unnecessary complicated one. I'm very curious to see other solutions. Anyway check everything I do, see if it makes sense.
n.sim <- 100 #number of simulations
n.times <- numeric(n.sim) #place holder for number of times each kid will sit
#at least once in the same table with each one of the other kids
# PRIME NUMBRS FOR KIDS ASSIGNMENTS
kids <- c(2, 3, 5, 7,
11, 13, 17, 19,
23, 29, 31, 37)
for (i in 1:n.sim){
table.1 <- matrix(0, 10, 4) #initiate table 1 10 days 4 kids
table.2 <- matrix(0, 10, 4) #initiate table 2 10 days 4 kids
table.3 <- matrix(0, 10, 4) #initiate table 3 10 days 4 kids
#populate Tables for 10 days
for (j in 1:10){
positions <- sample(kids, replace=FALSE)
table.1[j,] <-positions[1:4]
table.2[j,] <-positions[5:8]
table.3[j,] <-positions[9:12]
} # next j
Table <- data.frame(Table.1=apply(table.1, 1, prod),
Table.2=apply(table.2, 1, prod),
Table.3 =apply(table.3, 1, prod))
n.times[i] <- sum(diff(sort(rowSums(Table)))==0)
} # next i
Est.Prob <- sum(n.times)/n.sim
Now to explain what is going on in here.
First I assign kids first 12 prime numbers. Primes are really nice, if sums of primes are equal then all primes composing those sums are equal. Same true for products. I wanted to represent ordering by single number - so this is why I used primes. In Table variable I create a summary that holds all kids ordering for each table for 10 days. numbers are 'unique' since they are product of primes so if we encounter sum of these numbers at some other day, it means that kids are all the same at all tables. I'm doing this so I just have to compare singe number instead of 4 numbers (4 kids at table). Difference of sorted sum equal to zero will tell us how many times this happens in 10 days sum(diff(sort(rowSums(Table)))==0). I might be forgetting some of my number theory and some other math. But I think it is correct, or I hope it is :-)
Once again, I'm sure there is a simple solution. This is just first thing that came to my mind, and I didn't explore other ways of solving it.
To make things fancy to to show convergence with increase of sample size
rm(list=ls())
# number of simulations
n.sims <- c(10, 50, 100, 250, 500, 1000, 1500, 2000, 3000, 5000, 7000, 10000)
Est.Prob <- numeric(length(n.sims))
kids <- c(2, 3, 5, 7,
11, 13, 17, 19,
23, 29, 31, 37)
for (i in 1:length(n.sims)){
n.sim <- n.sims[i]
n.times <- numeric(n.sim)
for (j in 1:n.sim){
table.1 <- matrix(0, 10, 4)
table.2 <- matrix(0, 10, 4)
table.3 <- matrix(0, 10, 4)
for (k in 1:10){
positions <- sample(kids, replace=FALSE)
table.1[k,] <-positions[1:4]
table.2[k,] <-positions[5:8]
table.3[k,] <-positions[9:12]
} #next k
Table <- data.frame(Table.1=apply(table.1, 1, prod),
Table.2=apply(table.2, 1, prod),
Table.3 =apply(table.3, 1, prod))
n.times[j] <- sum(diff(sort(rowSums(Table)))==0)
} # next j
Est.Prob[i] <- sum(n.times)/n.sims[i]
} # next i
plot(Est.Prob~n.sims, type='o', pch=21, col='blue', bg='yellow',
ylab='Estimated Probability', xlab='Sample Size', ylim=c(0,0.05),
Mike White
06-23-2009, 04:30 AM
My attempt is shown below. I have used the combn function to get the permutaions of pairs at each table, these are then put in a matrix representing the number of meetings between each pair. If any pairs have not met after 10 days the corresponding matrix entry will be zero. The number of failures is then counted by checking the matrix for zeros after each simulation.
My results are slightly different so I may have made an error somewhere.
rm(list=ls())
# set seed so that same results can be reproduced
set.seed(1)
# give children names A to M
children<-LETTERS[1:12]
# set number of days
days<-10
# initiate variable to count days when children do not all meet each other
failures<-0
# set number of simulations
N<-1000
# set up loop for each simulation
for(n in 1:N){
# for each simulation set up matrix to record meetings between children
mat<-matrix(data=0, nrow=length(children), ncol=length(children))
colnames(mat)<-children
rownames(mat)<-children
# set diagonals to 1 - all children meet themselves!
diag(mat)<-1
for ( d in 1: days){
# select children for each table - could also use one sample of 12 and take 1:4, 5:8 and 9:12 for the tables
table1<-sample(children,4, replace=F)
table2<-sample(children[!children %in% table1], 4, replace=F)
table3<-children[!children %in% c(table1, table2)]
# determine pair combinations for Table 3 and put in upper and lower triangles of matrix
x<-combn(table1,2)
for (i in 1:ncol(x)){
mat[x[1,i],x[2,i]]<-mat[x[1,i],x[2,i]]+1
mat[x[2,i],x[1,i]]<-mat[x[2,i],x[1,i]]+1
}
# determine pair combinations for Table 3 and put in upper and lower triangles of matrix
x<-combn(table2,2)
for (i in 1:ncol(x)){
mat[x[1,i],x[2,i]]<-mat[x[1,i],x[2,i]]+1
mat[x[2,i],x[1,i]]<-mat[x[2,i],x[1,i]]+1
}
# determine pair combinations for Table 3 and put in upper and lower triangles of matrix
x<-combn(table3,2)
for (i in 1:ncol(x)){
mat[x[1,i],x[2,i]]<-mat[x[1,i],x[2,i]]+1
mat[x[2,i],x[1,i]]<-mat[x[2,i],x[1,i]]+1
} #
} # end for d
# check if any pairs of children have not met
failures<-failures+1*any(mat==0)
} # N
# estimate probability of all meeting each other
failures
# [1] 947
est.prob<-(N-failures)/N
est.prob
# [1] 0.053
Mike White
06-23-2009, 06:52 AM
Hi Tart
I have looked at your solution and have realised that our answers differ because we have different interpretations of the question!
I think that the question is ambiguous. I took the question to mean what is the probability of each of the 12 kids over 10 days will eventually have sat at any table with all the other 11 other kids.
Tart
06-23-2009, 10:38 AM
each kid will sit at least once in the same table with each one of the other kids ?
I interpreted it as one of the arrangements repeated itself, in 10 days, with only difference in table, ie. Table1 - 1 2 3 4 Table2 - 5 6 7 8, Table3 9 10 11 12
same arrangement would repeat it self, 5 6 7 8 - 1 2 3 4- 9 10 11 12 but without specific table in mind.
WeeG
06-24-2009, 03:43 AM
hey !
thanks for trying ! I didn't have time to check your answers yet, I will do it later on today, just one thing about the question ( I saw it wasn't clear ). I need the probability that each one of the kids will sit at least once in the same table with each one of the other kids. At first he can sit with kid 1,3 and 5, on the next day with 2,7,8, and so on, after the time is over, he will sit at least once with each one of his classmates.
mp83
06-25-2009, 05:47 AM
I always hate those type of problems, combinatorics always gave me a headache
Yeah, I know how you feel...
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# math
posted by .
1/4v=5
Use the substitution method to solve this problem.
• math -
1/4v=5
v = 5 / (1/4)
v = 5 * 4
v = 20
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2. ### Substitution Method-Plz help
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## Simple interest
#### Simple interest
1. The sum required to earn a monthly interest of Rs 400 at 10 % per annum at simple interest is
1. Rs 2000
2. Rs 12000
3. Rs 24000
4. Rs 48000
1. Simple Interest =(Principal × Rate × Time)/100
##### Correct Option: D
Total interest needed in a year = Rs 400 × 12
= Rs 4800
Principal = (100 × SI)/R × T
where, R = Rate
T = Time
SI= Simple Interest
1. In what time will the simple interest on Rs 400 at 10% per annum be the same as the simple interest on Rs 1000 for 4 year at 4 % per annum?
1. 2 yrs
2. 3 yrs
3. 4 yrs
4. 6 yrs
1. Simple Interest = (Principal × Rate × Time)/100
##### Correct Option: C
Here , P= Rs 1000
T= 4 yrs
R= 4 %
where, P= Principal
T= Time
R= Rate
Since , Simple Interest on Rs 1000=(1000 × 4 × 4)/100
= Rs 160
now, simple interest=Rs 160
P = Rs 400
R = 10 %
then, T=(100 × SI)/P × R
= (100 × 160)/(400 × 10)
= 4 yr
1. At what rate percent per annum will a sum of money double in 8 yr?
1. 12 %
2. 12.5 %
3. 13 %
4. 15 %
1. Amount A = 2 × P
where , P = Principal
Rate R = (100 × SI)/(P × T)
where , SI= Simple Interest
T= Time
##### Correct Option: B
Let Sum = P, Then SI=P
As Amount A = 2 × P
where , P = Principal
Rate R = (100 × SI)/(P × T)
= (100 × P)/(P × 8) %
= 12.5 %
where , SI= Simple Interest
T= Time
1. The amount instalment will discharge on debit of Rs 3220 due in 4 year at 10 % simple interest?
1. 500
2. 600
3. 700
4. None of these
1. Let the amount instalment be Rs ' x '
Then According to question,
(Amount of 'x' for 3 yrs) + (Amount of 'x' for 2 yrs) + (Amount of 'x' for 1 yrs) + x =3220
##### Correct Option: C
Let the amount instalment be Rs ' x '
Then According to question,
(Amount of 'x' for 3 yrs) + (Amount of 'x' for 2 yrs) + (Amount of 'x' for 1 yrs) + x =3220
or, [x+(x × 10 × 3)/100] + [x+(x × 10 × 2)/100] + [x+(x × 10 × 1)/100] + x=3220
⇒ 4x+ (30x/100)+(20x/100)+(10x/100)=3220
⇒ 460x=322000
⇒ x=Rs 700
∴ Each Instalment= Rs 700
1. A certain sum amounts to Rs 1586 in 2 year and Rs 1729 in 3 year. Find the rate and the sum.
1. 8 %
2. 9 %
3. 10 %
4. 11 %
1. Simple Interest in 1 year= Rs (1729 - 1586)
= Rs 143
And R= (100 × SI)/(P × T)
where, R = rate
SI= Simple Interest
P = Principal
T= Time
##### Correct Option: D
Simple Interest in 1 year= Rs (1729 - 1586)
= Rs 143
now, SI in 2 year = Rs 286
Principal P= Rs(1586 - 286)
= 1300
And R= (100 × SI)/(P × T)
= (100 × 143)/(1300 × 1)
= 11 %
where, R = rate
SI= Simple Interest
P = Principal
T= Time
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# How many ways are there to build a tower of 5 cubes height, out of red, yellow, blue, and green cubes, such that:
How many ways are there to build a tower of 5 cubes height, out of red, yellow, blue, and green cubes, such that at least one of each pair of adjacent cubes is green or blue?
Hey everyone. I first thought about solving this using a recursion relation but then realised it might be better to solve it using the Inclusion-Exclusion principle.
My attempt: Define $A_i$ - there is no blue/green cube among the pair of cubes: cube $i$ and cube $i+1$. $1\le i\le 4$
We are looking for $|\bigcap_{i=1}^4 \overline {A_i}|=|\overline {\bigcup_{i=1}^4 A_i}|$
$|\overline{\bigcup_{i=1}^4 A_i}|$= $4^5-4(2^2)+6(2^3)-4(2^4)+2^5$
where $4^5$ is the number of ways to build the tower without any restraints, etc. Is this correct, am I doing something wrong? Edit: Yes this is indeed not correct and yes I am indeed wrong :)
• Where does the number $4(2^2)$ come from? – Y. Forman Feb 6 '18 at 15:03
• @Y.Forman $4C1$ times $2^2$- for each $A_i$ there are 2 choices available (either red or yellow) for each cube in places i, i+1. Is this not correct? – Noy Perel Feb 6 '18 at 15:06
• Have you checked the result? My instinct tells me that it might be incorrect ;) I mean, compare $4^5$ with your final result... – 57Jimmy Feb 6 '18 at 15:10
• @NoyPerel See the first bullet point in my answer. – Y. Forman Feb 6 '18 at 15:12
• @57Jimmy Yeah, it makes no sense then. oops :( – Noy Perel Feb 6 '18 at 15:13
The approach is good, but two things should be fixed:
• The number of ways to color cubes outside of $i,i+1$ is not considered. E.g., you claim there are $2^2$ ways to color cases where $i,i+1$ are not green/blue. There are $2$ choices for each of cubes $i,i+1$ but also four choices for each of the other three cubes, so there should be $2^24^3$ ways of coloring this case.
• There are two ways in which two pairs of cubes can be colored without green/blue: three consecutive cubes colored not green/blue (your $+6\dots$ term), and two separate pairs colored without green/blue. You don't account for the second type.
• Thank you very much for your help. I still end up with a result that doesn't make sense though so I guess I still haven't grasp the idea. I ended up with the expression $4^5-4(2^2 4^3)+6(2^3 4^2+2^4 4)-4(2^4 4)+2^5$ – Noy Perel Feb 6 '18 at 15:39
• @NoyPerel There aren't $6$ ways to have two separate pairs, there are only $4$. – Y. Forman Feb 6 '18 at 15:42
• Thank you! So is it $4^5-4(2^2 4^3)+(6(2^3 4^2)+4(2^4 4))-4(2^4 4)+2^5$ then? – Noy Perel Feb 6 '18 at 15:53
• @NoyPerel You might also have to fix the three-pairs term, since three pairs can also occur (partially) separately – Y. Forman Feb 6 '18 at 15:54
• @NoyPerel I may have misled you... of the 6 double pairs, there are 3 consecutive triples and 3 separate pairs. Of the 4 triple pairs, there are 2 consecutive quadruples and 2 triple/double setups. The answer should be $4^5 - 42^24^3 +(32^34^2+32^44)-(22^44+22^5)+2^5$ – Y. Forman Feb 6 '18 at 18:08
Here is another approach: Denote by $a_n$ the number of admissible towers of height $n$. Then we have the recursion $$a_0=1, \quad a_1=4,\qquad a_n=2a_{n-1}+4a_{n-2}\quad(n\geq2)\ .$$ (In order to build an admissible tower of height $n$ begin with a blue or a green cube, and erect on it an admissible tower of height $n-1$; or begin with a red or a yellow cube, top it by a blue or a green cube, and erect on them an admissible tower of height $n-2$.)
The recursion can be solved using the "Master Theorem". One obtains $$a_n={5+3\sqrt{5}\over 10}\bigl(1+\sqrt{5}\bigr)^n +{5-3\sqrt{5}\over 10}\bigl(1-\sqrt{5}\bigr)^n\ .$$ In particular $a_5=416$.
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# Math
posted by on .
An architect designs a house that is 12 m wide. The rafters holding up the roof are equal length and meet at an angle of 68°. The rafters extend 0.6 m beyond the supporting wall. How long are the rafters?
Hi - I was wondering if you could help me with this question. I'm not sure what they mean by rafters extending beyond the support wall and how I would be able to figure out the length if it extends.
• Math - ,
base of right triangle = 12/2 = 6 m
angle at peak = 68/2 = 34 deg
we need the hypotenuse then add .6 m to each rafter
sin 34 = 6/L
L = 10.7
11.3 meters per rafter
• Math - ,
Thank you for clarifying! Also appreciate how quickly you helped me. Very good math skills :)
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# What is the domain of $x^{2x}$
What is the domain of $$f(x)=x^{2x}$$?
If $$f(x)=(x^2)^x$$then $$f$$ is defined for every real number but if $$f(x)=(x^x)^2$$ then $$f$$ is only defined and "nice" (excluding the negative $$-p/q$$ fractions) for positive real numbers.
Should we say $$f(x)=e^{2x\log(x)}$$ is only defined for positive $$x$$?
Thanks
• For $f(x)=(x^x)^2$, the function is defined for all real numbers except 0. And for $f(x) = e^{2xlog(x)}$, the domain is only positive numbers. – harshit54 Nov 17 '18 at 13:36
• @harshit54 Really? What is $f(x)=(x^x)^2$ for $x=-\frac14$? – Servaes Nov 17 '18 at 13:38
• @Servaes Okay, sorry. So it's defined for all positive reals, and negative integers. – harshit54 Nov 17 '18 at 14:02
I would say that it's defined only for positive numbers. Let's look at a simpler problem: what is the domain of $$x^\frac12$$? I can say "I could always write it as $$(x^2)^\frac14$$." The issue is order of operations. Unless you have parantheses, you need to calculate the exponent first. See for example https://en.wikipedia.org/wiki/Order_of_operations#Serial_exponentiation
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output.to from Sideway
Draft for Information Only
# Content
`Equilibrium of Rigid Body Free-Body Free-Body Diagram`
# Equilibrium of Rigid Body
A rigid body is in static equilibrium state when the rigid body remains stationary and is not accelerating in any way. In other words, a rigid body is in equilibrium when both the resultant force and the resultant moment are zero for all the forces and couples acting on it. Through system of forces transformation, an equivalent system of forces can be reduced to a force vector and a couple vector. Therefore the resultant force and the resultant couple vectors acting on the rigid body are equal to zero. That is:
and
Or in terms of rectangular components:
and
In other words, there is no translational and rotational motion when an object is in static equilibrium state.
## Free-Body
In engineering static application, a rigid body is in equilibrium when the systems of external forces cause no translational and rotational motion to the rigid body since the resultant force and resultant memont on the rigid body are zero. For example, a hanging mass from the ceiling
When solving the equilibrium problem of the considered body, in order to focus only on the interested body instead of the whole system, the interested body should be isolated from other bodies as a free body. According to Newton's third law, each applied force will have a reaction force of equal in magnitude and opposite in direction acting on the interacting bodies. Therefore, all the surrounding uninterested objects can be stripped away by replacing with the corresponding reaction force acting on the interesed body. For example, the interested objects are the mass and string.
Therefore, a free body diagram can be drawn by adding a reaction force to replace the force due to the ceiling, while the body force due to the body mass remains unchange.
Since the hanging string is also not an interested body, Imply the free body diagram of the body mass is :
## Free-Body Diagram
Important steps of constructing the free-body diagram are:
1. Selecting the interested bodys from the whole system, so that the free body of interest can be detached away from the ground, connections, supports and all other bodies.
2. In general, there are always forces exerted on the free body by those stripped away bodies. These forces are considered as external forces and are represented by forces acting on the free body.
3. Besides, the weight of the free body should also be considered and be included as external forces acting on the free body.
4. But for forces exerted on each other between various parts of the free body, these forces are considered as internal forces because the free body is considered as a rigid body also.
5. For known external forces, such as weight of free body, applied forces, both the magnitudes and directions of each known external forces should be marked at the corresponding position on the free-body diagram.
6. For those unknown external forces due to stripping away bodies, such as ground, connections, supports and all other bodies from the free body, these forces are usually the reaction forces to opposite the motion of the free body in order to constrain the free body to remain in the same static position. Each known external forces should also be marked at the corresponding position on the free-body diagram. Although the magnitudes of unknown external forces are unknown, the reaction directions of each unknown external forces at the supports and connections can usually determined by its type of support or connection.
7. The pointing direction of an unknown external forces after the avaialable reaction are determined can be assumed in any direction because the sense of the direction will confirm whether the assumption is correct or not.
8. Although, the profile of the free body can usually be neglected, the free-body diagram should include dimensions for calculating the moments of forces.
9. Finally, including a coordinate axis is also necessary for determining the sign and angle of forces
ID: 120200059 Last Updated: 2/15/2012 Revision: 0 Ref:
References
1. I.C. Jong; B.G. rogers, 1991, Engineering Mechanics: Statics and Dynamics
2. F.P. Beer; E.R. Johnston,Jr.; E.R. Eisenberg, 2004, Vector Mechanics for Engineers: Statics
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This presentation is the property of its rightful owner.
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# GBK Geometry PowerPoint PPT Presentation
GBK Geometry. Jordan Johnson. Today’s plan. Greeting Hand in Problem Write-up Lesson: Isometries, Congruence, and Symmetry Study Guide Clean-up. On parallelograms. Roger Shepard, 1981. Isometries / Rigid Motions.
GBK Geometry
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## GBK Geometry
Jordan Johnson
### Today’s plan
• Greeting
• Hand in Problem Write-up
• Lesson: Isometries, Congruence, and Symmetry
• Study Guide
• Clean-up
### Isometries / Rigid Motions
• A transformation is an isometry (or rigid motion) iff it preserves distances and angles.
1
2
3
### Definitions
• Transformations definition of congruence:
• Two figures are congruent iff there is an isometry in which one is the image of the other.
### Definition
• A glide reflection is the composite of a translation and a reflection in a line parallel to the direction of the translation.
• In other words: Reflect, then translate, or vice versa.
### 3-fold Rotational Symmetry
Flag of the Isle of Man
### Translational Symmetry
Other neat examples: autologlyphs and ambigrams.
Saturated
Unsaturated
### Summary
• Each isometry has an associated symmetry:
• Reflection symmetry with respect to a line:
• Figure coincides with its reflection image through the line
• Rotation symmetry:
• Figure coincides with its image when rotated less than 360°
• n-fold symmetry, if the smallest such rotation is 360°⁄n
• Translation symmetry:
• Figure coincides with its translation image
### Homework
• Problem for final
• Study guide
• Any make-up portfolio work
• Recommended review for Unit 5:
• Proof problems:
• Two reflections across parallel lines= translation perpendicular to the lines of reflection.
• Prove that the 3 transformations are isometries.
• Asgs #37-38.
### Clean-up / Reminders
• Pick up all trash / items.
• Push in chairs (at front and back tables).
• See you tomorrow!
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## Convert 897 Hectares to Square Yards
To calculate 897 Hectares to the corresponding value in Square Yards, multiply the quantity in Hectares by 11959.900463011 (conversion factor). In this case we should multiply 897 Hectares by 11959.900463011 to get the equivalent result in Square Yards:
897 Hectares x 11959.900463011 = 10728030.715321 Square Yards
897 Hectares is equivalent to 10728030.715321 Square Yards.
## How to convert from Hectares to Square Yards
The conversion factor from Hectares to Square Yards is 11959.900463011. To find out how many Hectares in Square Yards, multiply by the conversion factor or use the Area converter above. Eight hundred ninety-seven Hectares is equivalent to ten million seven hundred twenty-eight thousand thirty point seven one five Square Yards.
## Definition of Hectare
The hectare (symbol: ha) is an SI accepted metric system unit of area equal to 100 ares (10,000 m2) and primarily used in the measurement of land as a metric replacement for the imperial acre. An acre is about 0.405 hectare and one hectare contains about 2.47 acres. In 1795, when the metric system was introduced, the "are" was defined as 100 square metres and the hectare ("hecto-" + "are") was thus 100 "ares" or 1⁄100 km2. When the metric system was further rationalised in 1960, resulting in the International System of Units (SI), the are was not included as a recognised unit. The hectare, however, remains as a non-SI unit accepted for use with the SI units, mentioned in Section 4.1 of the SI Brochure as a unit whose use is "expected to continue indefinitely".
## Definition of Square Yard
The square yard is an imperial unit of area, formerly used in most of the English-speaking world but now generally eplaced by the square metre, however it i still in widespread use in the US., Canada and the U.K. It isdefined as the area of a quare with sides of one yard (three feet, thirty-six inches, 0.9144 metres) in length. There is no universally agreed symbol but the following are used: square yards, square yard, square yds, square yd, sq yards, sq yard, sq yds, sq yd, sq.yd., yards/-2, yard/-2, yds/-2, yd/-2, yards^2, yard^2, yds^2, yd^2, yards², yard², yds², yd².
## Using the Hectares to Square Yards converter you can get answers to questions like the following:
• How many Square Yards are in 897 Hectares?
• 897 Hectares is equal to how many Square Yards?
• How to convert 897 Hectares to Square Yards?
• How many is 897 Hectares in Square Yards?
• What is 897 Hectares in Square Yards?
• How much is 897 Hectares in Square Yards?
• How many yd2 are in 897 ha?
• 897 ha is equal to how many yd2?
• How to convert 897 ha to yd2?
• How many is 897 ha in yd2?
• What is 897 ha in yd2?
• How much is 897 ha in yd2?
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# The moving pointer and 0.999…. A useful cognitive root
Posted by: Gary Ernest Davis on: November 1, 2010
Debates arise periodically as to whether 0.999… is equal to 1. Here, the the 9’s never end in the expression “0.999…” .
To paraphrase at least one U.S. President, the answer depends on what we mean by “equal”.
#### Equality
Equality takes some explaining in a mathematical setting.
As an example that is familiar to almost everyone, the fractions $\frac{1}{2}$ and $\frac{500}{1000}$ are commonly regarded as equal and we write $\frac{1}{2}=\frac{500}{1000}$.
Yet $\frac{1}{2}$ is not identical with $\frac{500}{1000}$. There are least two ways in which this is so. First, we do not see a $500\textrm{ or } 1000$ in the fraction $\frac{1}{2}$. The form of these two fractions is different. Secondly, 1 part of an object – such as a cupcake – that has been divided into 2 equal pieces is quite different from 500 pieces of a cupcake that has been divided into 1000 equal pieces.
So what do we mean when we say that $\frac{1}{2}$ and $\frac{500}{1000}$ are equal?
We mean that $1\times 1000=2\times 500$.
More generally, two fractions $\frac{a}{b}, \frac{c}{d}$ are equal when $a\times d = b\times c$. Here, we take it for granted that we know what we mean by equality of whole numbers.
#### Decimals
Fractions can be written as – possibly infinite – decimal strings using the digits $0, 1, \ldots 9$.
The basic idea of a decimal string $x=0.a_1a_2a_3\ldots$ is that multiplication by 10 moves the digits to the left: $10\times x = a_1+0.a_2a_3a_4\ldots$. The reason for this is that we are thinking – in the back of our minds – that $x=0.a_1a_2a_3\ldots$ should mean $\frac{a_1}{10}+\frac{a_2}{100}+\frac{a_3}{1000}+\ldots$, whatever a possibly infinite sum might mean.
We avoid thinking about infinite sums by just focusing on the decimal strings themselves.
How could we write a fraction such as $\frac{1}{3}$ as a decimal string: $\frac{1}{3}=0.a_1a_2a_3\ldots$?
Using the basic property of decimal strings, described above, we would have $10\times\frac{1}{3}=a_0+0.a_2a_3a_4\ldots$. But $10\times\frac{1}{3} = 3+\frac{1}{3}$ so we would have $a_1=3 \textrm{ and } 0.a_2a_3a_4\ldots =\frac{1}{3}$. Now we are right back where we started, so we will have $a_n=3\textrm{ for all } n$.
In other words, using the basic property of decimal strings that multiplication by 10 moves the digits to the left, we have the decimal representation $\frac{1}{3}=0.333\ldots$. Here the “3”s go on forever, without end.
This is a cute trick, but what could it mean? We cannot resort to infinite sums if we do not know what real numbers are, so we need another interpretation to guide our intuition. This is not so much a matter of mathematical logic, which is well-covered by any number of ways of constructing the real numbers, as it is of an intuitive understanding of equality of real numbers, represented as decimal strings. Our aim is to increase, not decrease, understanding for students and teachers.
#### A decimal string as a sequence of instructions
Let’s think about a decimal string $0.a_1a_2a_3\ldots$ as a sequence of instructions: each of the digits $a_n$ is instructing us to move a pointer to a certain place on a line. We will use the decimal representation $0.333\ldots \textrm{ for } \frac{1}{3}$ to illustrate what we mean by a sequence of instructions for moving a pointer on a line.
Imagine a pointer pointing at 0 on the number line:
Here the decimal strings $0.1, 0.2, \ldots 0.9$ have been placed on the line at spots that divide the numbers between 0 and 1 into 10 equal-sized intervals. These decimal strings represent the numbers $\frac{1}{10},\frac{2}{10}, \ldots , \frac{9}{10}$.
The first digit – namely 3 – in the decimal string for $\frac{1}{3}$ indicates that we should move the pointer somewhere between the $0.3 \textrm{ and} 0.4$ spots:
To figure out where between 0.3 and 0.4 the pointer goes, we divide the space between 0.3 and 0.4 into 10 equal size intervals. We label the new endpoints of these intervals $0.31, 0.32, 0.33, 0.34, 0.35, 0.36, 0.37, 0.38, 0.39$.
The second digit – namely 3, again – in the decimal string for $\frac{1}{3}$ indicates that we should move the pointer somewhere between the $0.33 \textrm{ and} 0.34$ spots, as in the diagram.
Again, this does not pin down the pointer exactly. So we take the interval between 0.33 and 0.34 and divide it into 10 equal sized intervals. The new labels for the endpoints of these intervals will be $0.331, 0.332, 0.333, 0.334, 0.335, 0.336, 0.337, 0.338, 0.339$.
We read the third digit- 3, again – in the decimal string for $\frac{1}{3}$ and move the pointer somewhere between the $0.333 \textrm{ and} 0.334$ spots.
This process of moving the pointer will never end, because there is a never ending sequence of 3’s in the decimal string for $\frac{1}{3}$. The pointer never stops: it never actually points at $\frac{1}{3}$ at any stage we move it.
In terms of the moving pointer, how can we interpret $\frac{1}{3}=0.333\ldots$?
One way to answer this question is to say that as we move the pointer according to the instructions encoded in the decimal string $0.333\ldots$, the distance between the pointer and $\frac{1}{3}$ becomes progressively closer to 0.
If we had placed the pointer at the left hand end of an interval at each step, the distance between the pointer and $\frac{1}{3}$ would reduce by $\frac{1}{10}^{th}$ each time.
That it is one way to think about the “=” in $\frac{1}{3}=0.333\ldots$.
This is what is known as a “cognitive root” in the learning and teaching of mathematics:
• It is an idea that is based on something relatively easy to grasp.
• It is not the whole story – more sophisticated mathematical thought adds to, and enriches the idea.
• The idea is not thrown away as a student encounters more sophisticated mathematics: it is simply built upon.
The same idea of “equals” applies to $0.999\ldots = 1$.
We imagine a pointer located at 0 and move it to 0.9. The distance between the pointer and 1 is $\frac{1}{10}$.
We take the next digit in the decimal string – 9 – and move the pointer to 0.99. The distance between the pointer and 1 is now $\frac{1}{100}$. Each time we move the pointer according to the instruction encoded in the next digit in the decimal string the distance between the pointer and 1 is reduced by $\frac{1}{10}^{th}$.
That is the sense in which we mean “=” in $0.999 \ldots = 1$.
It all depends, to paraphrase a former US President again, what we mean by “equals.”
For a related discussion, see James Tanton’s video “Point Nine Forever”:
James rightly points out, in different terminology, that if the decimal string $x = 0.999\ldots$ is to represent a number at all, then, by our fundamental operation of multiplying the decimal string by 10 we have $10\times x = 9+0.999\ldots = 9+x \textrm{ so } 9\times x = 9 \textrm { and therefore } x=1$
### 7 Responses to "The moving pointer and 0.999…. A useful cognitive root"
Wonderful article!
Thanks. Odd the link is broken. It seemed OK when I clicked on it. Here’s the URL: http://www.warwick.ac.uk/staff/David.Tall/themes/cognitive-roots.html
“So what do we mean when we say that 1/2 and 500/1000 are equal?”
Just being completely picky, as usual Gary, I would never say that! Equivalent, maybe, but “equal” no!
Re: 0.999999999999… It’s like asking whether the sum of (1/2)^n for all values of n from 1 to infinity is equal to 1… (of course it tends to 1 as n tends to infinity, but that is different!).
Further, what if 1/3 were represented using binary instead of decimal fractions…
What is usually missed in most of these discussions is that recurring decimals can be represented by rational fractions, whereas non-recurring cannot. That, to me, is the more interesting mathematics!
Colin
Colin, you may say equivalent, but most elementary teachers say “equal”.
Equality is a very strange notion. No two distinct things can be equal, else they would not be distinct. So almost all “equality” is equivalence. The only distinguishing thing about equality is that it is a symmetric, transitive, reflexive relationship: that is, an equivalence relation.
I agree with you about the non-recurring decimals, but that’s a story for another post. Would you like to write it?
jfW767 Very true! Makes a change to see someone spell it out like that. :)
Here’s another way to think about it. Two real numbers are equal if their difference is 0. Working one digit at a time, the difference between these two numbers is easily seen to be less than 0.1, less than 0.01, less than 0.001, and so on. Hence, it is smaller than 1/10^n for any positive integer n, and smaller than any positive number. Hence, the difference is 0, unless you want to try to argue that 0.99999… > 1, which is shown to be false merely by looking at the ones digit of each number.
My experience with undergraduates is that they believe in infinitesimals. Many of them would not accept that just because a non-negative number is less then $\epsilon$ for arbitrarily small $\epsilon$ it must be 0.
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# math
posted by .
4. y/2+5=-12
6. x/3-9=0
7. 14+h/5=2
THIS IS IMPROTANT Y/2, X/3 AND H/5 ARE FRACTIONS. it doesnt mean divide its fractions that's why i need help plse tell me how you did it so i can figure out the rest of my problmes
• math -
Well, actually fractions and divide are pretty similar.
now for example
(y/2) + 5 = -12
(y/2) +0 = -17
multiply both sides by 2
y = -34
• math -
thx you so much i was so confused
• math -
4. y/2 + 5 = -12 (use the inverse operation to get rid of the 5)
y/2+5(-5)= -12 (-5)
y/2=-17 (isolate the variable, aka, multiply both sides by 2, to get your answer of...)
y= -34
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# Chapter 3 ATE I: Binary treatment
library(lmtest)
library(sandwich)
library(grf)
library(glmnet)
library(splines)
library(ggplot2)
library(reshape2)
## 3.1 Notation and definitions
Let’s establish some notation. Each data point will be defined by the triple $$(X_i, W_i, Y_i)$$. The vector $$X_i$$ represents covariates that are observed for individual $$i$$. Treatment assignment is indicated by $$W_i \in \{0, 1\}$$, with 1 representing treatment, and 0 representing control. The scalar $$Y_i$$ is the observed outcome, and it can be real or binary. Each observation is drawn independently and from the same distribution. We’ll be interested in assessing the causal effect of treatment on outcome.
A difficulty in estimating causal quantities is that we observe each individual in only one treatment state: either they were treated, or they weren’t. However, it’s often useful to imagine that each individual is endowed with two random variables $$(Y_i(1), Y_i(0))$$, where $$Y_i(1)$$ represents the value of this individual’s outcome if they receive treatment, and $$Y_i(0)$$ represents their outcome if they are not treated. These random variables are called potential outcomes. The observed outcome $$Y_i$$ corresponds to whichever potential outcome we got to see: $$$Y_i \equiv Y_i(W_i) = \begin{cases} Y_i(1) \qquad \text{if }W_i = 1 \text{ (treated)} \\ Y_i(0) \qquad \text{if }W_i = 0 \text{ (control)}\\ \end{cases}$$$
Since we can’t observe both $$Y_i(1)$$ and $$Y_i(0)$$, we won’t be able to make statistical claims about the individual treatment effect $$Y_i(1) - Y_i(0)$$. Instead, our goal will be to estimate the average treatment effect (ATE): $$$\tag{3.1} \tau := \mathop{\mathrm{E}}[Y_i(1) - Y_i(0)].$$$
Here, when we refer to the randomized setting we mean that we have data generated by a randomized control trial. The key characteristic of this setting is that the probability that an individual is assigned to the treatment arm is fixed. In particular, it does not depend on the individual’s potential outcomes: $$$\tag{3.2} Y_i(1), Y_i(0) \perp W_i.$$$ This precludes situations in which individuals may self-select into or out of treatment. The canonical failure example is a job training program in which workers enroll more often if they are more likely to benefit from treatment because in that case $$W_i$$ and $$Y_i(1) - Y_i(0)$$ would be positively correlated.
When condition (3.2) is violated, we say that we are in an observational setting. This is a more complex setting encompassing several different scenarios: sometimes it’s still possible to estimate the ATE under additional assumptions, sometimes the researcher can exploit other sources of variation to obtain other interesting estimands. Here, we will focus on ATE estimation under the following assumption:
$$$\tag{3.3} Y_i(1), Y_i(0) \perp W_i \ | \ X_i.$$$
In this document, we call this assumption unconfoundeness, though it is also known as no unmeasured confounders, ignorability or selection on observables. It says that all possible sources of self-selection, etc., can be explained by the observable covariates $$X_i$$. Continuing the example above, it may be that older or more educated are more likely to self-select into treatment; but when we compare two workers that have the same age and level of education, etc., there’s nothing else that we could infer about their relative potential outcomes if we knew that one went into job training and the other did not.
As we’ll see below, a key quantity of interest will be the treatment assignment probability, or propensity score $$e(X_i) := \mathop{\mathrm{P}}[W_i = 1 | X_i]$$. In an experimental setting this quantity is usually known and fixed, and in observational settings it must be estimated from the data. We will often need to assume that the propensity score is bounded away from zero and one. That is, there exists some $$\eta > 0$$ such that $$$\tag{3.4} \eta < e(x) < 1 - \eta \qquad \text{for all }x.$$$ This assumption is known as overlap, and it means that for all types of people in our population (i.e., all values of observable characteristics) we can find some portion of individuals in treatment and some in control. Intuitively, this is necessary because we’d like to will be comparing treatment and control at each level of the covariates and then aggregate those results.
As a running example, in what follows we’ll be using an abridged version of a public dataset from the General Social Survey (GSS) (Smith, 2016). The setting is a randomized control trial. Individuals were asked about their thoughts on government spending on the social safety net. The treatment is the wording of the question: about half of the individuals were asked if they thought government spends too much on “welfare” $$(W_i = 1)$$, while the remaining half was asked about “assistance to the poor” $$(W_i = 0)$$. The outcome is binary, with $$Y_i = 1$$ corresponding to a positive answer. In the data set below, we also collect a few other demographic covariates.
# Read in data
n <- nrow(data)
# Treatment: does the the gov't spend too much on "welfare" (1) or "assistance to the poor" (0)
treatment <- "w"
# Outcome: 1 for 'yes', 0 for 'no'
outcome <- "y"
covariates <- c("age", "polviews", "income", "educ", "marital", "sex")
## 3.2 Difference-in-means estimator
Let’s begin by considering a simple estimator that is available in experimental settings. The difference-in-means estimator is the sample average of outcomes in treatment minus the sample average of outcomes in control. $$$\tag{3.5} \widehat{\tau}^{DIFF} = \frac{1}{n_1} \sum_{i:W_i = 1} Y_i - \frac{1}{n_0} \sum_{i:W_i = 0} Y_i \qquad \text{where} \qquad n_w := |\{i : W_i = w \}|$$$
Here’s one way to compute the difference-in-means estimator and its associated statistics (t-stats, p-values, etc.) directly:
# Only valid in the randomized setting. Do not use in observational settings.
Y <- data[,outcome]
W <- data[,treatment]
ate.est <- mean(Y[W==1]) - mean(Y[W==0])
ate.se <- sqrt(var(Y[W == 1]) / sum(W == 1) + var(Y[W == 0]) / sum(W == 0))
ate.tstat <- ate.est / ate.se
ate.pvalue <- 2*(pnorm(1 - abs(ate.est/ate.se)))
ate.results <- c(estimate=ate.est, std.error=ate.se, t.stat=ate.tstat, pvalue=ate.pvalue)
print(ate.results)
## estimate std.error t.stat pvalue
## -0.347115545 0.004895638 -70.903028958 0.000000000
Or, alternatively, via the t.test function:
fmla <- formula(paste(outcome, '~', treatment)) # y ~ w
t.test(fmla, data=data)
##
## Welch Two Sample t-test
##
## data: y by w
## t = 70.903, df = 20840, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 0.3375197 0.3567114
## sample estimates:
## mean in group 0 mean in group 1
## 0.4398290 0.0927135
We can also compute the same quantity via linear regression, using the fact that $$$Y_i = Y_i(0) + W_i \left( Y_i(1) - Y_i(0) \right),$$$
so that taking expectations conditional on treatment assignment, $$$\mathop{\mathrm{E}}[Y_i | W_i] = \alpha + W_i \tau \qquad \text{where} \qquad \alpha := \mathop{\mathrm{E}}[Y_i(0)]$$$
[Exercise: make sure you understand this decomposition. Where is the unconfoundedness assumption (3.2) used?]
This result implies that we can estimate the ATE of a binary treatment via a linear regression of observed outcomes $$Y_i$$ on a vector consisting of intercept and treatment assignment $$(1, W_i)$$.
# Do not use! standard errors are not robust to heteroskedasticity! (See below)
fmla <- formula(paste0(outcome, '~', treatment))
ols <- lm(fmla, data=data)
coef(summary(ols))[2,]
## Estimate Std. Error t value Pr(>|t|)
## -0.347115545 0.004732343 -73.349622060 0.000000000
The point estimate we get is the same as we got above by computing the ATE “directly” via mean(Y[W==1]) - mean(Y[W==0]). However, the standard errors are different. This is because in R the command lm does not compute heteroskedasticity-robust standard errors. This is easily solvable.
# Use this instead. Standard errors are heteroskedasticity-robust.
# Only valid in randomized setting.
fmla <- formula(paste0(outcome, '~', treatment))
ols <- lm(fmla, data=data)
coeftest(ols, vcov=vcovHC(ols, type='HC2'))[2,]
## Estimate Std. Error t value Pr(>|t|)
## -0.347115545 0.004895638 -70.903028958 0.000000000
The difference-in-means estimator is a simple, easily computable, unbiased, “model-free” estimator of the treatment effect. It should always be reported when dealing with data collected in a randomized setting. However, as we’ll see later in this chapter, there are other estimators that have smallest variance. Therefore,
## 3.3 Turning an experimental dataset into an observational one
In what follows we’ll purposely create a biased sample to show how the difference in means estimator fails in observational settings. We’ll focus on two covariates: age and political views (polviews). Presumably, younger or more liberal individuals are less affected by the change in wording in the question. So let’s see what happens when we make these individuals more prominent in our sample of treated individuals, and less prominent in our sample of untreated individuals.
# Probabilistically dropping observations in a manner that depends on x
# copying old dataset, just in case
data.exp <- data
# defining the group that we will be dropped with some high probability
grp <- ((data$w == 1) & # if treated AND... ( (data$age > 45) | # belongs an older group OR
(data$polviews < 5) # more conservative )) | # OR ((data$w == 0) & # if untreated AND...
(
(data$age < 45) | # belongs a younger group OR (data$polviews > 4) # more liberal
))
# Individuals in the group above have a small chance of being kept in the sample
prob.keep <- ifelse(grp, .15, .85)
keep.idx <- as.logical(rbinom(n=nrow(data), prob=prob.keep, size = 1))
# Dropping
data <- data[keep.idx,]
Let’s see what happens to our sample before and after the change. This next figure shows the original dataset. Each observation is represented by a point on either the left or right scatterplots. An outcome of $$Y_i=1$$ is denoted by a blue circle, and $$Y_i=0$$ is denoted by a red square, but let’s not focus on the outcome at the moment. For now, what’s important to note is that the covariate distributions for treated and untreated populations are very similar. This is what we should expect in an experimental setting under unconfoundedness.
X <- model.matrix(formula("~ 0 + age + polviews"), data.exp) # old 'experimental' dataset
W <- data.exp$w Y <- data.exp$y
par(mfrow=c(1,2))
for (w in c(0, 1)) {
plot(X[W==w,1] + rnorm(n=sum(W==w), sd=.1), X[W==w,2] + rnorm(n=sum(W==w), sd=.1),
pch=ifelse(Y, 23, 21), cex=1, col=ifelse(Y, rgb(1,0,0,1/4), rgb(0,0,1,1/4)),
bg=ifelse(Y, rgb(1,0,0,1/4), rgb(0,0,1,1/4)), main=ifelse(w, "Treated", "Untreated"), xlab="age", ylab="polviews")
}
On the other hand, this is what the modified dataset looks like. The treated population is much younger and more liberal, while the untreated population is older and more conservative.
X <- model.matrix(formula("~ 0 + age + polviews"), data)
W <- data$w Y <- data$y
par(mfrow=c(1,2))
for (w in c(0, 1)) {
plot(X[W==w,1] + rnorm(n=sum(W==w), sd=.1), X[W==w,2] + rnorm(n=sum(W==w), sd=.1),
pch=ifelse(Y, 23, 21), cex=1, col=ifelse(Y, rgb(1,0,0,1/4), rgb(0,0,1,1/4)),
bg=ifelse(Y, rgb(1,0,0,1/4), rgb(0,0,1,1/4)), main=ifelse(w, "Treated", "Untreated"), xlab="age", ylab="polviews")
}
As we would expect the difference-in-means estimate is biased toward zero because in this new dataset we mostly treated individuals for which we expect the effect to be smaller.
# Do not use in observational settings.
# This is only to show how the difference-in-means estimator is biased in that case.
fmla <- formula(paste0(outcome, '~', treatment))
ols <- lm(fmla, data=data)
coeftest(ols, vcov=vcovHC(ols, type='HC2'))[2,]
## Estimate Std. Error t value Pr(>|t|)
## -2.887039e-01 8.630364e-03 -3.345211e+01 2.645275e-231
Note that the dataset created above still satisfies unconfoundedness (3.3), since discrepancies in treatment assignment probability are described by observable covariates (age and polviews). Moreover, it also satisfies the overlap assumption (3.4), since never completely dropped all treated or all untreated observations in any region of the covariate space. This is important because, in what follows, we’ll consider different estimators of the ATE that are available in observational settings under unconfoundedness and overlap.
## 3.4 Direct estimation
Our first estimator is suggested by the following decomposition of the ATE, which is possible due to unconfoundedness (3.3).
$$$\mathop{\mathrm{E}}[Y_i(1) - Y_i(0)] = \mathop{\mathrm{E}}[\mathop{\mathrm{E}}[Y_i | X_i, W_i=1]] - \mathop{\mathrm{E}}[\mathop{\mathrm{E}}[Y_i | X_i, W_i=0]]$$$
[Exercise: make sure you understand this. Where is the unconfoundedness assumption used?]
The decomposition above suggests the following procedure, sometimes called the direct estimate of the ATE:
1. Estimate $$\mu(x, w) := E[Y_i|X_i = x,W_i=w]$$, preferably using nonparametric methods.
2. Predict $$\hat{\mu}(X_i, 1)$$ and $$\hat{\mu}(X_i, 0)$$ for each observation in the data.
3. Average out the predictions and subtract them. $$$\widehat{\tau}^{DM} := \frac{1}{n} \sum_{i=1}^{n} \hat{\mu}(X_i, 1) - \hat{\mu}(X_i, 0)$$$
# Do not use! We'll see a better estimator below.
# Fitting some model of E[Y|X,W]
fmla <- as.formula(paste0(outcome, "~ ", paste("bs(", covariates, ", df=3)", "*", treatment, collapse="+")))
model <- lm(fmla, data=data)
# Predicting E[Y|X,W=w] for w in {0, 1}
data.1 <- data
data.1[,treatment] <- 1
data.0 <- data
data.0[,treatment] <- 0
muhat.treat <- predict(model, newdata=data.1)
muhat.ctrl <- predict(model, newdata=data.0)
# Averaging predictions and taking their difference
ate.est <- mean(muhat.treat) - mean(muhat.ctrl)
print(ate.est)
## [1] -0.3423029
This estimator allows us to leverage regression techniques to estimate the ATE, so the resulting estimate should have smaller root-mean-squared error. However, it has several disadvantages that make it undesirable. First, its properties will rely heavily on the model $$\hat{\mu}(x, w)$$ being correctly specified: it will be an unbiased and/or consistent estimate of the ATE provided that $$\hat{\mu}(x, w)$$ is an unbiased and/or consistent estimator of $$\mathop{\mathrm{E}}[Y|X=x, W=w]$$. In practice, having a well-specified model is not something we want to rely upon. In general, it will also not be asymptotically normal, which means that we can’t easily compute t-statistics and p-values for it.
A technical note. Step 1 above can be done by regressing $$Y_i$$ on $$X_i$$ using only treated observations to get an estimate $$\hat{\mu}(x, 1)$$ first, and then repeating the same to obtain $$\hat{\mu}(x, 0)$$ from the control observations. Or it can be done by regression $$Y_i$$ on both covariates $$(X_i, W_i)$$ together and obtaining a function $$\hat{\mu}(x, w)$$. Both have advantages and disadvantages, and we refer to Künzel, Sekhon, Bickel, Yu (2019) for a discussion.
## 3.5 Inverse propensity-weighted estimator
To understand this estimator, let’s first consider a toy problem. Suppose that we’d like to estimate the average effect of a certain learning intervention on students’ grades, measured on a scale of 0-100. We run two separate experiments in schools of type A and B. Suppose that, unknown to us, grades among treated students in schools of type A are approximately distributed as $$Y_i(1) | A \sim N(60, 5^2)$$, whereas in schools of type B they are distributed as $$Y_i(1) | B \sim N(70, 5^2)$$. Moreover, for simplicity both schools have the same number of students. If we could treat the same number of students in both types of schools, we’d get an unbiased estimate of the population mean grade among treated individuals: $$(1/2)60 + (1/2)70 = 75$$.
However, suppose that enrollment in the treatment is voluntary. In school A, only 5% enroll in the study, whereas in school B the number is 40%. Therefore, if we take an average of treated students’ grades without taking school membership into account, school B’s students would be over-represented, and therefore our estimate of treated student’s grades would be biased upward.
# Simulating the scenario above a large number of times
A.mean <- 60
B.mean <- 70
pop.mean <- .5 * A.mean + .5 * B.mean # both schools have the same size
# simulating the scenario about a large number of times
sample.means <- replicate(1000, {
school <- sample(c("A", "B"), p=c(.5, .5), size=100, replace=TRUE)
treated <- ifelse(school == 'A', rbinom(100, 1, .05), rbinom(100, 1, .4))
grades <- ifelse(school == 'A', rnorm(100, A.mean, 5), rnorm(100, B.mean, 5))
mean(grades[treated == 1]) # average grades among treated students, without taking school into account
})
hist(sample.means, freq=F, main="Sample means of treated students' grades", xlim=c(55, 75), col=rgb(0,0,1,1/8), ylab="", xlab="")
abline(v=pop.mean, lwd=3, lty=2)
legend("topleft", "truth", lwd=3, lty=2, bty="n")
To solve this problem, we can consider each school separately, and then aggregate the results. Denote by $$n_A$$ the number of students from school $$A$$, and $$n_{A,1}$$ denote the number of treated students in that school. Likewise, denote by $$n_B$$ and $$n_{B,1}$$ the same quantities for school $$B$$. Then our aggregated means estimator can be written as: $$$\tag{3.6} \overbrace{ \left( \frac{n_{A}}{n} \right) }^{\text{Proportion of A}} \underbrace{ \frac{1}{n_{A, 1}} \sum_{\substack{i \in A \\ i \text{ treated}}} Y_i }_{\text{avg. among treated in A}} + \overbrace{ \left( \frac{n_{B}}{n} \right) }^{\text{Proportion of B}} \underbrace{ \frac{1}{n_{B, 1}} \sum_{\substack{i \in B \\ i \text{ treated}}} Y_i }_{\text{avg. among treated in B}}.$$$
# simulating the scenario about a large number of times
agg.means <- replicate(1000, {
school <- sample(c("A", "B"), p=c(.5, .5), size=100, replace=TRUE)
treated <- ifelse(school == 'A', rbinom(100, 1, .05), rbinom(100, 1, .4))
grades <- ifelse(school == 'A', rnorm(100, A.mean, 5), rnorm(100, B.mean, 5))
# average grades among treated students in each school
mean.treated.A <- mean(grades[(treated == 1) & (school == 'A')])
mean.treated.B <- mean(grades[(treated == 1) & (school == 'B')])
# probability of belonging to each school
prob.A <- mean(school == 'A')
prob.B <- mean(school == 'B')
prob.A * mean.treated.A + prob.B * mean.treated.B
})
hist(agg.means, freq=F, main="Aggregated sample means of treated students' grades", xlim=c(50, 75), col=rgb(0,0,1,1/8), ylab="", xlab="")
abline(v=pop.mean, lwd=3, lty=2)
legend("topright", "truth", lwd=3, lty=2, bty="n")
Next, we’ll manipulate the expression (3.6). This next derivation can be a bit overwhelming, but please keep in mind that we’re simply doing algebraic manipulations, as well as establishing some common and useful notation. First, note that we can rewrite the average for school A in (3.6) as $$$\tag{3.7} \frac{1}{n_{A, 1}} \sum_{\substack{i \in A \\ i \text{ treated}}} Y_i = \frac{1}{n_A} \sum_{\substack{i \in A \\ i \text{ treated}}} \frac{1}{(n_{A,1} / n_A)} Y_i = \frac{1}{n_A} \sum_{i \in A} \frac{W_i}{(n_{A,1} / n_A)} Y_i,$$$
where $$W_i$$ is the treatment indicator. Plugging (3.7) back into (3.6), $$$\left( \frac{n_{A}}{n} \right) \frac{1}{n_A} \sum_{i \in A} \frac{W_i}{(n_{A,1} / n_A)} Y_i + \left( \frac{n_{B}}{n} \right) \frac{1}{n_B} \sum_{i \in B} \frac{W_i}{(n_{B,1} / n_B)} Y_i.$$$
Note that the $$n_A$$ and $$n_B$$ will cancel out. Finally, if we denote the sample proportion of treated students in school $$A$$ by $$\hat{e}(A_i) \approx n_{A,1}/n_A$$ and similar for school $$B$$, (3.6) can be written compactly as $$$\tag{3.8} \frac{1}{n} \sum_{i=1}^{n} \frac{W_i}{\hat{e}(X_i)} Y_i,$$$ where $$X_i \in \{A, B\}$$ denotes the school membership. Quantity $$\hat{e}(X_i)$$ is an estimate of the probability of treatment given the control variable $$e(X_i) = \mathop{\mathrm{P}}[W_i=1 | X_i ]$$, also called the propensity score. Because (3.8) is a weighted average with weights $$W_i/\hat{e}(X_i)$$, when written in this form we say that it is an inverse propensity-weighted estimator (IPW).
The IPW estimator is also defined when there are continuous covariates. In that case, we must estimate the assignment probability given some model, for example using logistic regression, forests, etc. With respect to modeling, the behavior of the IPW estimator is similar to the direct estimator: if $$\hat{e}(X_i)$$ is an unbiased estimate of $$e(X_i)$$, then (3.8) is an unbiased estimate of $$E[Y(1)]$$; and we can show that if $$\hat{e}(X_i)$$ is a consistent estimator of $$e(X_i)$$, then (3.8) is consistent for $$E[Y(1)]$$.
When the treatment propensity $$\hat{e}(X_i)$$ is small, the summands in (3.8) can be very large. In particular, if $$\hat{e}(x)$$ is exactly zero, then this estimator is undefined. That is why, in addition to requiring conditional unconfoundedness (3.3), it also requires the overlap condition (3.4). In any case, when overlap is small (i.e., $$\hat{e}(x)$$ is very close to zero for many $$x$$), IPW becomes an unattractive estimator due to its high variance. We’ll see in the next section an improved estimator that builds upon IPW but is strictly superior and should be used instead.
Finally, we just derived an estimator of the average treated outcome, but we could repeat the argument for control units instead. Subtracting the two estimators leads to the following inverse propensity-weighted estimate of the treatment effect: $$$\tag{3.9} \widehat{\tau}^{IPW} := \frac{1}{n} \sum_{i=1}^{n} Y_i \frac{W_i}{\widehat{e}(X_i)} - \frac{1}{n} \sum_{i=1}^{n} Y_i \frac{(1 - W_i) }{1-\widehat{e}(X_i)}.$$$
The argument above suggests the following algorithm:
1. Estimate the propensity scores $$e(X_i)$$ by regressing $$W_i$$ on $$X_i$$, preferably using a non-parametric method.
2. Compute the IPW estimator summand: $Z_i = Y_i \times \left(\frac{W_i}{\hat{e}(X_i)} - \frac{(1-W_i)}{(1-\hat{e}(X_i)} \right)$
3. Compute the mean and standard error of the new variable $$Z_i$$
# Available in randomized settings and observational settings with unconfoundedness+overlap
# Estimate the propensity score e(X) via logistic regression using splines
fmla <- as.formula(paste0("~", paste0("bs(", covariates, ", df=3)", collapse="+")))
W <- data[,treatment]
Y <- data[,outcome]
XX <- model.matrix(fmla, data)
logit <- cv.glmnet(x=XX, y=W, family="binomial")
e.hat <- predict(logit, XX, s = "lambda.min", type="response")
# Using the fact that
z <- Y * (W/e.hat - (1-W)/(1-e.hat))
ate.est <- mean(z)
ate.se <- sd(z) / sqrt(length(z))
ate.tstat <- ate.est / ate.se
ate.pvalue <- 2*(pnorm(1 - abs(ate.est/ate.se)))
ate.results <- c(estimate=ate.est, std.error=ate.se, t.stat=ate.tstat, pvalue=ate.pvalue)
print(ate.results)
## estimate std.error t.stat pvalue
## -3.393831e-01 1.385475e-02 -2.449579e+01 4.503613e-122
## 3.6 Augmented inverse propensity-weighted (AIPW) estimator
The next augmented inverse propensity-weighted estimator (AIPW) of the treatment effect is available under unconfoundedness (3.3) and overlap (3.4): \tag{3.10} \begin{aligned} \widehat{\tau}^{AIPW} &:= \frac{1}{n} \sum_{i=1}^{n} \hat{\mu}^{-i}(X_i, 1) - \hat{\mu}^{-i}(X_i, 0) \\ &+ \frac{W}{\widehat{e}^{-i}(X_i)} \left( Y_i - \hat{\mu}^{-i}(X_i, 1) \right) - \frac{1-W}{1-\widehat{e}^{-i}(X_i)} \left( Y_i - \hat{\mu}^{-i}(X_i, 0) \right) \end{aligned}
Let’s parse (3.10). Ignoring the superscripts for now, the first two terms correspond to an estimate of the treatment effect obtained via direct estimation. The next two terms resemble the IPW estimator, except that we replaced the outcome $$Y_i$$ with the residuals $$Y_i - \hat{\mu}(X_i, W_i)$$. At a high level, the last two terms estimate and subtract an estimate of the bias of the first.
The superscripts have to do with a technicality called cross-fitting that is required to prevent a specific form of overfitting and allow certain desirable asymptotic properties to hold. That is, we need to fit the outcome and propensity models using one portion of the data, and compute predictions on the remaining portion. An easy way of accomplishing this is to divide the data into K folds, and then estimate $$K$$ models that are fitted on $$K-1$$ folds, and then compute predictions on the remaining Kth fold. The example below does that with $$K=5$$. Note that this is different from cross-validation: the goal of cross-validation is to obtain accurate estimates of the loss function for model selection, whereas the goal of cross-fitting is simply to not use the same observation to fit the model and produce predictions.
The AIPW estimator (3.10) has several desirable properties. First, it will be unbiased provided that either $$\hat{\mu}(X_i, w)$$ or $$\widehat{e}(X_i, w)$$ is unbiased. Second, it will be consistent provided that either $$\hat{\mu}(X_i, w)$$ or $$\widehat{e}(X_i, w)$$ is consistent. This property is called double robustness (DR), because only one of the outcome or propensity score models needs to be correctly specified, so the estimator is “robust” to misspecification in the remaining model. Under mild assumptions, the AIPW estimator is also asymptotically normal and efficient — that is has the smallest asymptotic variance among a large class of estimators that includes the IPW estimator above. This is an attractive property because smaller variance implies smaller confidence intervals, and more power to test hypotheses about this parameter. In fact, it even has smaller variance than the difference-in-means estimators in the randomized setting. Therefore, we recommend the following. In randomized settings, report the difference-in-means estimator (for simplicity) along with AIPW-based estimates (for efficiency); in observational settings with unconfoundedness and overlap, report AIPW-based estimates.
The next snippet provides an implementation of the AIPW estimator where outcome and propensity models are estimated using generalized linear models with splines (via glmnet and splines packages).
# Available in randomized settings and observational settings with unconfoundedness+overlap
# A list of vectors indicating the left-out subset
n <- nrow(data)
n.folds <- 5
indices <- split(seq(n), sort(seq(n) %% n.folds))
# Preparing data
W <- data[,treatment]
Y <- data[,outcome]
# Matrix of (transformed) covariates used to estimate E[Y|X,W]
fmla.xw <- formula(paste("~ 0 +", paste0("bs(", covariates, ", df=3)", "*", treatment, collapse=" + ")))
XW <- model.matrix(fmla.xw, data)
# Matrix of (transformed) covariates used to predict E[Y|X,W=w] for each w in {0, 1}
data.1 <- data
data.1[,treatment] <- 1
XW1 <- model.matrix(fmla.xw, data.1) # setting W=1
data.0 <- data
data.0[,treatment] <- 0
XW0 <- model.matrix(fmla.xw, data.0) # setting W=0
# Matrix of (transformed) covariates used to estimate and predict e(X) = P[W=1|X]
fmla.x <- formula(paste(" ~ 0 + ", paste0("bs(", covariates, ", df=3)", collapse=" + ")))
XX <- model.matrix(fmla.x, data)
# (Optional) Not penalizing the main effect (the coefficient on W)
penalty.factor <- rep(1, ncol(XW))
penalty.factor[colnames(XW) == treatment] <- 0
# Cross-fitted estimates of E[Y|X,W=1], E[Y|X,W=0] and e(X) = P[W=1|X]
mu.hat.1 <- rep(NA, n)
mu.hat.0 <- rep(NA, n)
e.hat <- rep(NA, n)
for (idx in indices) {
# Estimate outcome model and propensity models
# Note how cross-validation is done (via cv.glmnet) within cross-fitting!
outcome.model <- cv.glmnet(x=XW[-idx,], y=Y[-idx], family="gaussian", penalty.factor=penalty.factor)
propensity.model <- cv.glmnet(x=XX[-idx,], y=W[-idx], family="binomial")
# Predict with cross-fitting
mu.hat.1[idx] <- predict(outcome.model, newx=XW1[idx,], type="response")
mu.hat.0[idx] <- predict(outcome.model, newx=XW0[idx,], type="response")
e.hat[idx] <- predict(propensity.model, newx=XX[idx,], type="response")
}
# Commpute the summand in AIPW estimator
aipw.scores <- (mu.hat.1 - mu.hat.0
+ W / e.hat * (Y - mu.hat.1)
- (1 - W) / (1 - e.hat) * (Y - mu.hat.0))
# Tally up results
ate.aipw.est <- mean(aipw.scores)
ate.aipw.se <- sd(aipw.scores) / sqrt(n)
ate.aipw.tstat <- ate.aipw.est / ate.aipw.se
ate.aipw.pvalue <- 2*(pnorm(1 - abs(ate.aipw.tstat)))
ate.aipw.results <- c(estimate=ate.aipw.est, std.error=ate.aipw.se, t.stat=ate.aipw.tstat, pvalue=ate.aipw.pvalue)
print(ate.aipw.results)
## estimate std.error t.stat pvalue
## -3.341320e-01 9.564019e-03 -3.493636e+01 1.939058e-252
Here’s another example of AIPW-based estimation using random forests via the package grf. The function average_treatment_effect computes the AIPW estimate of the treatment effect, and uses forest-based estimates of the outcome model and propensity scores (unless those are passed directly via the arguments Y.hat and W.hat). Also, because forests are an ensemble method, cross-fitting is accomplished via out-of-bag predictions – that is, predictions for observation $$i$$ are computed using trees that were not constructed using observation $$i$$.
# Available in randomized settings and observational settings with unconfoundedness+overlap
# Input covariates need to be numeric.
XX <- model.matrix(formula(paste0("~", paste0(covariates, collapse="+"))), data=data)
# Estimate a causal forest.
forest <- causal_forest(
X=XX,
W=data[,treatment],
Y=data[,outcome],
#W.hat=..., # In randomized settings, set W.hat to the (known) probability of assignment
num.trees = 100)
forest.ate <- average_treatment_effect(forest)
print(forest.ate)
## estimate std.err
## -0.33233019 0.01273536
Finally, even when overlap holds, when the propensity score is small AIPW can have unstable performance. In that case, other estimators of the average treatment effect are available. For example, see the approximate residual balancing (ARB) method in Athey, Imbens and Wager (2016) and its accompanying R package balanceHD.
## 3.7 Diagnostics
Here we show some basic diagnostics and best practices that should always be conducted. Although failing these tests should usually cause concern, passing them does not mean that the analysis is entirely free from issues. For example, even if our estimated propensity scores are bounded away from zero and one (and therefore seem to satisfy the overlap assumption), they may still be incorrect – for example, if we rely on modeling assumptions that are simply not true. Also, although we should expect to pass all the diagnostic tests below in randomized settings, it’s good practice to check them anyway, as one may find problems with the experimental design.
### 3.7.1 Assessing balance
As we saw in Section 3.3, in observational settings the covariate distributions can be very different for treated and untreated individuals. Such discrepancies can lead to biased estimates of the ATE, but as hinted in Section 3.5, one would hope that by up- or down-weighting observations based on inverse propensity weights their averages should be similar. In fact, we should also expect that averages of basis functions of the covariates should be similar after reweighting. This property is called balance. One way to check it is as follows. Given some variable $$Z_i$$ (e.g., a covariate $$X_{i1}$$, or an interaction between covariates $$X_{i1}X_{i2}$$, or a polynomial in covariates $$X_{i1}^2$$, etc), we can check the absolute standardized mean difference (ASMD) of $$Z_i$$ between treated and untreated individuals in our data, $$$\frac{|\bar{Z}_1 - \bar{Z}_0|}{\sqrt{s_1^2 + s_0^2}},$$$ where $$\bar{Z}_1$$ and $$\bar{Z}_0$$ are sample averages of $$Z_i$$, and $$s_1$$ and $$s_0$$ are standard deviations of $$Z_i$$ for the two samples of treated and untreated individuals. Next, we can check the same quantity for their weighted counterparts $$Z_{i}W_i/\hat{e}(X_i)$$ and $$Z_i(1-W_i)/(1-\hat{e}(X_i))$$. If our propensity scores are well-calibrated, the ASMD for the weighted version should be close to zero.
# Here, adding covariates and their interactions, though there are many other possibilities.
fmla <- formula(paste("~ 0 +", paste(apply(expand.grid(covariates, covariates), 1, function(x) paste0(x, collapse="*")), collapse="+")))
# Using the propensity score estimated above
e.hat <- forest$W.hat XX <- model.matrix(fmla, data) W <- data[,treatment] pp <- ncol(XX) # Unadjusted covariate means, variances and standardized abs mean differences means.treat <- apply(XX[W == 1,], 2, mean) means.ctrl <- apply(XX[W == 0,], 2, mean) abs.mean.diff <- abs(means.treat - means.ctrl) var.treat <- apply(XX[W == 1,], 2, var) var.ctrl <- apply(XX[W == 0,], 2, var) std <- sqrt(var.treat + var.ctrl) # Adjusted covariate means, variances and standardized abs mean differences means.treat.adj <- apply(XX*W/e.hat, 2, mean) means.ctrl.adj <- apply(XX*(1-W)/(1-e.hat), 2, mean) abs.mean.diff.adj <- abs(means.treat.adj - means.ctrl.adj) var.treat.adj <- apply(XX*W/e.hat, 2, var) var.ctrl.adj <- apply(XX*(1-W)/(1-e.hat), 2, var) std.adj <- sqrt(var.treat.adj + var.ctrl.adj) # Plotting par(oma=c(0,4,0,0)) plot(-2, xaxt="n", yaxt="n", xlab="", ylab="", xlim=c(-.01, 1.3), ylim=c(0, pp+1), main="Standardized absolute mean differences") axis(side=1, at=c(-1, 0, 1), las=1) lines(abs.mean.diff / std, seq(1, pp), type="p", col="blue", pch=19) lines(abs.mean.diff.adj / std.adj, seq(1, pp), type="p", col="orange", pch=19) legend("topright", c("Unadjusted", "Adjusted"), col=c("blue", "orange"), pch=19) abline(v = seq(0, 1, by=.25), lty = 2, col = "grey", lwd=.5) abline(h = 1:pp, lty = 2, col = "grey", lwd=.5) mtext(colnames(XX), side=2, cex=0.7, at=1:pp, padj=.4, adj=1, col="black", las=1, line=.3) abline(v = 0) Note above how in particular age and polviews – the variables we chose to introduce imbalance in Section 3.3 – are far from balanced before adjustment, as are other variables that correlate with them. In addition to the above, we can check the entire distribution of covariates (and their transformations). The next snippet plots histograms for treated and untreated individuals with and without adjustment. Note how the adjusted histograms are very similar for untreated (red) and treated (green) samples – that is what we should expect. # change this if working on different data set XX <- model.matrix(fmla, data)[,c("age", "polviews", "age:polviews", "educ")] W <- data[,treatment] pp <- ncol(XX) plot.df <- data.frame(XX, W = as.factor(W), IPW = ifelse(W == 1, 1 / e.hat, 1 / (1 - e.hat))) plot.df <- reshape(plot.df, varying = list(1:pp), direction = "long", v.names = "X", times = factor(colnames(XX), levels = colnames(XX))) ggplot(plot.df, aes(x = X, fill = W)) + geom_histogram(alpha = 0.5, position = "identity", bins=30) + facet_wrap( ~ time, ncol = 2, scales = "free") + ggtitle("Covariate histograms (unajusted)") ggplot(plot.df, aes(x = X, weight=IPW, fill = W)) + geom_histogram(alpha = 0.5, position = "identity", bins=30) + facet_wrap( ~ time, ncol = 2, scales = "free") + ggtitle("Covariate histograms (adjusted)") There exist other types of balance checks. We recommend checking out the R package cobalt. This vignette is a good place to start. ### 3.7.2 Assessing overlap It’s also important to check the estimated propensity scores. If they seem to cluster at zero or one, we should expect IPW and AIPW estimators to behave very poorly. Here, the propensity score is trimodal because of our sample modification procedure in Section 3.3: some observations are untouched and therefore remain with assignment probability $$0.5$$, some are dropped (kept) with probability 15% (85%). e.hat <- forest$W.hat
hist(e.hat, main="Estimated propensity scores (causal forest)", breaks=100, freq=FALSE, xlab="", ylab="", xlim=c(-.1, 1.1))
When overlap fails, the methods described above will not produce good estimates. In that case, one may consider changing the estimand and targeting the average treatment effect on the treated (ATT) $$\mathop{\mathrm{E}}[Y_i(1) - Y_i(0) | W_i=1]$$, or trimming the sample and focusing on some subgroup $$G_i$$ with bounded propensity scores $$\mathop{\mathrm{E}}[Y_i(1) - Y_i(0) | G_i]$$, as discussed in
Crump, Hotz, Imbens and Mitnik (Biometrika, 2009).
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# Triangle - math word problems
#### Number of problems found: 1034
• Perimeter of circle
Calculate the circumference of described circle to the triangle with sides 9,12,15 cm.
• Isosceles triangle
What are the angles of an isosceles triangle ABC if its base is long a=5 m and has an arm b=4 m.
• Park
In the park is marked diamond shaped line connecting locations A, D, S, C, B, A. Calculate its length if |AB| = 108 m, |AC| = 172.8 m.
• Rectangle diagonal
The rectangle, one side of which is 5 cm long, is divided by a 13 cm diagonal into two triangles. Calculate the area of one of these triangles in cm2.
• Hexagon area
The center of the regular hexagon is 21 cm away from its side. Calculate the hexagon side and its area.
• Triangular prism
Calculate the surface of a regular triangular prism with a bottom edge 8 of a length of 5 meters and an appropriate height of 60 meters and prism height is 1 whole 4 meters.
• Ballistic curve
The ballistic grenade was fired at a 45° angle. The first half ascended, the second fall. How far and how far it reached if his average speed was 1200km/h, and 12s took from the shot to impact.
• Angles ratio
In a triangle ABC true relationship c is less than b and b is less than a. Internal angles of the triangle are in the ratio 5:4:9. The size of the internal angle beta is:
• Diagonal
Calculate the length of the diagonal of the rectangle ABCD with sides a = 8 cm, b = 7 cm.
• Square diagonal
Calculate the length of diagonal of the square with side a = 23 cm.
• Six-sided polygon
In a six-sided polygon. The first two angles are equal, the third angle is twice (the equal angles), two other angles are trice the equal angle, while the last angle is a right angle. Find the value of each angle.
• Internal angles
One internal angle of the triangle JAR is 25 degrees. The difference is the size of the two other is 15°. Identify the size of these angles.
Ladder 10 meters long is staying against the wall so that its bottom edge is 6 meters away from the wall. What height reaches ladder?
• Circle section
Equilateral triangle with side 33 is inscribed circle section whose center is in one of the vertices of the triangle and the arc touches the opposite side. Calculate: a) the length of the arc b) the ratio betewwn the circumference to the circle sector and
• Decagon
Calculate the area and circumference of the regular decagon when its radius of a circle circumscribing is R = 1m
• Cube cut
The cube ABCDA'B'C'D ' has an edge of 12cm. Calculate the area of diagonal cut B DD'B '.
A ship travels 84 km on a bearing of 17°, and then travels on a bearing of 107° for 135 km. Find the distance of the end of the trip from the starting point, to the nearest kilometer.
• Diagonals of diamond
Find the area and circumference of the diamond ABCD with 15m and 11m diagonals.
• Dig water well
Mr. Zeman digging a well. Its diameter is 120 cm, and plans to 3.5 meters deep. How long (at least) must be a ladder, after which Mr. Zeman would have eventually come out?
• Angles of elevation
From points A and B on level ground, the angles of elevation of the top of a building are 25° and 37° respectively. If |AB| = 57m, calculate, to the nearest meter, the distances of the top of the building from A and B if they are both on the same side of
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We will send a solution to your e-mail address. Solved examples are also published here. Please enter the e-mail correctly and check whether you don't have a full mailbox.
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HW-Sec 15-Prob 1
# HW-Sec 15-Prob 1 - object is the closer to the lens its...
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Robb Page 19-Apr-2010 Physics 140 Professor Stoddard Chapter 15 – Problem 1 Ques. Your camera has a 35mm focal length lens. When you take a photograph of a distant mountain, how far from that lens is the real image? This requires an understanding of three values: Object distance – which is the distance between the lens and the subject or object. Image distance – which is the distance between the lens and the real image it forms. Focal length – which relates the object distance and the image distance. This relationship is referred to as the Lens Equation , and essentially means “the farther away an
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Unformatted text preview: object is, the closer to the lens its image forms” (pg. 524). As an equation, it can written as such: = + According to the textbook, the image distance for a distant object is equal to the focal length of the lens (pg. 524). Therefore, a 35mm lens (for a traditional film camera*) is a small (wide-angle) lens, and the real image formed will be bright and small and equal to the focal length of 35mm. = – = = 35mm * Note, a 35mm digital lens is roughly equivalent to a 50mm traditional (“normal”) film lens. 1 focal length 1 object distance 1 image distance 1 i 1 f 1 o 1 35...
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# Type 8: Parametric and Vector Questions
The parametric/vector equation questions only concern motion in a plane.
In the plane, the position of a moving object as a function of time, t, can be specified by a pair of parametric equations $x=x\left( t \right)\text{ and }y=y\left( t \right)$ or the equivalent vector $\left\langle x\left( t \right),y\left( t \right) \right\rangle$. The path is the curve traced by the parametric equations or the tips of the position vector. .
The velocity of the movement in the x- and y-direction is given by the vector $\left\langle {x}'\left( t \right),{y}'\left( t \right) \right\rangle$. The vector sum of the components gives the direction of motion. Attached to the tip of the position vector this vector is tangent to the path pointing in the direction of motion.
The length of this vector is the speed of the moving object. $\text{Speed }=\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}+{{\left( {y}'\left( t \right) \right)}^{2}}}$. (Notice that this is the same as the speed of a particle moving on the number line with one less parameter: On the number line $\text{Speed}=\left| v \right|=\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}}$.)
The acceleration is given by the vector $\left\langle {{x}'}'\left( t \right),{{y}'}'\left( t \right) \right\rangle$.
What students should know how to do:
• Vectors may be written using parentheses, ( ), or pointed brackets, $\left\langle {} \right\rangle$, or even $\vec{i},\vec{j}$ form. The pointed brackets seem to be the most popular right now, but all common notations are allowed and will be recognized by readers.
• Find the speed at time t$\text{Speed }=\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}+{{\left( {y}'\left( t \right) \right)}^{2}}}$
• Use the definite integral for arc length to find the distance traveled $\displaystyle \int_{a}^{b}{\sqrt{{{\left( {x}'\left( t \right) \right)}^{2}}+{{\left( {y}'\left( t \right) \right)}^{2}}}}dt$. Notice that this is the integral of the speed (rate times time = distance).
• The slope of the path is $\displaystyle \frac{dy}{dx}=\frac{{y}'\left( t \right)}{{x}'\left( t \right)}$. See this post for more on finding the first and second derivatives with respect to x.
• Determine when the particle is moving left or right,
• Determine when the particle is moving up or down,
• Find the extreme position (farthest left, right, up, down, or distance from the origin).
• Given the position find the velocity by differentiating; given the velocity find the acceleration by differentiating.
• Given the acceleration and the velocity at some point find the velocity by integrating; given the velocity and the position at some point find the position by integrating. These are just initial value differential equation problems (IVP).
• Dot product and cross product are not tested on the BC exam, nor are other aspects.
When this topic appears on the free-response section of the exam there is no polar equation question and vice versa. When not on the free-response section there are one or more multiple-choice questions on parametric equations.
Free-response questions:
• 2012 BC 2
• 2016 BC 2
Multiple-choice questions from non-secure exams
• 2003 BC 4, 7, 17, 84
• 2008 BC 1, 5, 28
• 2012 BC 2
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## 1 thought on “Type 8: Parametric and Vector Questions”
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https://www.elitedigitalstudy.com/12260/reduce-to-the-standard-form-11-4i-21i3-4i5i
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Reduce to the standard form $$( \dfrac{1}{1-4i}-\dfrac{2}{1+i})(\dfrac{3-4i}{5+i})$$ to the standard form
Asked by Pragya Singh | 1 year ago | 50
##### Solution :-
$$( \dfrac{1}{1-4i}-\dfrac{2}{1+i})(\dfrac{3-4i}{5+i})$$
$$(\dfrac{1}{1-4i}-\dfrac{2}{1+i})[\dfrac{3-4i}{5+i}]$$
$$[\dfrac{1+i-2+8i}{1+i-4i-4i^2}] [\dfrac{3-4i}{5+i}]$$
$$[\dfrac{-1+9i}{5-3i}] [\dfrac{3-4i}{5+i}]$$
$$[\dfrac{-3+4i+27i-36i^2}{25+5i-15i-3i^2}]$$
$$\dfrac{33+31i}{2(14-5i)}$$
$$\dfrac{33+31i}{2(14-5i)}\times \dfrac{(14+5i)}{(14+5i)}$$
[On multiplying numerator and denominator by(14 + 5i)]
$$\dfrac{462+165i+434i+155i^2}{2[(14)^2-(5i)^2]}$$
$$\dfrac{307+599i}{2(196-25i^2)}$$
$$\dfrac{307+599i}{2(221)}$$ = $$\dfrac{307}{442}+\dfrac{599i}{442}$$
This is the required standard form
Answered by Abhisek | 1 year ago
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#### Solve the quadratic equations by factorization method only 6x2 – 17ix – 12 = 0
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#### Solve the quadratic equations by factorization method only x2 + (1 – 2i)x – 2i = 0
Solve the quadratic equations by factorization method only x2 + (1 – 2i)x – 2i = 0
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A263725 Smallest prime q > prime(n+3) such that the number p = prime(n)^2 + prime(n+1)^2 + prime(n+2)^2 + prime(n+3)^2 + q^2 is also prime. 1
13, 17, 37, 31, 31, 37, 41, 41, 43, 47, 59, 61, 89, 79, 71, 79, 79, 89, 97, 109, 127, 107, 109, 109, 113, 139, 131, 139, 151, 149, 157, 157, 173, 181, 173, 191, 191, 193, 197, 223, 199, 211, 233, 239, 229, 233, 263, 257, 263, 271, 271, 277, 271, 281, 281, 293, 293, 311, 349, 317, 353, 331, 353, 353, 359, 419, 359, 419, 379, 419, 397, 401, 431, 409, 409, 433, 461, 443, 487, 449, 541, 487, 463, 569, 479, 467, 487, 491, 503 (list; graph; refs; listen; history; text; internal format)
OFFSET 2,1 COMMENTS The corresponding primes p form A263724. The prime q exists for all n > 1 under Schinzel's Hypothesis H; see Sierpinski (1988), p. 221. REFERENCES W. Sierpinski, Elementary Theory of Numbers, 2nd English edition, revised and enlarged by A. Schinzel, Elsevier, 1988. LINKS Wikipedia, Schinzel's Hypothesis H EXAMPLE The primes 373 = 3^2 + 5^2 + 7^2 + 11^2 + 13^2, 653 = 5^2 + 7^2 + 11^2 + 13^2 + 17^2, and 1997 = 7^2 + 11^2 + 13^2 + 17^2 + 37^2 lead to a(2) = 13, a(3) = 17, and a(4) = 37. MATHEMATICA Table[k = 4; While[! PrimeQ[Sum[Prime[n + j]^2, {j, 0, 3}] + Prime[n + k]^2], k++]; Prime[n + k], {n, 2, 90}] CROSSREFS Cf. A263724. Sequence in context: A126808 A053009 A069485 * A174056 A307894 A322472 Adjacent sequences: A263722 A263723 A263724 * A263726 A263727 A263728 KEYWORD nonn AUTHOR Jonathan Sondow, Oct 24 2015 STATUS approved
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compoly - Maple Help
compoly
determine a possible composition of a polynomial
Calling Sequence compoly(r) compoly(r, x) compoly(r, {x, ... })
Parameters
r - polynomial x - variable on which the composition will be made
Description
• The function compoly returns a pair $p\left(x\right)$, $x=q\left(x\right)$ such that subs(x=q(x), p(x)) is equal to r, the input polynomial. If such a pair cannot be found, it returns FAIL. $p\left(x\right)$ and $q\left(x\right)$ are nonlinear polynomials and $q\left(x\right)$ has a low-degree in x greater or equal to 1.
• When compoly is called without additional arguments, it is equivalent to being called with a second argument which is indets(r). When the second argument is a set with more than one variable, then a multivariate composition is attempted. In this case, the result has the same form, but the second polynomial is multivariate, that is, $p\left(x\right)$, $x=q\left(x,...\right)$. For the multivariate case, the second polynomial, $q\left(x\right)$, may be of degree 1.
• Note that the composition may not be unique. In particular, if $p\left(x\right)$, $x=q\left(x,...\right)$ is a composition, then we can find another $p\left(x\right)$ such that replacing $q\left(x,...\right)$ by $cq\left(x,\dots \right)+b$ will also result in a valid composition. This non-determinacy is eliminated by selecting c and b such that the q polynomial has integer content 1 and its independent term is 0.
Examples
> $\mathrm{compoly}\left({x}^{6}-9{x}^{5}+27{x}^{4}-27{x}^{3}-2{x}^{2}y+6xy+1,x\right)$
${{x}}^{{3}}{-}{2}{}{x}{}{y}{+}{1}{,}{x}{=}{{x}}^{{2}}{-}{3}{}{x}$ (1)
> $\mathrm{compoly}\left({x}^{4}-3{x}^{3}-x+5,x\right)$
${\mathrm{FAIL}}$ (2)
> $\mathrm{compoly}\left({x}^{2}+2xy-7x+{y}^{2}-7y+16\right)$
${{x}}^{{2}}{-}{7}{}{x}{+}{16}{,}{x}{=}{x}{+}{y}$ (3)
> $\mathrm{compoly}\left({x}^{4}+4{x}^{3}{y}^{3}+6{x}^{2}{y}^{6}+4x{y}^{9}+{y}^{12}+x+{y}^{3}-1,\left\{x,y\right\}\right)$
${{x}}^{{4}}{+}{x}{-}{1}{,}{x}{=}{{y}}^{{3}}{+}{x}$ (4)
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https://oeis.org/A086140
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A086140 Primes p such that three (the maximum number) primes occur between p and p+12. 29
5, 7, 11, 97, 101, 1481, 1867, 3457, 5647, 15727, 16057, 16061, 19417, 19421, 21011, 22271, 43777, 43781, 55331, 79687, 88807, 101107, 144161, 165701, 166841, 195731, 201821, 225341, 247601, 257857, 266677, 268811, 276037, 284737, 326141, 340927 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS p+12 must be a prime. - Harvey P. Dale, Jun 11 2015 A086140 is the union of A022006 and A022007. By merging the two b-files I have extended the current b-file up to n=10000 (nearly n=20000 would have been possible). I add a comparison (see Links) between the frequency of prime quintuplets and an asymptotic approximation, which is unproven but likely to be true, and based on a conjecture first published by Hardy and Littlewood in 1923. Twins, triples and quadruplets are treated as well. - Gerhard Kirchner, Dec 07 2016 LINKS Harvey P. Dale and Gerhard Kirchner, Table of n, a(n) for n = 1..10000 (first 1000 terms from Harvey P. Dale) Gerhard Kirchner, Comparison with an assumed asymptotic distribution EXAMPLE There are two types of prime quintuplets, and both are represented in this sequence. (11, 13, 17, 19, 23) is a prime quintuplet of the form (p, p+2, p+6, p+8, p+12), so 11 is in the sequence, and (97, 101, 103, 107, 109) is a prime quintuplet of the form (p, p+4, p+6, p+10, p+12), so 97 is in the sequence. - Michael B. Porter, Dec 19 2016 MATHEMATICA cp[x_, y_] := Count[Table[PrimeQ[i], {i, x, y}], True] {d=12, k=0}; Do[s=Prime[n]; s1=Prime[n+1]; If[PrimeQ[s+d]&&Equal[cp[s+1, s+d-1], 3], k=k+1; Print[s]], {n, 1, 100000}] (* Second program: *) Transpose[Select[Partition[Prime[Range[30000]], 5, 1], #[[5]]-#[[1]] == 12&]][[1]] (* Harvey P. Dale, Jun 11 2015 *) CROSSREFS Cf. A031930, A046133, A086139, A086136, A022006, A022007, A001359 (twins), A007529 (triples), A007530 (quadruplets). Sequence in context: A091509 A027728 A218275 * A104387 A133761 A057659 Adjacent sequences: A086137 A086138 A086139 * A086141 A086142 A086143 KEYWORD nonn AUTHOR Labos Elemer, Jul 29 2003 STATUS approved
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Last modified June 16 14:16 EDT 2021. Contains 345057 sequences. (Running on oeis4.)
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https://cs.stackexchange.com/questions/64106/how-to-find-the-minimal-dfa-for-the-language
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# How to find the minimal DFA for the language?
Design a minimal DFA for the language which contains numbers in the form of binary strings starting with $101_2$ and is divisible by $100_{10}$?
I think the numbers will be 700,1400,2100 ... in the form of binary strings which are starting with $101_2$ and are even divisible by $100_{10}$.
100 is divisible by 25 and 4.
For divisible by 4 - Minimum 3 state DFA is possible For divisible by 25 - Min 25 state DFA is possible.
Hence, for divisible by $100_{10}$ - Minimum 100 states DFA is possible .
So, for starting with $101_2$, I have to add 3 extra states along with one dead state.
So, total states in minimal DFA = 100 + 3 + 1 = 104
Is my approach correct or am I missing something?
• So, i thought to post it here, as maybe it is not a field for automata. Oct 1, 2016 at 19:32
• divisible by $100_{10}$ or $100_2$? Oct 1, 2016 at 20:25
• Please do not post the same question on multiple sites. Each community should have an honest shot at answering without anybody's time being wasted. Thank you!
– D.W.
Oct 2, 2016 at 0:20
• In your list of numbers, you appear to have missed 1300, 1500. By my reckoning, the list begins 700, 1300, 1400, 1500, 2600, 2700, 2800, 2900, 3000, 5200, 5300, 5400, ... Oct 2, 2016 at 23:44
Your approach towards estimating the number of states is just fine. however the number of states looks quite intimidating and it may be cumbersome to minimize the DFA later.
In such cases constructing the automaton through intersection is easier. for example as we know the regular languages are closed under the intersection operation we can devise 3 different automata to accept :
1. String divisible by 25
2. String divisible by 4
3. Strings beginning with 101
And then arrive at the final automaton by taking intersections of the 3 machines.
### How this helps?
Many of the states originally present in the different machines gradually become unreachable from the initial state in the final automaton and thus the number of states reduce or the DFA is minimized.
• Before going any further with a question i would like to make sure the machine you require tests divisibility with 100 in base 10 or 100 in base 2 Oct 2, 2016 at 16:10
• "Many of the states originally present in the different machines gradually become unreachable from the initial state in the final automaton" - Are you sure? How do you know? I'm quite skeptical: I don't think any of the states become unreachable.
– D.W.
Oct 2, 2016 at 23:21
Your initial approach is right but can be improved.
First, a binary number is a multiple of $4$ if and only if it ends with $00$. Thus the conditions "starting with $101$" and "multiple of $4$" together define the language $L = 101\{0,1\}^*00$. The minimal DFA of $L$ is the following: You now have to compute the intersection of $L$ with the language $K = \{ \text{binary strings representing the multiples of$25$}\}$. As you observed, the minimal DNA of $K$ has $25$ states, say $0, \ldots, 24$, corresponding to the possible values modulo $25$ of a natural number. A DFA accepting $L \cap K$ is obtained by taking the accessible states in the product of the previous automata. The initial state is $(1, 0)$ and the first 4 transitions are the following (since, for instance $101_2 = 5_{10}$). Starting from state $4$, you may have to consider all states $(4,n)$, $(5,n)$ and $(6,n)$, where $0 \leqslant n \leqslant 24$. This gives an upperbound of $3 \times 25 + 3 = 78$ for the minimal DFA of $L \cap K$ (plus one if you want to add the sink state).
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Welcome to my math blog! The purpose of this blog is to help you stay informed about our learning and experiences that have taken place during our math class. I have also included links your child (and you) may want to use in order to supplement math learning in 5th grade.
## Monday, April 3, 2017
### STEAM: Geometric Sewing (Line Designs)
Every year after taking our Math STAAR test, I like to give my kids a little mathematical art therapy! It begins with the Kaleidoscopic Octagon, which emphasizes rotational and reflectional symmetry. Then, we move into geometric sewing. Basically, this is line design with string.
Line designs form a basis for mathematical understanding of geometric shapes and relationships of points, segments, and angles. Each of the line segments is really a tangent for each of the curves being formed. But because of what we focus on, we often see the curves. For example, some of the curves that can be created are circles, parabolas, ellipses, hyperbolas, spirals, and some lesser known curves called cardioids, limacons, and deltoids. Yet in each case they were created with angles of different sizes, regular and irregular polygons, and a lot of segments and points.
http://www.teachersnetwork.org/dcs/math/stringart/
Today, we began by preparing our sewing surface. I gave each student an 8"x 8" piece of tagboard. I also gave them each a pattern from a book that is no longer in print titled "String Art...10 Basic Designs." In fact, it is from so long ago the patterns were originally to be run on a ditto machine! The patterns may be old, but the excitement is always new!
Once we have cut out the pattern and taped it onto our tagboard, I gave each student a stick pin and and old mouse pad. This may be my students' favorite part... we put the tagboard down onto the mouse pad and use the stick pin to poke a hole into every dot on the pattern. I have learned, after YEARS of practice, that pre-poking the holes leads to fewer injuries with needles!
Next, it was time to teach my students how to thread a needle. There is one rule I have during this entire project... I DO NOT THREAD NEEDLES!
It takes a while to learn how to thread a needle, but I have a very cool way of teaching my students how to thread a needle that involves using tape to make an aglet. I learned about aglets while watching Phineas and Ferb. They even have an A-G-L-E-T Song!
Then, we created our aglet:
Once we had our needle on our thread, the time was up! AARGH! We sew tomorrow!
STEAM is an educational approach to learning that uses Science, Technology, Engineering, the Arts and Mathematics as access points for guiding student inquiry, dialogue, and critical thinking. The end results are students who take thoughtful risks, engage in experiential learning, persist in problem-solving, embrace collaboration, and work through the creative process. These are the innovators, educators, leaders, and learners of the 21st century! (
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https://studysoup.com/tsg/10070/physics-principles-with-applications-6-edition-chapter-5-problem-12pe
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Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 5 - Problem 12pe
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# an object is to rest on an incline without slipping, the
ISBN: 9780130606204 3
## Solution for problem 12PE Chapter 5
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Problem 12PE
an object is to rest on an incline without slipping, the friction must be equal the component of the weight of the object parallel to the incline. This requires greater and greater friction for steeper slopes. Show that the maximum angle of an incline above the –1? horizontal for which an object will not slide down is ? = tan? ?? . Yos?may use the result of the previous problem. Assu? me that a ? = 0 and that static friction has reached its maximum value.
Step-by-Step Solution:
Step-by-step solution Step 1 of 4 The free body diagram for the inclined plane is given below, the coefficient of static friction
Step 2 of 4
Step 3 of 4
##### ISBN: 9780130606204
This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6. The full step-by-step solution to problem: 12PE from chapter: 5 was answered by , our top Physics solution expert on 03/03/17, 03:53PM. Since the solution to 12PE from 5 chapter was answered, more than 410 students have viewed the full step-by-step answer. The answer to “an object is to rest on an incline without slipping, the friction must be equal the component of the weight of the object parallel to the incline. This requires greater and greater friction for steeper slopes. Show that the maximum angle of an incline above the –1? horizontal for which an object will not slide down is ? = tan? ?? . Yos?may use the result of the previous problem. Assu? me that a ? = 0 and that static friction has reached its maximum value.” is broken down into a number of easy to follow steps, and 86 words. This full solution covers the following key subjects: Friction, Object, Incline, maximum, Greater. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204.
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https://essayprince.net/math-20-question-multiple-choice-quiz/
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# Math 20 Question Multiple Choice Quiz
Write my research paper
Nathaniel buys a pair of jeans costing \$45.30. Using the table below to find the sales tax on this item, what is the total purchase price? Amount of SaleTax 44.70 ? 44.892.92 44.90 ? 45.092.93 45.10 ? 45.292.94 45.30 ? 45.492.95 a.\$48.35 b.\$48.25 c.\$47.00 d.\$48.02 Eva purchases a bicycle costing \$150.95. State taxes are 5.5% and local sales taxes are 2.4%. The store charges \$30 for assembly. What is the total purchase price? (Round your answer to the nearest cent if necessary)
a.\$155.95 b.\$192.88 c.\$161.41 d.\$168.25 Georgia bought outdoor furniture for a total purchase price of \$2,150.75. State taxes were 5.75%. Find the amount of the sales tax. (Round your answer to the nearest cent if necessary)
a.\$123.67 b.\$116.94 c.\$109.91 d.\$154.08 Spiros purchases a Rolex watch costing \$10,600. State taxes are 4.5% and the federal excise tax is 11%. What is the total purchase price? a.\$11,752 b.\$11,320 c.\$11,128 d.\$12,243 Al purchases a speedboat costing \$24,500. State taxes are 5.5% and federal excise tax is 13%. What is the total purchase price? (Round your answer to the nearest cent if necessary)
a.\$25,442.50 b.\$27,128.50 c.\$29,032.50 d.\$26,240.00 You have some property that has an assessed value of \$87,500. If the tax rate is \$8.79 per \$100 of assessed value, calculate the tax due. (Round your answer to the nearest cent if necessary)
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http://www.slideshare.net/acsteele/probability-3944024
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# probability
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### probability
1. 1. Probability Ms. Steele’s Fourth Grade Class May 3rd, 2010
2. 2. What is probability? <ul><ul><li>Probability is defined as the likelihood that an event will occur, expressed as the ratio of the number of favorable outcomes in the set of outcomes divided by the total number of possible outcomes </li></ul></ul>
3. 3. What does that mean? <ul><ul><li>Probability tells you the chances of a certain event, such as getting struck by lightning or choosing a blue cube out of a bag full of cubes. </li></ul></ul>
4. 4. For example: <ul><li>You have a bag filled with colored cubes. In this bag you have: </li></ul><ul><ul><li>3 yellow cubes </li></ul></ul><ul><ul><li>4 red cubes </li></ul></ul><ul><ul><li>1 blue </li></ul></ul><ul><ul><li>6 green cubes </li></ul></ul><ul><ul><li>Out of the cubes in the bag, what are the chances of us picking a yellow cube? A red cube? A blue cube? Or a green cube? </li></ul></ul>
5. 5. What happens when we don’t replace the item? <ul><ul><li>What do you think happens when you don’t replace the last item you chose? </li></ul></ul><ul><ul><li>What would be the probability of choosing each type of cube? </li></ul></ul>
6. 6. Let’s see what happens.. This bag is filled with different types of candy. What are the types of candy we have and how many are there of each?
7. 7. Our Experiment
8. 8. We have 7 Twix, 8 Skittles, and 8 Reese’s Peanut Butter Cups. How many do we have in total?
9. 9. What is the probability of.. <ul><ul><li>Selecting a Twix out of our bag of candy? </li></ul></ul><ul><ul><ul><ul><li>(Hint: there are 7 Twix in our bag) </li></ul></ul></ul></ul>7/23 Or 7 out of 23
10. 10. Let’s See What Happens if We Don’t Replace the Candy! <ul><ul><li>What is the probability of the volunteer selecting one type of candy? </li></ul></ul><ul><ul><li>What is the probability of the volunteer selecting two different types of candy? </li></ul></ul><ul><ul><li>What is the probability of the volunteer selecting every other type of candy except for the first one they selected? </li></ul></ul>
11. 11. Thank you for participating! Enjoy the candy!
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http://www.slideshare.net/IzzatNajmi/specific-heat-capacity
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# specific heat capacity
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### Transcript
• 1. Present by: ‘izzat najmi bin ibrahimNoor fatinah binti muhd rusli Punitha A/P Nagappan
• 2. DEFINITITION A materials heat capacity is a measure of how much energy must be exchanged between an object and its environment to produce a change in temperature. In other words, it is a measure of an objects "capacity" to hold "heat." Materials with high heat capacities, such as water, require large amounts of energy to produce a small temperature change. Heat Capacity Heat capacity is the amount of heat required to increase the temperature of an object by 1 oC (or 1 K).
• 3. TYPE OF HEAT CAPACITY Molar heat capacity Specific heat Heat capacity
• 4. FORMULA OF SPECIFIC HEAT CAPACITY
• 5. EXAMPLEHow much thermal energy is required to raise the temperature of a 2 kg aluminium block from 25 C to 30 C? [The specific heat capacity of aluminium is 900 Jkg-1 oC-1]Answer:Mass, m = 2kg Specific heat capacity, c = 900 Jkg-1 oC-1 Temperature change, θ = 30 - 25 = 5 oCThermal energy required, Q = mcθ = (2)(900)(5) = 9000J.
• 6. CONVERSION OF ELECTRICAL ENERGY INTOTHERMAL ENERGY
• 7. EXAMPLE A lead shot of mass 5g is placed at the bottom of a vertical cylinder thatis 1m long and closed at both ends. The cylinder is inverted so that the shotfalls 1 m. By how much will the temperature of the shot increase if thisprocess is repeated 100 times? [The specific heat capacity of lead is130Jkg-1K-1] Answer: m = 5gh = 1m × 100 = 100mg = 10 ms-2c = 130Jkg-1K-1θ=? In this case, the energy conversion is from potential energy to heatenergy. We assume that all potential energy is converted into heat energy.Therefore mgh = mcθgh = cθ
• 8. CONVERSION OF GRAVITATIONAL ENERGY INTOTHERMAL ENERGY
• 9. CONVERSION OF KINETIC ENERGY INTOTHERMAL ENERGY
• 10. MIXING 2 LIQUID
• 11. EXAMPLEExample 4What will be the final temperature if 500 cm3 of water at 0 �C is added to 200cm3 of water at 90 �C? [Density of water = 1gcm-3]Answer:The density of water is 1g/cm3, which means the mass of 1 cm3 of water is equal to 1g.Let the final temperature = θ m1 = 500g = 0.5kg c1 = c θ1 = θ - 0 = θ m2 = 200g = 0.2kg c2 = c θ2 = 90 - θm1c1θ1 = m2c2θ2 (0.5) c ( θ ) = (0.2) c ( 90 - θ ) 0.5 θ = 18 - 0.2 θ 0.5 θ + 0.2 θ = 18 0.7 θ = 18 θ = 25.71 oC
• 12. THE END
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# How To Find Lateral Surface Area
## Understanding Lateral Surface Area and Its Importance
The lateral area formula is a crucial concept in geometry and is used to find the lateral area of any solid object. It specifically calculates the area of the non-base faces of a three-dimensional figure. Lateral area formulas are essential in determining the lateral surface area of various geometric shapes such as cuboids, cubes, cylinders, cones, and spheres. Understanding these formulas is vital for accurate calculations and real-world applications.
### Lateral Area Formulas for Different Geometric Solids
The lateral area formula varies for different types of objects. Each geometric solid has its specific lateral area formula, which is used to calculate its lateral surface area. Let’s explore some of the common lateral area formulas for different geometric shapes:
### Cuboid
The lateral surface area of a cuboid can be calculated using the formula:
$S_{lat} = 2h(l + w)$
Where $$h$$ is the height, $$l$$ is the length, and $$w$$ is the width of the cuboid.
### Cube
For a cube, the lateral surface area can be found using the formula:
$S_{lat} = 4a^2$
Where $$a$$ represents the length of the side of the cube.
### Cylinder
The lateral surface area of a cylinder is determined by the formula:
$S_{lat} = 2\pi rh$
Here, $$r$$ denotes the radius and $$h$$ represents the height of the cylinder.
### Cone
For a cone, the lateral surface area can be calculated using the formula:
$S_{lat} = \pi r \sqrt{r^2 + h^2}$
Where $$r$$ is the radius and $$h$$ is the slant height of the cone.
### Sphere
The lateral surface area of a sphere is given by the formula:
$S_{lat} = 4\pi r^2$
Here, $$r$$ represents the radius of the sphere.
## Importance of Understanding Lateral Area Formulas
Understanding lateral area formulas is crucial for various real-world applications. Whether it’s calculating the material required to construct a container, determining the amount of paint needed to cover a curved surface, or estimating the wrapping paper needed for a gift, the knowledge of lateral area formulas is indispensable. Additionally, in fields such as architecture, engineering, and design, the ability to accurately calculate lateral surface area is essential for creating and constructing three-dimensional structures.
## How to Find Lateral Surface Area
When it comes to finding the lateral surface area of different geometric solids, it’s essential to understand the specific formulas for each shape. By following the correct formula and substituting the given values, one can accurately calculate the lateral surface area of the desired object. Let’s take a closer look at the process of finding the lateral surface area for a few common geometric shapes:
### Finding Lateral Surface Area of a Cuboid
To find the lateral surface area of a cuboid, use the formula $$S_{lat} = 2h(l + w)$$, where $$h$$ is the height, and $$l$$ and $$w$$ are the length and width, respectively. Substitute the given values into the formula and perform the necessary calculations to obtain the lateral surface area of the cuboid.
### Finding Lateral Surface Area of a Cylinder
For a cylinder, the lateral surface area can be determined using the formula $$S_{lat} = 2\pi rh$$, where $$r$$ is the radius and $$h$$ is the height. By substituting the known values into the formula and performing the required arithmetic, one can find the lateral surface area of the cylinder.
### Finding Lateral Surface Area of a Cone
When calculating the lateral surface area of a cone, use the formula $$S_{lat} = \pi r \sqrt{r^2 + h^2}$$, where $$r$$ represents the radius and $$h$$ is the slant height. Substituting the given values into the formula and evaluating the expression will yield the lateral surface area of the cone.
## Common FAQs About Finding Lateral Surface Area
### 1. Why is the lateral area important in geometry?
The lateral area is crucial in geometry as it helps in determining the surface area of three-dimensional objects, which is essential for various real-world applications such as construction, design, and manufacturing.
### 2. Can the lateral area formula be used for irregular shapes?
No, the lateral area formula is specific to regular geometric solids with defined formulas for each shape. For irregular shapes, other methods such as approximation or decomposition into regular shapes may be used to find the lateral surface area.
### 3. How does understanding lateral area formulas benefit professionals in architecture and engineering?
For professionals in architecture and engineering, understanding lateral area formulas is essential for accurately calculating the surface area of structures, which is crucial for material estimation, structural design, and construction planning.
### 4. What are some real-world examples where knowledge of lateral area formulas is applied?
Real-world examples include determining the amount of wallpaper needed to cover a curved wall, calculating the surface area of cylindrical tanks for storing liquids, and estimating the material required to construct geometrically shaped objects such as pillars and beams.
### 5. Are there any online tools available for calculating lateral surface area?
Yes, there are various online calculators and tools that allow users to input the dimensions of a geometric solid and obtain the calculated lateral surface area based on the specific formula for that shape.
### 6. How can students improve their understanding of lateral area formulas?
Students can enhance their understanding by practicing solving problems related to finding lateral surface area for different geometric shapes, seeking guidance from teachers or online resources, and visualizing the three-dimensional objects to comprehend the concept better.
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# How would you solve this linear system, $Ax=b?$
I am trying to learn linear algebra on my own and am going through some old exams I found. I'm not sure how to solve this one. My initial thought was to find the inverse of $$A$$ and multiply $$b$$ with it, but apparently since it's not a square matrix I can't take the inverse of $$A$$. I also considered trying to combine rows to simplify and find some variables that way that I could use to solve the rest of the system but wasn't successful with that. I'd appreciate some guidance from the community.
Solve the linear system $$Ax=b$$ where
$$A= \begin{bmatrix} 3 & -4 & -3 & 1 & 2 & 1 \\ 0 & 2 & 1 & -1 & -3 & -1 \\ 0 & 0 & 0 & 0 & 3 & 1 \\ \end{bmatrix}$$
and
$$b= \begin{bmatrix} -1 \\ 4 \\ 6 \\ \end{bmatrix}$$ and write your solution in vector form.
$$A= \begin{bmatrix} 3 & -4 & -3 & 1 & 2 & 1 \\ 0 & 2 & 1 & -1 & -3 & -1 \\ 0 & 0 & 0 & 0 & 3 & 1 \\ \end{bmatrix}$$
and
$$b= \begin{bmatrix} -1 \\ 4 \\ 6 \\ \end{bmatrix}$$
The matrix $A$ is already in row echelon form. The pivot columns are the first column, second column, and the fifth columns. They are the columns that contain leading non-zero entries of each row.
Assign parameter $r$,$s$, and $t$ to the third, fourth, and sixth columns respectively.
$$x_3=r$$
$$x_4=s$$ $$x_6=t$$
The last row tells us that $3x_5+t=6$. Hence, we have
$$x_5=\frac{6-t}{3}$$
Try to express $x_1$ and $x_2$ in $r,s$, and $t$ too.
You have only three equations and six unknowns, as the $x$ vector must be six long to be able to multiply it by $A$. The rows of $A$ are linearly independent, so there will be a three dimensional space of solutions. You will be able to choose three variables at whim, then calculate the other three. The last row insists that one of the variables you choose must be $x_5$ or $x_6$ and one of the free variables must be the other.
You're right that you can't solve this system of equations by finding an inverse of $A$. No such matrix exists because $A$ isn't square! In general, you can solve this problem by performing Gaussian elimination on the augmented matrix $\left(A \mid b\right)$ and as long as there are no equations of the form $c = 0$ in the reduced row-echelon form of $(A\mid b)$, where $c$ is itself nonzero, you will have a solution, or perhaps infinitely many.
Consider the augmented matrix $(A \mid b)$ whose first $6$ columns are the columns of $A$ and whose seventh column is $b$. Add row $3$ to row $2$ and replace row $2$ with the result. Proceed until $(A \mid b)$ is in reduced row-echelon form. That should be enough to get you started.
It is clear that in that equation there are more variables than equations, so the system is indetermined.
However, you may use a least squares aproximation to solve it in the sense of minimizing the 2-norm error. Mathematically this means:
$\bar{x}=argmin_x ||Ax-b||_2$
The answer to this optimización problem is $\bar{x}=A^{\dagger}b$ where $A^{\dagger}$ is th so called Moore-Penrose pseudoinverse.
This ovbiously is not the only solución to the equations, however, it gives you an aproximation that might be useful that minimizes the 2-norm error.
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Lesson Objectives
• Demonstrate an understanding of how to factor
• Demonstrate an understanding of how to multiply two binomials using FOIL
• Learn how to factor a trinomial using the AC method
• Learn how to factor a trinomial using the reverse FOIL method
## How to Factor Trinomials when the Leading Coefficient is not 1
In our last lesson, we learned how to factor trinomials of the form:
ax2 + bx + c, a = 1
This means we covered the easy case scenario in which we factored:
x2 + bx + c
Having a coefficient of 1 for our x2 part meant that the first term of each binomial was x:
(x + __)(x + __)
To factor a trinomial of this form, we only needed to figure out two integers that would sum to b (the coefficient of x) and give a product of c (constant). As an example, suppose we want to factor:
x2 + 5x + 6
We could set up our known part as:
(x + __)(x + __)
Now we just need two integers that sum to 5 and give a product of 6. Two such integers would be 2 and 3. We can place the integers in the blanks in any order:
x2 + 5x + 6 = (x + 2)(x + 3)
### Factoring Trinomials using the AC Method
What can we do when the coefficient on the x2 part is not 1? This is the harder scenario and often gives students some grief when starting out. There are many techniques available for this type of problem, but we will only focus on two such solutions. The first is known as factoring using the AC method or grouping. Essentially we are expanding our trinomial into a four-term polynomial and then using the factoring by grouping method we learned a few lessons ago.
• Factor out any common factor other than 1 or (-1)
• Look for two integers whose product is ac (a multiplied by c) and whose sum is b
• Use these two integers to rewrite the middle term so that we have a four-term polynomial
• Factor the four-term polynomial using the grouping method
Let's take a look at an example.
Example 1: Factor each.
12x2 + 27x + 6
Step 1) Factor out any common factor other than 1 or (-1)
In this case, we can factor out a 3:
3(4x2 + 9x + 2)
Now we will only focus on what is inside of the parentheses.
Step 2) Look for two integers whose product is ac and whose sum is b:
a = 4, b = 9, c = 2
a • c » 4 • 2 = 8
We need to find two integers whose sum is 9 and whose product is 8. Let's think about the factors of 8:
1, 8
2, 4
1 • 8 = 8
1 + 8 = 9
Our two integers are 1 and 8.
Step 3) Use these two integers to rewrite the middle term so that we have a four-term polynomial.
3(4x2 + 8x + 1x + 2)
Notice how we changed 9x into (8x + 1x). This is legal since (8x + 1x = 9x).
Step 4) Factor the four-term polynomial using the grouping method:
3(4x2 + 8x + 1x + 2)
3(4x(x + 2) + 1(x + 2))
3(x + 2)(4x + 1)
Example 2: Factor each.
8x2 + 14x - 30
Step 1) Factor out any common factor other than 1 or (-1)
In this case, we can factor out a 2:
2(4x2 + 7x - 15)
Now we will only focus on what is inside of the parentheses.
Step 2) Look for two integers whose product is ac and whose sum is b:
a = 4, b = 7, c = -15
a • c » 4 • -15 = -60
We need to find two integers whose sum is 7 and whose product is -60. Let's think about the factors of 60 and then work out the signs:
1, 60
2, 30
3, 20
4, 15
5, 12
6, 10
Looking through our factors, think about the difference between each pair. We can then work out the sign afterward. If we subtract 12 - 5, we get 7. We have found our two integers (12, -5):
-5 • 12 = -60
-5 + 12 = 7
Step 3) Use these two integers to rewrite the middle term so that we have a four-term polynomial.
2(4x2 + 12x - 5x - 15)
Notice how we changed 7x into (12x - 5x). This is legal since (12x - 5x = 7x).
Step 4) Factor the four-term polynomial using the grouping method:
2(4x2 + 12x - 5x - 15)
2(4x(x + 3) + (-5)(x + 3))
2(x + 3)(4x - 5)
### Factoring Trinomials using Reverse FOIL
Alternatively, when we factor a trinomial and the leading coefficient is not 1, we can use the reverse FOIL method. The reverse FOIL method involves undoing the FOIL process through trial and error. Let's first multiply two binomials together using FOIL:
(5x + 1)(x + 5)
F » 5x • x = 5x2
O » 5x • 5 = 25x
I » 1 • x = x
L » 1 • 5 = 5
We know that we can combine like terms and obtain:
5x2 + 26x + 5
How could we reverse this process? Let's start with the first term 5x2. This is the result of the F in FOIL. This means first times first. 5 is a prime number and can only come from 5 times 1. This means we can set up our first terms as:
(5x + __)(x + __)
Now we think about the last term of 5. We know the two blank spots need to multiply together to give us 5. Again, since 5 is a prime number, we know the only possibilities are 5 and 1. We need to check the outer and inner products, these need to sum to 26x:
Trial 1:
(5x + 5)(x + 1)
Outer » 5x
Inner » 5x
O + I = 5x + 5x = 10x
10x ≠ 26x
Since we did not successfully obtain the correct middle term, we try another configuration:
(5x + 1)(x + 5)
Outer » 25x
Inner » 1x
O + I = 25x + 1x = 26x
We have the correct middle term, therefore we have found our correct factorization. Let's take a look at another example.
Example 3: Factor each.
7x2 - 15x - 18
Step 1) Let's begin by looking out our first term (7x2). 7 is a prime number and can only come from 1 times 7:
(7x + __)(x + __)
Step 2) Let's think about our last term (-18). We will look at the factors of +18 and then play with the signs:
1, 18
2, 9
3, 6
Step 3) Any combination of one positive and one negative would give us a negative product. Let's think about the possibilities:
We want to achieve a middle term of (-15x). This comes from the O + I in the FOIL process. In the outer step, 7x would multiply one factor and in the inner step, 1x would multiply the other. Let's just work with positives for a moment. We begin with the factors 1 and 18:
7 • 1 = 7
7 • 18 = 126
1 • 1 = 1
1 • 18 = 18
There's no way to make the factors 1 and 18 work no matter the signs, let's move on to the next pair of factors, 2 and 9:
7 • 9 = 63
7 • 2 = 14
1 • 9 = 9
2 • 9 = 18
Again, no way to make the factors 2 and 9 work no matter the signs, let's move on to the next pair of factors, 3 and 6:
7 • 3 = 21
7 • 6 = 42
1 • 3 = 3
1 • 6 = 6
The coefficient of the middle term, -15, can come from (-21 + 6). This means we need to multiply (7x • -3) and (1x • 6). We will factor our trinomial as:
7x2 - 15x - 18 = (7x + 6)(x - 3)
#### Skills Check:
Example #1
Factor each. $$20x^{2}+ 132x + 72$$
Please choose the best answer.
A
$$20(x + 3)(x - 6)$$
B
$$4(5x - 3)(x + 6)$$
C
$$2(5x + 3)(x + 6)$$
D
$$4(x + 11)(3x + 12)$$
E
$$4(5x + 3)(x + 6)$$
Example #2
Factor each. $$3x^{4}+ 14x^{3}- 24x^{2}$$
Please choose the best answer.
A
$$3(x + 2)(x - 4)$$
B
$$x^{2}(3x - 4)(x - 6)$$
C
$$3x^{2}(x + 1)(x - 8)$$
D
$$x^{2}(3x - 4)(x + 6)$$
E
$$x^{4}(3x - 4)(x + 8)$$
Example #3
Factor each. $$2x^{2}+ 13x + 21$$
Please choose the best answer.
A
$$(2x + 3)(x + 7)$$
B
$$(x + 11)(2x + 3)$$
C
$$(2x + 7)(x + 3)$$
D
$$4(5x - 2)(x - 9)$$
E
$$4(5x + 2)(x - 9)$$
Congrats, Your Score is 100%
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+0
# Given: y= -3x – 4
0
359
1
+153
Given: y= -3x – 4
a. Identify the slope: ___________ b. Identify the Y intercept: _____________
c. Graph the line
Nov 3, 2017
#1
+7705
+1
This equation is a straight line which is in slope-intercept form.
You can observe easily that slope = -3, y-intercept = -4
the graphing? do it yourself.
Nov 4, 2017
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# Thread: Why is 'a' proportional to 'bc', not 'root bc'?
1. Hey, this is getting on my nerves. I have been thinking over this very fundamental thing of Mathematics for quite a long time without finding any possible solution. Here it goes.
Say, 'a' is proportional to 'b', and 'a' is again proportional to 'c', then we may infer a conclusion from this that 'a' is proportional to 'bc'.
Fine! This is true indeed. Every now and then we use this conclusion in Physics. So what I want now is a mathematical proof of this assertion, which I have been trying to find with no success at all. Please help.
2.
3. Originally Posted by J Rahman
Hey, this is getting on my nerves. I have been thinking over this very fundamental thing of Mathematics for quite a long time without finding any possible solution. Here it goes.
Say, 'a' is proportional to 'b', and 'a' is again proportional to 'c', then we may infer a conclusion from this that 'a' is proportional to 'bc'.
Fine! This is true indeed. Every now and then we use this conclusion in Physics. So what I want now is a mathematical proof of this assertion, which I have been trying to find with no success at all. Please help.
It is wrong.
a is proportional to a
a is still proportional to a
a is NOT proportional to a^2
4. No no, you dint get my point. Let me give you an example. The physicists say, F is proportional to q1q2, and F is proportional to 1/r^2, therefore F is proportional to q1q2/r^2. My question is, how do they make this combination? There are many other similar examples in Physics where they make use of this combination principle. I want the mathematical proof -- how do they derive the "conclusion" from the "given".
5. Originally Posted by J Rahman
No no, you dint get my point. Let me give you an example. The physicists say, F is proportional to q1q2, and F is proportional to 1/r^2, therefore F is proportional to q1q2/r^2. My question is, how do they make this combination? There are many other similar examples in Physics where they make use of this combination principle. I want the mathematical proof -- how do they derive the "conclusion" from the "given".
It is still wrong, for exactly the reason that I gave you.
You are the one not getting the point.
If F is proportional to q1q2 and proportional to 1/r*2 then even if F =constant x q1q2/r^2 it is proportional to q1q2 for r fixed or to 1/r^2 for q1q2 fixed but it is only proportional to q1q2/r^2 if that ratio is fixed.
6. Thank you sir. Thank you very much. I am such a brainless moron. It was obvious. I am clear now. Thank you once again.
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https://business-accounting.net/overview-for-matrix-plot/
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Overview for Matrix Plot
Scatter graph method
She reviews department records to determine the activity levels for those same months. After gathering data from these two places, the accountant has all the information she needs to perform the analysis. Thus, the scatter diagram method is the simplest device to study the degree of relationship between the variables by plotting the dots for each pair of variable values given. The chart on which the dots are plotted is also called as a Dotogram.
A correlation of -0.97 is a strong negative correlation while a correlation of 0.10 would be a weak positive correlation. A low Pearson correlation coefficient does not mean that no relationship exists between the variables. To check for nonlinear relationships graphically, create a scatterplot or use simple regression. One method of estimating mixed costs is by preparing a scatter graph.
Remember, correlation strength is measured from -1.00 to +1.00. The correlation coefficient often expressed as r, indicates a measure of the direction and strength of a relationship between two variables. When the r value is closer to +1 or -1, it indicates that there is a stronger linear relationship between the two variables.
Scatter plot
You can reduce this potential problem by collecting information at other activity levels and affirming the fixed and variable relationships at these other levels. The result could be that the furthest data points are thrown out, resulting in a more reliable high-low analysis. Where the correlation coefficient is 0 this indicates there is no relationship between the variables (one variable can remain constant while the other increases or decreases).
What are the 3 types of scatter plots?
A scattergraph has a horizontal x-axis that represents production activity, a vertical y-axis that represents cost, data that are plotted as points on the graph, and a regression line that runs through the dots to represent the relationship between the variables.
A regression line can be used to statistically describe the trend of the points in the scatter plot to help tie the data back to a theoretical ideal. This regression line expresses a mathematical relationship between the independent and dependent variable. Depending on the software used to generate the regression line, you may also be given a constant that expresses the ‘goodness of fit’ of the curve. That is to say, to what degree of certainty can we say this line truly describes the trend in the data. The correlational constant is usually expressed as R2 (R-squared).
In the case that there are a few outliers (data points that are located far away from the rest of the data) the line will adjust so that it represents those points as well. The scatter diagram graphs pairs of numerical data, with one variable on each axis, to look for a relationship between them.
Understanding the Scattergraph Method
If the correlation coefficient has a negative value (below 0) it indicates a negative relationship between the variables. This means that the variables move in opposite directions (ie when one increases the other decreases, or when one decreases the other increases).
For example, weight and height, weight would be on y axis and height would be on the x axis. Correlations may be positive (rising), negative (falling), or null (uncorrelated). If the pattern of dots slopes from lower left to upper right, it indicates a positive correlation between the variables being studied.
The concept is useful in the analysis of pricing and the derivation of budgets. It could be used to determine the fixed and variable components of the costs associated with a product, product line, machine, store, geographic sales region, subsidiary, or customer.
A value of zero indicates that there is no relationship between the two variables. The scatter graph method is used to segregate mixed costs and is more accurate than the high-low mehod. By plotting relevant data points in a graph, the fixed and variable cost components can be determined. The high-low method is used to discern the fixed and variable portions of a mixed cost. The essential concept is to collect the cost at a high activity level and again at a low activity level, and then extract the fixed and variable cost components from this information.
Scatter plot matrices
• The scattergraph method is a visual technique for separating the fixed and variable elements of a semi-variable expense (also called a mixed expense) in order to estimate and budget for future costs.
• A scatter plot can suggest various kinds of correlations between variables with a certain confidence interval.
• For example, weight and height, weight would be on y axis and height would be on the x axis.
For each month, you need the total of the cost and the total of the activity. The activity for utilities could be kilowatts used, gallons used, hours used, etc. Plot on the scattergraph a regression line that represents the relationship between the various data points.
Another advantage of this method is that it only requires two sets of numbers to calculate the fixed and variable costs. The accountant reviews the financial transactions for the account over several months to obtain the total cost amount.
Anytime the correlation coefficient, denoted as r, is greater than zero, it’s a positive relationship. Conversely, anytime the value is less than zero, it’s a negative relationship.
The following plots show data with specific Spearman correlation coefficient values to illustrate different patterns in the strength and direction of the relationships between variables. The larger the absolute value of the coefficient, the stronger the relationship between the variables. A line of best fit is often useful to attempt to represent data with the equation of a straight line in order to predict values that may not be displayed on the plot. The line of best fit is determined by the correlation between the two variables on a scatter plot.
The accountant lists each set of data and identifies the high and low values. She divides the difference in dollars by the difference in activity to calculate the cost per unit of activity, or the variable activity. She multiplies the variable cost per unit by the number of activities to calculate the total variable cost. She subtracts the total variable cost from the total cost to determine the fixed cost. In cost accounting, the high-low method is a way of attempting to separate out fixed and variable costs given a limited amount of data.
If the pattern of dots slopes from upper left to lower right, it indicates a negative correlation. A line of best fit (alternatively called ‘trendline’) can be drawn in order to study the relationship between the variables. An equation for the correlation between the variables can be determined by established best-fit procedures.
A scatter plot is also very useful when we wish to see how two comparable data sets agree to show nonlinear relationships between variables. The ability to do this can be enhanced by adding a smooth line such as LOESS. Furthermore, if the data are represented by a mixture model of simple relationships, these relationships will be visually evident as superimposed patterns. Another disadvantage of the high-low method is the number of steps necessary to perform this analysis. The accountant needs to gather monthly data regarding the expense being analyzed and the unit of activity.
The total amount of fixed costs is assumed to be the same at both points of activity. The change in the total costs is thus the variable cost rate times the change in the number of units of activity. The costs associated with a product, product line, equipment, store, geographic sales region, or subsidiary, consist of both variable costs and fixed costs. To determine both cost components of the total cost, an analyst or accountant can use a technique known as the high-low method.
Whether this regression line should be linear or curved depends on what your hypothesis predicts the relationship is. When a curved line is used, it is typically expressed as either a second order (cubic) or third order (quadratic) curve. Higher order curves may follow the actual data points more closely, but rarely provide a better mathematical description of the relationship. The high-low method is used to calculate the variable and fixed cost of a product or entity with mixed costs. It considers the total dollars of the mixed costs at the highest volume of activity and the total dollars of the mixed costs at the lowest volume of activity.
A scatter graph splits the cost into fixed and variable portions. Creating a scatter graph can help you determine the future amounts of a cost. The data from a completed scatter graph provides the information you need to create a cost equation for each of your mixed costs. The cost equation will provide you with fixed and variable estimates for each mixed cost. You will need historical data to create the scatter graph; one year’s worth is the minimum to get a good approximation of the expense.
Using Common Stock Probability Distribution Methods
For a linear correlation, the best-fit procedure is known as linear regression and is guaranteed to generate a correct solution in a finite time. No universal best-fit procedure is guaranteed to generate a correct solution for arbitrary relationships.
A typical regression line has an upward slant, indicating that costs increase with unit volume. The regression line may also intercept the y axis above the zero cost level, indicating the presence of fixed costs that must be incurred even in the absence of any unit activity.
The high-low method involves taking the highest level of activity and the lowest level of activity and comparing the total costs at each level. The range of values for the correlation coefficient is -1.0 to 1.0. In other words, the values cannot exceed 1.0 or be less than -1.0 whereby a correlation of -1.0 indicates a perfectnegative correlation, and a correlation of 1.0 indicates a perfectpositive correlation.
The scattergraph method is a visual technique for separating the fixed and variable elements of a semi-variable expense (also called a mixed expense) in order to estimate and budget for future costs. A scatter plot can suggest various kinds of correlations between variables with a certain confidence interval.
If the variables are correlated, the points will fall along a line or curve. The better the correlation, the tighter the points will hug the line. This cause analysis tool is considered one of the seven basic quality tools. Because the data points represent real data collected in a laboratory setting rather than theoretically calculated values, they will represent all of the error inherent in such a collection process.
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Explanation
If they are 2C/5 apart they have either covered C/2-2C/5 = C/10 together, then they will be 2C/5 distance apart after 4 minutes and after every 40K + 4 minutes.
If they met and travelled 2C/5 distance away from each other, they will be 2C/5 distance apart after 20 + 16 minutes = 36 minutes or after every 40K + 36 minutes.
If they met and travelled 3C/5 distance away from each other, they will be 2C/5 distance apart after 20 + 24 minutes = 44 minutes or after every 40K + 44 minutes.
So, 68 is not a possible value of t.
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CC-MAIN-2021-39
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https://coefs.uncc.edu/nabyars/index-of-nabyarsintegration/index-of-nabyarsintegrationvolumes/index-of-nabyarsintegrationvolumesdisk/volumes-by-integration-5/
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# Volumes by Integration
Volumes by Integration: Disk Method
It would look something like this:
Now, imagine a very thin disk, as shown below. This disk can cover the entire volume by sweeping from bottom to top, changing its diameter as it moves.
The thickness of the disk is dy.
The volume of the differential disk is:
dV=pr2dy
Where in this case the thickness of the disk is dy and the radius is x.
So: dV=px2dy
Before we can integrate this, it needs to be in terms of y (since we have the dy. If we had dx, everything would need to be in terms of x.)
So, if y=x2, then x=y1/2. We can plug this in for x:
dV=p(y1/2)2dy = pydy.
This is easy to integrate.
Let’s look at another example using dx this time.
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CC-MAIN-2021-39
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https://engineering.stackexchange.com/questions/44605/friction-in-wedges-finding-the-mass-of-the-block-c-on-the-verge-of-slipping
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# Friction in Wedges: finding the mass of the block C on the verge of slipping
I need to solve for the mass of block C that is in the verge of slipping. All surfaces have a coefficient of static friction of 0.11. The cable connecting the wedge B and C is parallel with the incline. I have given for the mass of the wedges A and B.
I have established the FBD and its initial equations but I am not sure if it is correct.
For wedge A,
The initial equations are $$\sum F_x = F_1(cos20) + N_1(sin20) - F_2(cos40) - N_2(sin40) = 0$$ and $$\sum F_y = W_A - F_1(sin20) + N_1(cos20) - F_2(sin40) + N_2(cos40) = 0$$
For wedge B,
The initial equations are $$\sum F_x = F_3 + F_2(cos40) + N_2(sin40) + T(cos11) = 0$$ and $$\sum F_y = -W_B - T(sin11) + F_2(sin40) - N_2(cos40) = 0$$
For block C,
My initial equations are $$\sum F_x = -T - W_C(sin11) - F_4 = 0$$ and $$\sum F_y = N_4 - W_C(cos11) = 0$$
I will edit my illustration for my tension here. I think it is wrong.
Did I miss something in the cases of friction? Are those all I need to solve for the mass of block C?
• Is the first figure drawn by you ? If so, can you post the original figure also in the question?
– AJN
Jun 10, 2021 at 16:17
• The direction of We of block C. Can you plot that vector on the first diagram? Will it look like the one drawn in the last diagram? Same thing with the tension when drawn on the last diagram. Plot both in the first figure and compare to the last one. String is parallel to incline and so it is parallel to the base of block C. But it doesn't look like that in the last diagram???
– AJN
Jun 10, 2021 at 16:21
• Angle between Wb and T in B's FBD is 90-11 deg. I expected it to be 90+11 deg in the FBD for C block. But it looks like 90 deg. (I assume that Weight are all parallel).
– AJN
Jun 10, 2021 at 16:26
$$\sum Fy = 0$$, the normal force, $$N_4 = W_Ccos\theta$$
The shear friction force, $$S_4 = \mu N$$
$$\sum Fx = 0$$, the required tension, $$T = W_Csin\theta - S_4$$
You shall get the tension produced by the blocks "A" and "B", in the same manner as shown above, for which the magnitude shall be the same as $$"T"$$.
Note that the forces $$T, W,$$ and $$N$$ must act thru the centroid of mass "C". The shear friction force $$S$$ is a reactive force, its magnitude depends on the normal force and friction coefficient only, and always in the direction against the motion.
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https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12A_Problems
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GET READY FOR THE AMC 12 WITH AoPS
Learn with outstanding instructors and top-scoring students from around the world in our AMC 12 Problem Series online course.
# 2021 Fall AMC 12A Problems
2021 Fall AMC 12A (Answer Key)Printable versions: Wiki • Fall AoPS Resources • Fall PDF Instructions This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator). Figures are not necessarily drawn to scale. You will have 75 minutes working time to complete the test. 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
## Problem 1
What is the value of $\frac{(2112-2021)^2}{169}$?
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$
## Problem 2
Menkara has a $4 \times 6$ index card. If she shortens the length of one side of this card by $1$ inch, the card would have area $18$ square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by $1$ inch?
$\textbf{(A) }16\qquad\textbf{(B) }17\qquad\textbf{(C) }18\qquad\textbf{(D) }19\qquad\textbf{(E) }20$
## Problem 3
Mr. Lopez has a choice of two routes to get to work. Route A is $6$ miles long, and his average speed along this route is $30$ miles per hour. Route B is $5$ miles long, and his average speed along this route is $40$ miles per hour, except for a $\frac{1}{2}$-mile stretch in a school zone where his average speed is $20$ miles per hour. By how many minutes is Route B quicker than Route A?
$\textbf{(A)}\ 2 \frac{3}{4} \qquad\textbf{(B)}\ 3 \frac{3}{4} \qquad\textbf{(C)}\ 4 \frac{1}{2} \qquad\textbf{(D)}\ 5 \frac{1}{2} \qquad\textbf{(E)}\ 6 \frac{3}{4}$
## Problem 4
The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$(\textbf{A})\: 1\qquad(\textbf{B}) \: 3\qquad(\textbf{C}) \: 5 \qquad(\textbf{D}) \: 7\qquad(\textbf{E}) \: 9$
## Problem 5
Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$st pole along this road is exactly one mile ($5280$ feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?
$\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15$
## Problem 6
As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$. Point $F$ lies on $\overline{AD}$ so that $DE=DF$, and $ABCD$ is a square. What is the degree measure of $\angle AFE$?
$[asy] size(6cm); pair A = (0,10); label("A", A, N); pair B = (0,0); label("B", B, S); pair C = (10,0); label("C", C, S); pair D = (10,10); label("D", D, SW); pair EE = (15,11.8); label("E", EE, N); pair F = (3,10); label("F", F, N); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F); draw(A--B--C--D--cycle); draw(D--EE--F--cycle); label("110^\circ", (15,9), SW); [/asy]$
$\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$
## Problem 7
A school has $100$ students and $5$ teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are $50, 20, 20, 5,$ and $5$. Let $t$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let $s$ be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is $t-s$?
$\textbf{(A)}\ {-}18.5 \qquad\textbf{(B)}\ {-}13.5 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 13.5 \qquad\textbf{(E)}\ 18.5$
## Problem 8
Let $M$ be the least common multiple of all the integers $10$ through $30,$ inclusive. Let $N$ be the least common multiple of $M,32,33,34,35,36,37,38,39,$ and $40.$ What is the value of $\frac{N}{M}?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 37 \qquad\textbf{(D)}\ 74 \qquad\textbf{(E)}\ 2886$
## Problem 9
A right rectangular prism whose surface area and volume are numerically equal has edge lengths $\log_{2}x, \log_{3}x,$ and $\log_{4}x.$ What is $x?$
$\textbf{(A)}\ 2\sqrt{6} \qquad\textbf{(B)}\ 6\sqrt{6} \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 48 \qquad\textbf{(E)}\ 576$
## Problem 10
The base-nine representation of the number $N$ is $27{,}006{,}000{,}052_{\text{nine}}.$ What is the remainder when $N$ is divided by $5?$
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4$
## Problem 11
Consider two concentric circles of radius $17$ and $19.$ The larger circle has a chord, half of which lies inside the smaller circle. What is the length of the chord in the larger circle?
$\textbf{(A)}\ 12\sqrt{2} \qquad\textbf{(B)}\ 10\sqrt{3} \qquad\textbf{(C)}\ \sqrt{17 \cdot 19} \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 8\sqrt{6}$
## Problem 12
What is the number of terms with rational coefficients among the $1001$ terms in the expansion of $\left(x\sqrt[3]{2}+y\sqrt{3}\right)^{1000}?$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 166 \qquad\textbf{(C)}\ 167 \qquad\textbf{(D)}\ 500 \qquad\textbf{(E)}\ 501$
## Problem 13
The angle bisector of the acute angle formed at the origin by the graphs of the lines $y = x$ and $y=3x$ has equation $y=kx.$ What is $k?$
$\textbf{(A)} \ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)} \ \frac{1+\sqrt{7}}{2} \qquad \textbf{(C)} \ \frac{2+\sqrt{3}}{2} \qquad \textbf{(D)} \ 2\qquad \textbf{(E)} \ \frac{2+\sqrt{5}}{2}$
## Problem 14
In the figure, equilateral hexagon $ABCDEF$ has three nonadjacent acute interior angles that each measure $30^\circ$. The enclosed area of the hexagon is $6\sqrt{3}$. What is the perimeter of the hexagon? $[asy] size(10cm); pen p=black+linewidth(1),q=black+linewidth(5); pair C=(0,0),D=(cos(pi/12),sin(pi/12)),E=rotate(150,D)*C,F=rotate(-30,E)*D,A=rotate(150,F)*E,B=rotate(-30,A)*F; draw(C--D--E--F--A--B--cycle,p); dot(A,q); dot(B,q); dot(C,q); dot(D,q); dot(E,q); dot(F,q); label("C",C,2*S); label("D",D,2*S); label("E",E,2*S); label("F",F,2*dir(0)); label("A",A,2*N); label("B",B,2*W); [/asy]$ $\textbf{(A)} \: 4 \qquad \textbf{(B)} \: 4\sqrt3 \qquad \textbf{(C)} \: 12 \qquad \textbf{(D)} \: 18 \qquad \textbf{(E)} \: 12\sqrt3$
## Problem 15
Recall that the conjugate of the complex number $w = a + bi$, where $a$ and $b$ are real numbers and $i = \sqrt{-1}$, is the complex number $\overline{w} = a - bi$. For any complex number $z$, let $f(z) = 4i\hspace{1pt}\overline{z}$. The polynomial $$P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1$$ has four complex roots: $z_1$, $z_2$, $z_3$, and $z_4$. Let $$Q(z) = z^4 + Az^3 + Bz^2 + Cz + D$$ be the polynomial whose roots are $f(z_1)$, $f(z_2)$, $f(z_3)$, and $f(z_4)$, where the coefficients $A,$ $B,$ $C,$ and $D$ are complex numbers. What is $B + D?$
$(\textbf{A})\: {-}304\qquad(\textbf{B}) \: {-}208\qquad(\textbf{C}) \: 12i\qquad(\textbf{D}) \: 208\qquad(\textbf{E}) \: 304$
## Problem 16
An organization has $30$ employees, $20$ of whom have a brand A computer while the other $10$ have a brand B computer. For security, the computers can only be connected to each other and only by cables. The cables can only connect a brand A computer to a brand B computer. Employees can communicate with each other if their computers are directly connected by a cable or by relaying messages through a series of connected computers. Initially, no computer is connected to any other. A technician arbitrarily selects one computer of each brand and installs a cable between them, provided there is not already a cable between that pair. The technician stops once every employee can communicate with each other. What is the maximum possible number of cables used?
$\textbf{(A)}\ 190 \qquad\textbf{(B)}\ 191 \qquad\textbf{(C)}\ 192 \qquad\textbf{(D)}\ 195 \qquad\textbf{(E)}\ 196$
## Problem 17
For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$
## Problem 18
Each of $20$ balls is tossed independently and at random into one of $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$
## Problem 19
Let $x$ be the least real number greater than $1$ such that $\sin(x) = \sin(x^2)$, where the arguments are in degrees. What is $x$ rounded up to the closest integer?
$\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20$
## Problem 20
For each positive integer $n$, let $f_1(n)$ be twice the number of positive integer divisors of $n$, and for $j \ge 2$, let $f_j(n) = f_1(f_{j-1}(n))$. For how many values of $n \le 50$ is $f_{50}(n) = 12?$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$
## Problem 21
Let $ABCD$ be an isosceles trapezoid with $\overline{BC} \parallel \overline{AD}$ and $AB=CD$. Points $X$ and $Y$ lie on diagonal $\overline{AC}$ with $X$ between $A$ and $Y$, as shown in the figure. Suppose $\angle AXD = \angle BYC = 90^\circ$, $AX = 3$, $XY = 1$, and $YC = 2$. What is the area of $ABCD$?
$[asy] size(10cm); usepackage("mathptmx"); import geometry; void perp(picture pic=currentpicture, pair O, pair M, pair B, real size=5, pen p=currentpen, filltype filltype = NoFill){ perpendicularmark(pic, M,unit(unit(O-M)+unit(B-M)),size,p,filltype); } pen p=black+linewidth(1),q=black+linewidth(5); pair C=(0,0),Y=(2,0),X=(3,0),A=(6,0),B=(2,sqrt(5.6)),D=(3,-sqrt(12.6)); draw(A--B--C--D--cycle,p); draw(A--C,p); draw(B--Y,p); draw(D--X,p); dot(A,q); dot(B,q); dot(C,q); dot(D,q); dot(X,q); dot(Y,q); label("2",C--Y,S); label("1",Y--X,S); label("3",X--A,S); label("A",A,2*E); label("B",B,2*N); label("C",C,2*W); label("D",D,2*S); label("Y",Y,2*sqrt(2)*NE); label("X",X,2*N); perp(B,Y,C,8,p); perp(A,X,D,8,p); [/asy]$ $\textbf{(A)}\: 15\qquad\textbf{(B)} \: 5\sqrt{11}\qquad\textbf{(C)} \: 3\sqrt{35}\qquad\textbf{(D)} \: 18\qquad\textbf{(E)} \: 7\sqrt{7}$
## Problem 22
Azar and Carl play a game of tic-tac-toe. Azar places an in $X$ one of the boxes in a $3$-by-$3$ array of boxes, then Carl places an $O$ in one of the remaining boxes. After that, Azar places an $X$ in one of the remaining boxes, and so on until all boxes are filled or one of the players has of their symbols in a row—horizontal, vertical, or diagonal—whichever comes first, in which case that player wins the game. Suppose the players make their moves at random, rather than trying to follow a rational strategy, and that Carl wins the game when he places his third $O$. How many ways can the board look after the game is over?
$\textbf{(A) } 36 \qquad\textbf{(B) } 112 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 148 \qquad\textbf{(E) } 160$
## Problem 23
A quadratic polynomial with real coefficients and leading coefficient $1$ is called $\emph{disrespectful}$ if the equation $p(p(x))=0$ is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial $\tilde{p}(x)$ for which the sum of the roots is maximized. What is $\tilde{p}(1)$?
$\textbf{(A) } \frac{5}{16} \qquad\textbf{(B) } \frac{1}{2} \qquad\textbf{(C) } \frac{5}{8} \qquad\textbf{(D) } 1 \qquad\textbf{(E) } \frac{9}{8}$
## Problem 24
Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}.$ In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a.$ What is the sum of all possible values of $a$?
$\textbf{(A) } 24 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 66 \qquad \textbf{(E) } 84$
## Problem 25
Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$. There is a polynomial $$q(x) = c_3x^3+c_2x^2+c_1x+c_0$$such that $D(m) = q(m)$ for all odd integers $m\ge 5$. What is $c_1?$
$\textbf{(A)}\ {-}6\qquad\textbf{(B)}\ {-}1\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 11$
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## Welcome to H3 Maths
Blog Support for Growing Mathematicians
### Pi Day is getting closer!
March3
Yes, we are again (see the last Pi Post) heading towards Pi Day (March 14 or 3/14 in the USA). Of course, in Australia and New Zealand we are able to celebrate a day earlier (on 14/03) than our North American mathematicians! We wonder whether Google will produce a unique Pi Day logo?
by posted under Uncategorized | Comments Off on Pi Day is getting closer!
#### Post Support
What is x? Try substituting 1,2,3,4. I think you have it solved now!
—–
The pink triangle is one third of the area of the square. Use half base x height formula for area of the triangle, etc.
The coffee cup logic puzzle – Answer is Cup 5 as all the others have blocked pipes. 🙂
6×6 for the maximum dog pen area of 36 sq meters.
Oxford Exam Answer: According to Rebecca Cotton-Barratt, of Christ Church, this maths question tests abstract thinking”
“I’d initially ask the candidate what shape they think will be formed, and then ask them how they can test this hypothesis,” Cotton-Barratt says.
“They might initially try sketching the ladder at different stages – but ultimately what we want is something that we can generalise and that is accurate (you can’t be sure that your drawing is that accurate, particularly when you’re making a sketch on a whiteboard and don’t have a ruler). So eventually they will fall back on maths, and try to model the situation using equations.
“If they get stuck we would ask them what shape the ladder makes with the wall and floor, and they’ll eventually spot that at each stage the ladder is forming a right-angled triangle. Some might then immediately leap to Pythagoras’ Theorem and use that to find the answer (which is that it forms a quarter circle centred on the point where the floor meets the wall).Of course, Pythagoras could easily find the hypotenuse – it is the green line along the water! (Hint: the hypotenuse is always opposite the right angle!)
………………
Frustratingly there is no definitive answer to the riddle, leaving guessers with no choice but to continue scratching their heads.
Dr Kevin Bowman, course leader for Mathematics at the University of Central Lancashire said: ‘You can interpret it in many ways; one way is no more correct than another.
“There’s no ambiguity in the first equation; 3 apples is 30, so one apple is worth 10.
The Fruit Puzzle…
This isn’t the first mind-bending puzzle to sweep the internet in recent months. Earlier this year, National Geographic’s puzzle asking you to identify which direction a bus is travelling in left thousands of adults scratching their heads (see earlier post). One person suggests that, “because all the bananas aren’t the same, you could say that they all represent different amounts. You might even say that the two coconut pieces in the third equation are different sizes, and therefore add up to three quarters or even seven eighths when put together. In that sense, there are an infinite amount of possible answers.”
Dr Kevin Bowman, course leader for Mathematics at the University of Central Lancashire said: ‘You can interpret it in many ways; one way is no more correct than another.
“There’s no ambiguity in the first equation; 3 apples is 30, so one apple is worth 10.”
Another said, “1 apple equals 10, coconut equals 6 and banana bunch equals 4 so your answer is 20.”
————————————————————————————-
All exterior angles of one coin add up to 360 degrees. Since a coin has 12 sides, each exterior angle = 30 degrees. Two angles are formed between the two coins. Therefore, the angle formed is 60 degrees.
Quite an easy pattern in the Oct 10-11 Post. Subtract the first two numbers to get the first number in the right column; add the first two numbers in the left column to get the last two of the right column!
Parking Lot Puzzle: Turn your computer screen upside down (or stand on your head), then it becomes easy 🙂
In each row, adding gives the last 2 digits and subtracting gives the first.
The blue cherry picker has an extension arm that can’t be seen very well. This has placed the workers closer to the camera and created a strong false sense of scale simply because your eye assumes that the workers should be on the same plane as the base of the cherry picker!
Yes, it was Major General Stanley in the “Pirates of Penzance!” Check out the link in the picture.
The extra rope needed is exactly 2 x pi or 6.28m!
Christmas Teaser: Today is the 1st of January. Bill’s 8th birthday was yesterday, so the day before (December 30) he was still 7 years old. This December he will turn 9 and, next year, will be 10!
What did the math mother feed her new baby? Formula Milk!
What is a bubble? It is a thin sphere of liquid enclosing air (in most cases) or another gas.
Number of toes = 5170
How many Mathematicians to change a light bulb? Why, n+1 of course (one to hold the light)!
Jan 24, 2014: Assuming a free fall rate of 9.8m/sec/sec it would take just 4.06sec to fall 81m.
= 1 (see first line in the post)
Yes, the TV show with hints of Mathematics and Physics (along with the usual tensions of flatmates?) – did you choose 79?
Leonhard Euler (1707-1783) was an incredibly productive mathematician who published almost 900 books! He took an interest in Latin Squares – grids where each row and column each contains a member of a set of numbers. This forms the basis for Sudoku!
Trig Ratios post: yes, the Sine and Cosine ratios are the same when their angles add up to 90 degrees! This relationship can be expressed as: Sine A = Cosine (90-A) or Cosine A = Sine (90-A)
Good work in identifying this trig pattern. Now, here is a follow up questions which we will address in the next post. Does this pattern suggest that there is a link between Sine and Cosine ratios? Come on, come on… be quick with your answer…Yes, well done – of course there must be!!
Yes, zero is an Integer (which keeps to negative and positive integers apart).
Sam had to position himself to make sure that he 8 the chocolate!
There are 7 days in a week (i.e. Modulo 7). 490 days will be the same day that you chose, so the 491st day will be tomorrow!
Yes, C is the missing section – giving the same difference between numbers in the rows and columns.
That’s a mean looking crocodile! Unless, of course, you knew that it measured just 40cm – yes, just over a foot long!! The camera’s wide angle lens has distorted the image and this makes tiny croc look menacing!
Yes, the 100m time for Bolt works out to be 37kms/hr or 22mls/hr. Of course that is just the average time, not the max speed he reached!
Category 3 climbs last approximately 5 kilometres (3.1 miles), have an average grade of 5 percent, and ascend 150 metres (500 feet).
Category 2 climbs are the same length or longer at an 8 percent grade and ascend 500 metres (1,600 feet).
Category 1 climbs last 20 kilometres (12.4 miles) with an average 6 percent grade and ascend 1,500 metres (4,900 feet).
Category H climbs are the hardest including an altitude difference of at least 1,000 metres (3,280 feet) from start to finish and have an average grade of at least 7 percent.
….
Finding missing numbers is great fun and many readers are regular users of Sudoku. In the recent post (July 13) we find that the sum of the numbers in each row and column is 6, 12, ? Therefore, we need to get 18 as the sum in the final row and column. So, 9 is the missing number in order to complete the puzzle.
Great to see some recent posts on Calculus and we hope that some of our junior students (Years 6+) have a close look at these and develop an interest in this (more advanced) Mathematics.
Trend lines are a practical way to analyse the patterns of data over time and are particularly helpful in population, commerce and environmental change, such as the arctic ice post. The best way to find the answer to the question posed in this post is to click on the original article, copy the graph and paste into (e.g.) Word, using the landscape format. Then, using a ruler, carefully draw the same lines that I have shown in the post. This will help arrive at a more accurate answer. When you have the answer, post a comment to the blog and we can check it out to see if you are right (or close). Good luck Junior Mathematician!
1 year = 31 556 926 seconds
1729 – A rather dull number?
The mathematician G. H. Hardy was visiting the Indian mathematician Ramanujan while he was ill in hospital. Hardy was making small talk and remarked that 1729, the number of the taxi that brought him to the hospital, was a rather dull number. “No Hardy!” repled Ramanujan, “It is a very interesting number. It is the smallest number which can be expressed as the sum of two cubes in two different ways!” You see, even “dull” numbers have special properties!
#### Blog Diary
Dear Blog Diary,
Our night sky has always fascinated H3, and there have been some recent releases of amazing images from our nearby galaxies. The size and sheer complexity of our solar system is staggering and, mathematically, quite difficult to describe because the numbers are simply so big!
The fireworks background gives readers some idea of how students feel when they suddenly get a mathematical concept and can apply it with success. This is what excites learners to do well in their math studies. This is also what inspires teachers to want to help students have these "aha" moments! As the famous Winston Churchill said, "Never, Never, Never, Never, Never give up on your maths!" (Well, he almost said that).
The "x" factor - it was intriguing to see the TED talk post that explained why we use x to indicate an unknown quantity in Algebra. Hope our readers also enjoyed this view on what we take for granted in our everyday Mathematics.
Lewis and Clark explored routes to the American west...all the way to Oregon City where, today, there is a great museum to herald this famous migration period (see link in the post). So, the header image show canoes heading in which direction? East? How do you know? Should mathematicians expect every picture or drawing to point north? NO, of course not! So, to answer the post question - the canoes could be heading in ANY direction!
I had a discussion with a fellow teacher the other day that was along the lines of how sad it was that students today have lost a sense of fine craftsmanship when it comes to products and services. For example, old cameras were beautifully crafted and lasted, with regular servicing, for up to one or two generations. Today, with our "instant society" we are surrounding with products that have little permanency. The revival of fine architecture in the Art Deco movement is a recent highlighted post. In the same way, important mathematical proofs are timeless and give us all a better sense of something solid and permanent in our fragile world. I do hope that students who engage in Mathematics at any level also share this passion for numbers, patterns and proofs that are fixed and reliable signposts in a sea of turbulent ideas and rapid change.
Thanks to the positive feedback from Warren in Perth who wrote, "Congrats and good luck in your crusade to bring the joy and beauty of maths back to schools." See the Welcome page for the full comment. It is always great to have helpful ideas and feedback from blog readers. Again, thanks so much for taking the time to read H3 Maths.
It was in the news recently that Apple was looking to spend some \$97 billion - that's 97,000,000,000. At the rate of \$1000 a day, it would take an incredible 265,780 years to spend. That's an insane amount of money and it would be a good exercise to work out how this amount could help fix some of the big issues in the world today, such as the debt crisis in Europe, or Global Warming.
Being able to "roughly" work out an answer in Mathematics is called "Approximation". A good example of using this is in the little test post from the New York Times - looking at the rise in median house prices across a period of time. The answer is lower down in this column... :-)
Above is an algebraic expression with two sets of brackets, -
(x+1)(y-2). The brackets mean "multiply" so each bracket is a factor of an expanded algebraic expression. There are four parts to the bracketed factors, hence the term "quadratic" which comes up often in Year 9 and Year 10 (Freshman and Sophomore) grades. As a growing mathematician you will need to become competent with factorising and expanding algebraic terms.
Great to see so many visitors from 17 different countries - a Prime Number as well! Of course, there are more countries in our Visitor list but they did not show up on the new clustr map.
The blog about maths being all about language is really not entirely true...was just waiting for someone to comment! You see, Mathematics is also very much about shapes, patterns and trends, which were left of the list. In fact, maths is really about everything!! (Answer to median house prices = B)
Welcome to our first visitor from South Africa!
Numbers - they are the DNA of Mathematics and some recent posts will focus on the way that different number groups (called Number Sets) behave - very much like the different groups of people that you mix with (or not) at a party!
Making visual connections is an often forgotten focus in Mathematics yet is integral to most maths testing. I hope you enjoy the challenge of finding the right location for the van on Lombard Street! Your need a sense of orientation and scale but it is really not that difficult.
Welcome to our visitor from San Francisco, just after the San Fran posting! This is a great city, with so much architectural and cultural diversity as well as such a wonderful location.
Patterns - now here's a great subject to get your maths juices boiling! Show me a keen math student and I will guarantee that he or she is into patterns! Of course, the true-blue mathematician is also into random patterns - which we call "chaos" - and that is another great math topic to look at at some other (random) time! Do Zebra stripes count as random patterns? ;-)
The importance of a good breakfast is our focus for the weekmix!
Great to see a recent blog visitor from Gresham, Oregon. Great scenery around the Columbia River Gorge including the second highest waterfall in the USA. Home to some good mathematicians too!
A good friend and wonderful Mathematics teacher (now retired but used to live in Gresham too) send through this kind comment from the USA recently; ".. spent some time on your math blog and was very impressed. I am hoping that students are taking advantage of it. I was particularly impressed with your process of getting students to think mathematically and not just look at math as a hallway that is filled with hurdles called classroom exercises. The most exciting part of math is when you open a side door and explore other rooms that may lead to a maze of interrelated opportunities in math explorations." Many thanks!
A visitor reads our blog from the I-95 (see post). Is this a space-time warp from our Dr Who files or a wonky GPS?
Dear Blog,
Over 100 visitors for January. 100 visitors reminds me of the famous story regarding the great mathematician, Carl Friedrich Gauss. He started primary (elementary) school at age 7 and his genius became apparent when his teacher asked the class to add up (the sum) of all (integer) numbers from 1 through 100. Gauss did this almost instantly by noticing 1+100 = 101; 2+99 = 101, 3 + 98 = 101 for a total of 50 pairs. Therefore the total was 50 x 101 = 5050. He may have reached this mentally by doing 50x100=5000 + 50 = 5050? Whatever method, what a quick mathematical mind at such a young age! Yes, Gauss had a keen interest in how numbers worked and this is a key to doing well in Mathematics.
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Leinster Perfect Groups - Gonit Sora
06 Feb Leinster Perfect Groups
Download this post as PDF (will not include images and mathematical symbols).
Perfect number is an ancient object of study. A positive integer \$\$n\$\$ is said to be perfect if \$\$n\$\$ is equal to twice the sum of its divisors. For example, \$\$6=1+2+3+6=12\$\$. Hence, \$\$6\$\$ is a perfect number. The concept of prime number is first introduced by Greek mathematicians. Till now we have \$\$48\$\$ known perfect numbers.
Tom Leinster, University of Edinburgh studied some groups taking the analogues properties of perfect numbers. In his paper “Perfect number and groups” (arXiv:math/0104012vl[math.GR]1Apr 2001) he defined a class of groups and named them “Perfect Group”. He defined the group as : “A group G is said to be perfect if sum of the order of its normal subgroups is twice the order of the group.”
Originally Perfect Group implies some other class of group. Leinster asks his apology to name these groups as Perfect Groups. Probably the reason of naming to show the analogues properties with Perfect Numbers. From now on we will call such groups as Leinster Perfect Groups.
Let us look into some examples of Leinster Perfect Groups. We are well aware of the cyclic group \$\$mathbb{Z}_n\$\$. A group \$\$mathbb{Z}_n\$\$ is Leinster Perfect if and only if \$\$n\$\$ is a perfect number. So, \$\$mathbb{Z}_6, mathbb{Z}_{28}, mathbb{Z}_{496}\$\$ are all perfect groups.
The group of all isometries of a regular n-sided polygon is called Dihedral Group of order \$\$n\$\$ and mathematically it is defined as
\$\$D_{2n}=<a,b|a^n=e,b^2=e,aba^{-1}=b^{-1}>\$\$
Tom Leinster showed that when \$\$n\$\$ is an odd perfect number then \$\$D_{2n}\$\$ is a Leinster Perfect Group. Interestingly, till now we don’t have any example of odd perfect number. So, if you can give me an odd perfect number then I can give you the corresponding Leinster Perfect Dihedral Group.
Define \$\$D(G)=sum_N|N|\$\$, \$\$N\$\$ is a normal subgroup of \$\$G\$\$. We can show that \$\$D(G_1times G_2)=D(G_1)times D(G_2)\$\$ i.e. the function \$\$D\$\$ is multiplicative. Using the multiplicatively property of \$\$D\$\$, we can show that \$\$S_3times mathbb{Z}_5,A_5times mathbb{Z}_{15128},A_6times mathbb{Z}_{366776}\$\$ are also Leinster Perfect Groups. In his paper On Arithmetic Functions of Finite Groups, Prof. Ashish Kumar Das, NEHU defined the Generalized Quaternion group as:
\$\$Q_{4m}=<a,b|a^{2m}=1,b^2=a^m,bab^{-1}=a^{-1}>.\$\$
And further he gave some examples of Leinster Perfect group such as
\$\$Q_{12},Q_{20}times mathbb{Z}_{19},Q_{224}times A_5times mathbb{Z}_{43}times mathbb{Z}_{11}\$\$ etc.
Till now we have seen examples of even order Leinster Perfect groups only. In mathoverflow, Leinster asked whether there is any odd order Leinster Perfect Group and the only example (till now) was first discovered by Fransḉois Bruhault one day after the question was asked. This group is given by \$\$G=(mathbb {Z}_{127}Delta mathbb{Z}_7)timesmathbb{Z}_{3^4.11^2.19^2.113}\$\$ where \$\$Delta\$\$ denote the semi direct product.
Further research is going on this particular class of groups and following are some important results (till date) on Perfect Groups:
• Every Nilpotent Quotient of a Leinster group is cyclic.
• The Generalized Quaternion group \$\$Q_{4m}\$\$ is Leinster if and only if \$\$m=3\$\$.
• No finite semi-simple group is Leinster group.
• There is no Leinster group of order \$\$p^2q^2\$\$ where \$\$p\$\$ and \$\$q\$\$ are primes.
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# Double Integral Test question
1. Apr 6, 2005
### mkkrnfoo85
*This was accidently posted in the 'Calculus & Analysis' section. Moderators can delete that one. Sorry.*
I took a test today. I wanted to know if I set my limits up correctly and got the right answer, because I've been having problems with that. Okay, here is the question:
A space is bounded by x = 0, y = 0, xy-plane, and the plane: 3x + 2y + z = 6. Find the volume using a double integral.
So, this is how I went about the problem...
Since the space is bounded by the xy-plane, I set z = 0 for the plane. This gave me:
$$3x + 2y = 6, y = \frac{-3x+6}{2}$$
That's just an equation of a line, so I plotted that on the xy-plane.
At x = 0, y = 3 (0,3)
At y = 0, x = 2 (2,0)
This is the hard part for me...setting limits. I got:
$$0\leq x\leq 2$$
and
$$\frac{-3x+6}{2}\leq y\leq 0$$
(hopefully)
*sidenote: If it's wrong, would someone like to show me a simple strategy to setting limits? Also, if there's anything else you could do to help me set limits on integration, that would be really helpful.
To go on with the problem, my resulting double integral was:
$$\int_{x=0}^{x=2} \int_{y= \frac{-3x+6}{2}}^{y=0} (-3x-2y+6)dydx$$
integrating with respect to y first, I got:
$$\int_{x=0}^{x=2} (-\frac{9}{4} x^2 + 9x-18)dx$$
resulting in answer = 9
**Main questions:
1) Are my limits of integration set correctly?
2) Is my resulting answer correct?
Thanks for reviewing.
-Mark
Last edited: Apr 6, 2005
2. Apr 6, 2005
### mkkrnfoo85
bump. i would really like some help on this.
3. Apr 6, 2005
### asrodan
You have your y limits backward. The lowest possible value of y is zero and the value of y calculated from the line equation ranges from 0 to 3 for the allowed values of x.
You made a small mistake in the y integration, the -18 should be -9. Although the answer is still 9.
The answer with the correct y limit order is 6.
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# How do you find the CPM of a critical path?
## How do you find the CPM of a critical path?
5. Calculate the critical path
1. Step 1: Write down the start and end time next to each activity.
2. Step 2: Look at the end time of the last activity in the sequence to determine the duration of the entire sequence.
3. Step 3: The sequence of activities with the longest duration is the critical path.
### How do you calculate ES EF LS and LF?
Formulas for calculating Total Float and Free Float are as follows:
1. Total Float = LS – ES (it is also calculated by LF – EF)
2. Free Float = Lowest ES of successors – EF.
#### How do you calculate EF and ES?
To Calculate ES and EF times:
1. Label ES = zero at the start of the project.
2. Label EF = 0 + activity time for activities without any preceding activities.
3. Label ES for each activity, all of whose predecessors have been labeled = Maximum of {EF times of all immediately preceding activities including dummy activities}
What is ES EF and LS LF and Slack?
critical path is determined through the identification of event events connected by the activity activity with loose time zero or EF = LF to find out the most recent time in start or end (LS and LF), where ES (early start) is start time of the earliest activity, LS (late start) is start time of last activity, EF (early …
What is ES EF and LS LF and slack?
## How do you calculate earliest expected completion time?
The formula used for calculating Early Start and Early Finish dates:
1. Early Start of the activity = Early Finish of predecessor activity + 1.
2. Early Finish of the activity = Activity duration + Early Start of activity – 1.
### How do you calculate es?
Calculating Early Start (ES) and Early Finish (EF)
1. Early Start of the activity = Early Finish of predecessor activity + 1.
2. Early Finish of the activity = Activity duration + Early Start of activity – 1.
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# Decimals Worksheets
• Section 1 of the decimals worksheet contains 36 skills-based decimals questions, in 3 groups to support differentiation
• Section 2 contains 4 applied decimals questions with a mix of word problems and deeper problem solving questions
• Section 3 contains 4 foundation and higher level GCSE exam style decimals questions
• Answers and a mark scheme for all decimals questions are provided
• Questions follow variation theory with plenty of opportunities for students to work independently at their own level
• All division problems created by fully qualified expert secondary maths teachers
• Suitable for GCSE maths revision for AQA, OCR and Edexcel exam boards
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### Decimals at a glance
If a number has a part that is not whole, it is called a decimal number. We write the whole number part, followed by a decimal point, then tenths, hundredths, thousandths, and so on. It is important that students can recognise the decimal place value of each column on a place value chart – for example, know that 0.4 represents 4 tenths. Ordering decimals also requires students to apply this knowledge of place value.
Decimal arithmetic is similar to whole number arithmetic. Decimal addition follows the same method as integer addition, as does decimal subtraction. When multiplying decimals, we use powers of ten to convert our calculation into a whole-number multiplication, then reverse this process to get our decimal answer.
There are a couple of different methods for decimal division. If it is just the dividend that is a decimal, with an integer divisor, the standard algorithm for short or long division can be used. We have to ensure the decimal point is placed in the quotient (answer) directly above its position in the dividend.
If the divisor is a decimal, we write the division as a fraction and use the concept of equivalent fractions to convert to a whole-number calculation.
Looking forward, students can then progress to additional number worksheets, for example a percent worksheet, a rounding decimals worksheet or an
For more teaching and learning support on Number our GCSE maths lessons provide step by step support for all GCSE maths concepts.
## Do you have KS4 students who need more focused attention to succeed at GCSE?
There will be students in your class who require individual attention to help them succeed in their maths GCSEs. In a class of 30, it’s not always easy to provide.
Help your students feel confident with exam-style questions and the strategies they’ll need to answer them correctly with our dedicated GCSE maths revision programme.
Lessons are selected to provide support where each student needs it most, and specially-trained GCSE maths tutors adapt the pitch and pace of each lesson. This ensures a personalised revision programme that raises grades and boosts confidence.
Find out more
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# Coefficient of friction
• Nov 15th 2008, 10:38 AM
gracey
Coefficient of friction
Body A lies on a rough horizontal table and is connected to freely hanging bodies B and C by light inextensible strings passing over smooth pulleys.
A is 5kg B is 3kg and C is 2kg. When the system is released from rest, the body B descends with an acceleration of 0.28. Find the coefficient of friction between body A and the table.
Any help is much appreciated thanks
• Nov 15th 2008, 12:08 PM
skeeter
Let $T_1$ = tension between mass A and B
$T_2$ = tension between mass B and C
forces acting on mass A ...
$F_{net} = m_A a$
$T_1 - f_k = m_A a$
*** $T_1 - \mu m_A g = m_A a$
forces acting on mass B ...
$F_{net} = m_B a$
*** $T_2 + m_B g - T_1 = m_B a$
forces acting on mass C ...
$F_{net} = m_C a$
*** $m_C g - T_2 = m_C a$
solve the system of equations (***) for $\mu$
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# Line segments intersecting at the nine-point center
Can you provide a proof for the following claim:
Claim.Given any $$\triangle ABC$$. The tangent lines to the circumscribed circle of $$\triangle ABC$$ are constructed at vertices $$A$$,$$B$$,$$C$$ . Let point $$D$$ be the intersection point of tangent lines that passes through vertices $$A$$ and $$C$$ , point $$E$$ the intersection point of tangent lines that passes through vertices $$A$$ and $$B$$ , point $$F$$ the intersection point of tangent lines that passes through vertices $$B$$ and $$C$$ , and let $$H_1$$,$$H_2$$,$$H_3$$ be the orthocenters of $$\triangle ACD$$, $$\triangle AEB$$ and $$\triangle BFC$$ respectively. Then line segments $$AH_3$$,$$BH_1$$ and $$CH_2$$ concur at the nine-point center of $$\triangle ABC$$.
GeoGebra applet that demonstrates this claim can be found here.
I don't know how to start the proof. All I know is that nine-point center of the triangle lies in the middle of the line segment whose endpoints are orthocenter and circumcenter , but I don't know how to use that fact. Any hints are welcomed.
In fact we can say more : $$N_9$$ is the midpoint of segments $$AH_i$$.
First we show that $$BOCH_2$$ is a rhombus. $$\triangle BEC$$ is isosceles so $$H_2$$ lies on its symmetry axis. So $$BH_2C$$ is isosceles as is $$BOC$$. Hence $$OH_2$$ is perpendicular to $$BC$$ at its midpoint $$M$$. Also $$\angle OBC = 90 -A = 90 - \angle EBC = \angle H_2CB$$.
Next we show $$AHH_2O$$ is a parallelogram. $$AH \perp BC \Rightarrow AH || OH_2$$. It is known that $$2OM = AH$$. So $$AH = OH_2$$.
We conclude $$AH_2$$ is bisected at intersection of diagonals of parallelogram $$AHH_2O$$. But midpoint of $$OH$$ is $$N_9$$. We're done.
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1. Length of a curve?
I have the function $f(x) = x^2$, is there a way to measure the length of the line between $(0,0)$ and $(2,4)$ ?
2. Specify
Are you trying to find the distance between the two points or the length of the curved line along the two points?
3. Originally Posted by Quick
Are you trying to find the distance between the two points or the length of the curved line along the two points?
The distance of the curved line
4. Yes, that's possible with integration.
The arc length of a function f(x) between x = a and x = b is given by:
$\ell = \int\limits_a^b {\sqrt {1 + f'\left( x \right)^2 } dx}$
In your case (I omitted the calculation), the result is:
$
\ell = \int\limits_0^2 {\sqrt {1 + 4x^2 } dx} = \frac{1}{4}\ln \left( {\sqrt {17} + 4} \right) + \sqrt {17} \approx 4.647
$
5. Originally Posted by TD!
Yes, that's possible with integration.
The arc length of a function f(x) between x = a and x = b is given by:
$\ell = \int\limits_a^b {\sqrt {1 + f'\left( x \right)^2 } dx}$
In your case (I omitted the calculation), the result is:
$
\ell = \int\limits_0^2 {\sqrt {1 + 4x^2 } dx} = \frac{1}{4}\ln \left( {\sqrt {17} + 4} \right) + \sqrt {17} \approx 4.647
$
I havn't started integration yet, (still studying calculus) can you please explain how you derived the formula and how you evaluated it?
Thanks
6. Originally Posted by chancey
I havn't started integration yet, (still studying calculus) can you please explain how you derived the formula and how you evaluated it?
Thanks
If you haven't seen integration yet, you won't understand much of a short explanation. The evaluation isn't that easy either, at least not for a beginner.
Perhaps you could do some reading: here and/or here.
7. OK thanks, I'll give it a read
8. Originally Posted by chancey
I havn't started integration yet, (still studying calculus)
Integration is part of calculus, you may still be studying differentiantion.
RonL
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Home » What Equals 160 In Multiplication? Update
# What Equals 160 In Multiplication? Update
Let’s discuss the question: what equals 160 in multiplication. We summarize all relevant answers in section Q&A of website Countrymusicstop.com in category: MMO. See more related questions in the comments below.
## What equals in multiply?
The values you are multiplying are called the factors. The answer in a multiplication problem is called the product. You find the product when you multiply two or any number of factors.
## What multiply equals 150?
What are the factors of 150? The factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, and 150.
### A number when increased by 12 equals 160 times its reciprocal. find the number
A number when increased by 12 equals 160 times its reciprocal. find the number
A number when increased by 12 equals 160 times its reciprocal. find the number
## What can equal 15 multiplication?
1 x 15 = 15. 3 x 5 = 15. 5 x 3 = 15. 15 x 1 = 15.
## What can equal 180 in multiplication?
5 x 36 = 180. 9 x 20 = 180. 10 x 18 = 180. 12 x 15 = 180.
## What can you multiply to get 6?
List of Factor Pairs for 6
• 1 x 6 = 6.
• 2 x 3 = 6.
• 3 x 2 = 6.
• 6 x 1 = 6.
## What can u multiply to get 140?
The pairs of numbers that when multiplied together make 140 are the factors of 140. 1 × 140, 2 × 70 , 4 × 35 , 5 × 28 , 7 × 20, 10 × 14 are the pairs that make 140 and they are all the factors of 140.
## What times what gives you 392?
4 x 98 = 392. 7 x 56 = 392. 8 x 49 = 392. 14 x 28 = 392.
## What equals 112 when multiplied?
Factor pairs of 112 are the pairs of factors which, when multiplied together, result in a product equal to the number 112. (1 × 112), (2 × 56), (4 × 28), (7 × 16), (8 ×14) make 112.
## What can equal 48?
The factor pairs of the number 48 are: 1 x 48, 2 x 24, 3 x 16, 4 x 12, and 6 x 8.
## What do you multiply to get 24?
1 × 24 = 24. 2 × 12 = 24. 3 × 8 = 24. 4 × 6 = 24.
## What do you multiply to get 54?
1 x 54 = 54. 2 x 27 = 54. 3 x 18 = 54.
### Prime factorization of 160 and 198
Prime factorization of 160 and 198
Prime factorization of 160 and 198
## What times what can equal 56?
List of Factor Pairs for 56
• 1 x 56 = 56.
• 2 x 28 = 56.
• 4 x 14 = 56.
• 7 x 8 = 56.
• 8 x 7 = 56.
• 14 x 4 = 56.
• 28 x 2 = 56.
• 56 x 1 = 56.
## What multiplication makes 162?
Factor pairs are the pairs of two numbers that, when multiplied, give the number. Factor pairs of 162 are (1,162) (2,81) (3,54) (6,27) and (9,18).
## What multiplied equals 224?
32 × 7 = 224. 28 × 8 = 224. 14 × 16 = 224.
## What equals 270 when multiplied?
5 x 54 = 270. 10 x 27 = 270. 15 x 18 = 270. 18 x 15 = 270.
## What can you multiply to get 7?
Number 7 has infinite multiples as it can be multiplied with any whole number and we have infinite whole numbers. A multiple can be the common multiple of two or more numbers. Example: 20 is the common multiple of 2, 4, 5, 10,and 20.
First 20 Multiples of 7.
Multiplication Multiples of 7
7 × 20 140
## What gives you 8 in multiplication?
The multiples of 8 are the numbers that are generated when 8 is multiplied by any natural number.
8, 16, 24, 32,….., 72, 80, 88,….
8 × 6 = 48 8 multiplied by 6 to get 48
8 × 7 = 56 8 multiplied by 7 to get 56
8 × 8 = 64 8 multiplied by 8 to get 64
8 × 12 = 96 8 multiplied by 12 to get 96
17 thg 12, 2020
## What are the factors of 160?
The factors of 160 are 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 80 and 160.
## What numbers can be multiplied to get 120?
As you can see, all the factor pairs of 120 are equal to the number 120 if you multiply them together.
• 1 x 120 = 120.
• 2 x 60 = 120.
• 3 x 40 = 120.
• 4 x 30 = 120.
• 5 x 24 = 120.
• 6 x 20 = 120.
• 8 x 15 = 120.
• 10 x 12 = 120.
## Which number is a factor of 200?
The factors of 200 are 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, and 200.
## What can equal 98?
Thus we have found that the 6 factors of 98 are 1, 2, 7, 14, 49, and 98.
### 3 A natural number, when increased by 12, equals 160 times its reciprocal Find the number
3 A natural number, when increased by 12, equals 160 times its reciprocal Find the number
3 A natural number, when increased by 12, equals 160 times its reciprocal Find the number
## What can go into 196?
The factors of 196 are 1, 2, 4, 7, 14, 28, 49, 98, and 196.
## What times what gives you 192?
Factors of 192 in Pairs
3 × 64 = 192. 4 × 48 = 192. 6 × 32 = 192. 8 × 24 = 192.
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MATH
Math 1B - Fall 1999 - Bergman - Midterm 2 Make-up
# Math 1B - Fall 1999 - Bergman - Midterm 2 Make-up - 1
• Notes
• 1
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Unformatted text preview: 10/04/2001 THU 12:26 FAX 6434330 MOFFITT LIBRARY I001 George M. Bergman Fall 1999, Math 1B 2 November, 1999 961 Evans Hall Second Midterm — makeup exam 8:10—9:30 AM 1. (36 points, 6 points apiece) Find the following. If an expression is undefined, say so. (a) 2:22 5”/n5. (b) 23:19” + 2%”). (c) The set of all real numbers p such that 22:2 72—13 (In n)_2 converges. (d) The Maclaurin series for sin 7m. (6) The Taylor series for Ur3 centered at x = — 1. (f) The solution to the differential equation xy’ = y2+1 satisfying the initial condition 31(1) = 1- 2. (16 points) Let a and b be real numbers. Prove that 22:1 converges if ”03+ ”b and only if at least one of a and b is > 1. 3. (30 points, 6 points apiece) For each of the items listed below, give either an example of the situation described, or a brief reason why no such example exists. (If you give an example, you are not asked to show that it has the asserted property.) (a) A power series 2:0 an (Jr—1)" which converges only at x = 2. (b) A power series 2:121 an x” which converges for all xe [—1, 1] and no other x. (c) A power series 2:20 an (x+2)n which converges for all real numbers x. (d) A series 2320:] a which converges, but such that 211 [an| diverges. n (e) A series 211 an which diverges, but such that 2;} Ian| converges. 4. (18 points) (a) (7 points) Find the first three terms (i.e., the constant, linear, and ——x square terms) of the Taylor series for e centered at x: 2. (b) (11 points) Prove using the formula for the remainder (“Taylor’s Formula”) that for —x all x in the interval [1.5, 2.5], the sum of the above three terms approximates e to within 63/2/48. ...
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# Finding the largest integer that cannot be partitioned in a certain way
I want to use Mathematica to solve the problem:
Find the maximum $k$ such that $6x+9y+20z=k$ does not have a non-negative solution.
I tried FrobeniusSolve. But what is the elegant way to find the maximum?
I know the theoretical background of this problem, and I know other ways of getting the solution. But I want to see how this can be done elegantly in Mathematica.
• I'm not sure I understand this. If positive x, y, z are a solution for a given k, then nx, ny and nz are a solution for nk. So it seems to me there is no maximum k. – Sjoerd C. de Vries May 28 '13 at 20:35
• @SjoerdC.deVries it is expected that there are infinitely many k that have a solution. In fact the question seem to assume that for some N, all k>N have solutions. We wish to find the smallest such N. That is the same as to say, we wish to find the largest k that does not have a solution. I think. – Jacob Akkerboom May 28 '13 at 20:43
• The short answer to this question is to use FrobeniusNumber[{6, 9, 20}]. – Jacob Akkerboom May 28 '13 at 20:53
• mathworld.wolfram.com/CoinProblem.html if you want to implement something yourself. There is a downloadable notebook. – Jacob Akkerboom May 28 '13 at 20:56
• @JacobAkkerboom I guess we're reading different things in the question and I'm afraid I don't see any grounds for your interpretation based on the literal text of the question. The question asks for a maximum k and I feel it's pretty clear there can generally not be a maximum k. – Sjoerd C. de Vries May 28 '13 at 20:58
You should use FrobeniusNumber instead of FrobeniusSolve, since it serves this purpose
The Frobenius number of $a_{1}, ... a_{n}$,.is the largest integer b for which the Frobenius equation $a_{1} x_{1}+ ... a_{n} x_{n} = k$ has no non-negative integer solutions. The $a_{i}$ must be positive integers.
FrobeniusNumber[{6, 9, 20}]
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Nevertheless you can still get the result playing with FrobeniusSolve, there might be many possible ways, let's point out one of them using Cases with an appropriate replacement rule e.g.
Max @ Cases[ Table[{ k, FrobeniusSolve[{6, 9, 20}, k] != {}}, {k, 100}], {a_, False} -> a]
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In case of not knowing FrobeniusNumber, one can get an idea also with Reduce or Solve although these ways are not recommended for diophantine equations, see e.g. Finding the number of solutions to a diophantine equation.
• That looks good, I am glad you did it, I will go to sleep :). – Jacob Akkerboom May 28 '13 at 22:26
• I'd use FindInstance in preference to Reduce, for the particular task of deciding existence (of solutions). – Daniel Lichtblau May 28 '13 at 22:46
• @Artes Yeah, no surprise the quantifier elimination would need either infinite time or memory, and possibly both. – Daniel Lichtblau May 29 '13 at 0:00
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# Class 7 - Regression - Regression Analysis uses a...
• Notes
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This preview shows page 1 - 7 out of 26 pages.
Regression Analysis uses a mathematical model to describe the relationship between one variable and one or more other variables. or ]
Classic Formula for a line (used by Excel for graph trendline) Y = m X + b, m = Slope and b = Y Intercept Phenomenon or Population Linear Regression Notation Y = β 0 + β 1 X + ε (page 439), where β 0 = Y Intercept for the population regression line β 1 = Slope for the population regression line ε = Random error (This error term shows that Y values vary around the population regression line.) σ 2 ε = Variance( ε ) = Variance of the random errors Sample Regression Line for Simple Linear Regression b 0 = Y Intercept for the regression line fitted to the sample data, b 1 = Slope for the regression line fitted to the sample data, Line Fitted to Sample Data X b b Y + = 1 0 ˆ
Mult. Reg. Page 3 Multiple Linear Regression with k variables Phenomenon (population) Model for Y, Y = β 0 + β 1 X 1 + β 2 X 2 + ... + β k X k + ε (page 514) Sample Linear Regression Model with estimated coefficients Y-hat = b 0 +b 1 X 1 + b 2 X 2 + ... + b k X k (page 511)
(also denoted by Y-hat) Y-hat = b 0 + b 1 X (simple model) Y-hat = f[predictor variable(s)] Residual = Y - (Y-hat) = error estimate based on estimated regression model Page 523 SS(Error) = SSE = Sum of Squared Errors = Sum of Squared Residuals SS(Total) = Sum of Squared Deviations of Y values from the sample mean of Y SS(Total) = SST = SS(Y) , (9.10) on page 409 of Canavos & Miller, 1999 SS(Regression) = SSR = Sum of Squares attributable to the regression model SS(Total) = SS(Regression) + SS(Error) SST = SSR + SSE Method of least squares selects the regression model coefficients that minimize the value of SSE for a set of data. (Least Squares Estimates = b j ) Predicted Value of Y = Y ˆ
R-square = R 2 = Coefficient of Determination R-square = Proportion of the total variability that can be explained using the fitted regression model (page 204 & page 523) R-square = SS(regression) / SS(total) R 2 = SSR / SST = (SST - SSE) / SST = 1 - (SSE/SST) 2 1 ) ˆ ( Y Y SSE n i i = - = ( 29 n Y Y Y Y SST n i i n i i n i i 2 1 1 2 1 2 - = - = = = =
Estimation of the Variance of the Errors. (pages 203, 443 & 512) MSE will be used to denote the sample estimate of error variance for the Y values MSE represents Mean Square Error MSE = SSE / (degrees of freedom error) = SS(residual) / (degrees of freedom residual) Degrees of Freedom Total = df(Total) = n-1 Degrees of Freedom Regression = df(Reg) = number of predictor variables = k Degrees of Freedom Error = df(Error) = df(Total) - df(Regression) = n-k-1 df(Total) = df(Regression) + df(Error) In Excel Regression, Standard Error = Square Root of MSE.
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# Quants Quiz for SBI PO EXAM
1. The cost of five chairs and three tables is ₹ 3,110. The cost of one chair is ₹ 210 less than the cost of one table. What is the cost of two tables and two chairs?
• 1) ₹ 1,660
• 2) ₹ 1,860
• 3) ₹ 2,600
• 4) Cannot be determined
• 5) None of these
2. The ratio of the present ages of Ram, Rohan and Raj is 3 : 4 : 5. If the average of their present ages is 28 years then what would be the sum of the ages of Ram and Rohan together after 5 years?
• 1) 45 years
• 2) 55 years
• 3) 52 years
• 4) 59 years
• 5) None of these
3. The total area of a circle and a rectangle is 1166 sq cm. The diameter of the circle is 28 cm. What is the sum of the circumference of the circle and the perimeter of the rectangle if the length of the rectangle is 25 cm?
• 1) 186 cm
• 2) 182 cm
• 3) 184 cm
• 4) Cannot be determined
• 5) None of these
4. Raman scored 456 marks in an exam and Sita got 54 per cent marks in the same exam, which is 24 marks less than that of Raman. If the minimum pass marks in the exam is 34 per cent, then how many more marks did Raman score than the minimum pass marks?
• 1) 184
• 2) 196
• 3) 190
• 4) 180
• 5) None of these
5. The smallest angle of a triangle is equal to two-thirds of the smallest angle of a quadrilateral. The ratio of the angles of the quadrilateral is 3 : 4 : 5 : 6. The largest angle of the triangle is twice its smallest angle. What is the sum of the second largest angle of the triangle and the largest angle of the quadrilateral?
• 1) 160°
• 2) 180°
• 3) 190°
• 4) 170°
• 5) None of these
6. A 320-metre-long train moving at an average speed of 120 kmph crosses a platform in 24 seconds. A man crosses the same platform in 4 minutes. What is the speed of the man in metre/second?
• 1) 2.4
• 2) 1.5
• 3) 1.6
• 4) 2.0
• 5) None of these
7. The simple interest accrued on a certain principal is ₹ 7,200 in six years at the rate of 12 pcpa. What would be the compound interest accrued on that principal at the rate of 5 pcpa in 2 years?
• 1) ₹ 1,020
• 2) ₹ 1,055
• 3) ₹ 1,050
• 4) ₹ 1,025
• 5) None of these
8. The sum of the square of the first number and the cube of the second number together is 568. Also, the square of the second number is 15 less than the square of 8. What is the value of three-fifths of the first number? (assuming both the numbers are positive)
• 1) 18
• 2) 8
• 3) 9
• 4) 16
• 5) None of these
9. The sum of 8 consecutive odd numbers is 656. Also, the average of four consecutive even numbers is 87. What is the sum of the smallest odd number and the second largest even number?
• 1) 165
• 2) 175
• 3) 163
• 4) Cannot be determined
• 5) None of these
10. Seema purchased an item for ₹ 9,600 and sold it at a loss of 5 per cent. From that money she purchased another item and sold it at a gain of 5 per cent. What is her overall gain/loss?
• 1) Loss of ₹ 36
• 2) Profit of ₹ 24
• 3) Loss of ₹ 54
• 4) Profit of ₹ 36
• 5) None of these
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• 0
Newbie
# A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area Q.6
• 0
Find is the solution for class 10 ncert chapter surface areas and volumes . Find the best solution of the exercise 13.1 question number 6 give me the easiest and simplest solution of this question .A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area
Share
1. Two hemisphere and one cylinder are shown in the figure given below.
Here, the diameter of the capsule = 5 mm
∴ Radius = 5/2 = 2.5 mm
Now, the length of the capsule = 14 mm
So, the length of the cylinder = 14-(2.5+2.5) = 9 mm
∴ The surface area of a hemisphere = 2πr2 = 2×(22/7)×2.5×2.5
= 275/7 mm2
Now, the surface area of the cylinder = 2Ï€rh
= 2×(22/7)×2.5×9
(22/7)×45 = 990/7 mm2
Thus, the required surface area of medicine capsule will be
= 2×surface area of hemisphere + surface area of the cylinder
= (2×275/7) × 990/7
(550/7) + (990/7) = 1540/7 = 220 mm2
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ENCYCLOPEDIA 4U .com
Web Encyclopedia4u.com
# Group action
In mathematics, groups are often used to describe symmetries of objects. This is formalized by the notion of a group action: every element of the group "acts" like a bijective map (or "symmetry") on some set. In this case, the group is also called a transformation group of the set.
Table of contents 1 Definition 2 Examples 3 Types of actions 4 Orbits and stabilizers 5 Morphisms and isomorphisms between G-sets 6 Generalizations
### Definition
If G is a group and X is a set, then a (left) group action of G on X is a binary function G × X -> X (where the image of g in G and x in X is written as g.x) which satisfies the following two axioms:
1. g.(h.x) = (gh).x for all g, h in G and x in X.
2. e.x = x for every x in X; here e denotes the identity element of G.
From these two axioms, it follows that for every g in G, the function which maps x in X to g.x is a bijective map from X to X. Therefore, one may alternatively and equivalently define a group action of G on X as a group homomorphism G -> Sym(X), where Sym(X) denotes the group of all bijective maps from X to X.
If a group action G × X -> X is given, we also say that G acts on the set X or X is a G-set.
In complete analogy, one can define a right group action of G on X as a function X × G -> X by the two axioms (x.g).h = x.(gh) and x.e = x. In the sequel, we consider only left group actions.
### Types of actions
The action of G on X is called
• transitive if for any two x, y in X there exists an g in G such that g.x = y;
• simply transitive if for any two x, y in X there exists precisely one g in G such that g.x = y.
• faithful (or effective) if for any two different g, h in G there exists an x in X such that g.xh.x;
• free if for any two different g, h in G and all x in X we have g.xh.x;
Every free action on a
non-empty set is faithful. A group G that acts faithfully on a set X is isomorphic to a permutation group on X. An action is regular if and only if it is transitive and free.
### Orbits and stabilizers
If we define N = {g in G : g.x = x for all x in X}, then N is a normal subgroup of G and the factor group acts faithfully on X).x = g.x. The action of G on X is faithful if and only if N = {e}.
If Y is a subset of X, we write GY for the set { g.y : y in Y and g in G}. We call the subset Y invariant under G if GY = Y (which is equivalent to GYY). In that case, G also operates on Y. The subset Y is called fixed under G if g.y = y for all g in G and all y in Y. Every subset that's fixed under G is also invariant under G, but not vice versa.
Any operation of G on X defines an equivalence relation on X: two elements x and y are called equivalent if there exists a g in G with g.x = y. The equivalence class of x under this equivalence relation is given by the set Gx = { g.x : g in G } which is also called the orbit of x. The elements x and y are equivalent if and only if their orbits are the same: Gx = Gy. Every orbit is an invariant subset of X on which G acts transitively. The action of G on X is transitive if and only if all elements are equivalent, meaning that there is only one orbit. The set of all orbits is written as X/G.
For every x in X, we define Gx = { g in G : g.x = x }. This is a subgroup of G, and it is called the stabilizer of x or isotropy subgroup at x. The action of G on X is free if and only if all stabilizers consist only of the identity element.
There is a natural bijection between the set of all left cosets of the subgroup Gx and the orbit of x, given by hGx |-> h.x. Therefore, |Gx| = [G : Gx], and so
This result, known as the orbit-stabilizer theorem, is especially useful if G and X are finite, because then it can be employed for counting arguments. A related result is Burnside's lemma:
where r is the number of orbits, and Xg is the set of points fixed by g. This result too is mainly of use when G and X are finite, when it can be interpreted as follows: the number of orbits is equal to the average number of points fixed per group element.
### Morphisms and isomorphisms between G-sets
If X and Y are two G-sets, we define a morphism from X to Y to be a function f : X -> Y such that f(g.x) = g.f(x) for all g in G and all x in X. If such a function f is bijective, then its inverse is also a morphism, and we call f an isomorphism and the two G-sets X and Y are called isomorphic; for all practical purposes, they are indistinguishable in this case.
Some example isomorphisms:
• Every regular G action is isomorphic to the action of G on G given by left multiplication.
• Every free G action is isomorphic to G×S, where S is some set and G acts by left multiplication on the first coordinate.
• Every transitive G action is isomorphic to left multiplication by G on the set of left cosets of some subgroup H of G.
With this notion of morphism, the collection of all G-sets forms a category; this category is a topos.
### Generalizations
One often considers continuous group actions: the group G is a topological group, X is a topological space, and the map G × X → X is continuous with respect to the product topology of G × X. The space X is also called a G-space in this case. This is indeed a generalization, since every group can be considered a topological group by using the discrete topology. All the concepts introduced above still work in this context, however we define morphisms between G-spaces to be continuous maps compatible with the action of G. The above statements about isomorphisms for regular, free and transitive actions are no longer valid for continuous group actions.
One can also consider actions of monoids on sets, by using the same two axioms as above. This does not define bijective maps and equivalence relations however.
Instead of actions on sets, one can define actions of groups and monoids on objects of an arbitrary category: start with an object X of some category, and then define an action on X as a monoid homomorphism into the monoid of endomorphisms of X. If X has an underlying set, then all definitions and facts stated above can be carried over. For example, if we take the category of vector spaces, we obtain group representations in this fashion.
One can view a group G as a category with a single object in which every morphism is invertible. A group action is then nothing but a functor from G to the category of sets, and a group representation is a functor from G to the category of vector spaces. In analogy, an action of a groupoid is a functor from the groupoid to the category of sets or to some other category.
Content on this web site is provided for informational purposes only. We accept no responsibility for any loss, injury or inconvenience sustained by any person resulting from information published on this site. We encourage you to verify any critical information with the relevant authorities.
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# Point slope form of a Line
## Equation
$y-y_{1} \,=\, m(x-x_{1})$
It is an equation of a straight line when a straight line intercepts $x$-axis at a point with some slope.
### Proof
$P(x_{1}, y_{1})$ is a point on straight line and $Q(x, y)$ is any point on the same straight line. The straight line $\small \overleftrightarrow{PQ}$ has some slope in Cartesian coordinate system.
The slope of the line is denoted by $m$ in geometric system.
Draw a parallel line to horizontal axis from point $P$ and also draw a perpendicular line towards $x$-axis from point $Q$. The two lines are intersected at point $R$. Thus, a right triangle $\Delta RPQ$ is formed geometrically.
The angle of the $\Delta RPQ$ is theta ($\theta$) and then calculate the slope of the straight line.
$m \,=\, \tan{\theta}$
Evaluate tan of angle theta to express slope of the line in terms of coordinates of the line.
$\implies m \,=\, \dfrac{QR}{PR}$
$\implies m \,=\, \dfrac{OQ-OR}{OR-OP}$
$\implies m \,=\, \dfrac{y-y_{1}}{x-x_{1}}$
$\implies m(x-x_{1}) \,=\, y-y_{1}$
$\,\,\, \therefore \,\,\,\,\,\, y-y_{1} \,=\, m(x-x_{1})$
It is a linear equation and it is in terms of a point and slope. So, the equation of straight line is called as point-slope form equation of a straight line.
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Section1.3
# Section1.3 - STOR 155, Section 3 Tuesday, January 19, 2010...
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STOR 155, Section 3 Tuesday, January 19, 2010 IPS6e Section 1.3
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Section 1.3: Density curves and normal distributions Density curves Center and spread for density curves Normal distributions 68-95-99.7 rule for normal distributions Standardizing observations Calculations for normal distributions Using the standard normal table Inverse normal calculations Normal quantile plots
Density curves A density curve is any curve that is always on or above the horizontal axis has an area of 1 beneath it. Density curves are used as smooth approximations to the shapes of histograms. For many types of data, the histogram of a large dataset can be very well approximated by one kind of density curve or another.
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Histograms can be approximated by density curves Area under a density curve is always 1. Any histogram can be scaled so that the areas of its bars add to 1. Then we can look for a density curve that fits it well. 2
Histograms can be approximated by density curves If a histogram is scaled so that its total area is 1, then the area of any bar equals the proportion of the population with values represented by that bar. If the histogram is well approximated by a density curve, then this area is approximately the area under the curve corresponding to that bar. Proportion of scores under 6 equals area of blue bars. Proportion of scores under 6 approximately equals blue area under density curve. How to find areas under density curves? Would need calculus. BUT: For some special density curves (especially Normal curves), there are tables (or software) that give areas.
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Mean and median for density curves If a histogram is approximated by a density curve, then the median is the point that divides the area under the curve in half. the mean is the point at which the graph would balance if it were solid. Given a formula for a density curve, one would use calculus to find the mean and median, as well as the quartiles and the standard deviation. (Not in this course.)
A few kinds of density curves 10 12 14 16 18 20 0 0.05 0.1 0.15 0.2 0.25 0 0.2 0.4 0.6 0.8 1 0 0.5 1 1.5 2 0 2 4 6 8 10 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 15 20 25 30 35 40 45 0 0.02 0.04 0.06 0.08 0.1 Uniform (find areas by geometry) Triangular (find areas by geometry) ‘Gamma’ (not in this course) Normal Many real- world distributions are well approximated by normal density curves.
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density curves 10 12 14 16 18 20 0 0.05 0.1 0.15 0.2 0.25 Suppose the histogram of a large dataset is well approximated by the density curve shown (a uniform density curve). What proportion of the scores
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## This note was uploaded on 06/12/2010 for the course STOR 155 taught by Professor Andrewb.nobel during the Spring '08 term at UNC.
### Page1 / 38
Section1.3 - STOR 155, Section 3 Tuesday, January 19, 2010...
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General
General
Easy
Question
# Explain how you can use your graphing calculator to show that the rational expressions andare equivalent under a given domain. What is true about the graphat x = 0and Why?
Hint:
## The correct answer is: The graph at x = 0, will have a value y =7.
### Step 1 of 2:Simplify the expression Once we simplified it, we got it reduced as.Hence, they are equivalent.Step 2 of 2:The domains of both the expressions are set of real numbers.Considering the expression , it is clear that it’s a polynomial. Hence, the domain would be set of real numbers.For the expression, we could simplify the expression to which again is a polynomial.The graph at x = 0, will have a value y=7.
The domain of any polynomial is set of real numbers.
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Home > In Excel > Formula For Percentage Of Total In Excel
# Formula For Percentage Of Total In Excel
## Contents
Reply armaan khan says: April 27, 2016 at 4:31 am =(select total/no of subject) press ENTER ans is percentage Reply Venkataramanan V says: April 27, 2016 at 4:54 pm Hi I Example: To Calculate Selling Price: \$120 x 11% VAT (1.11) x 20% (Margin)(1.2) = \$159.84 To calculate backwards: \$159.84 / 11% VAT (0.89) / 20% (Margin)(0.8) = \$113.81 The cost price Reply Muralli says: March 5, 2015 at 10:36 am My total amount is for example \$3500,I want to know the amount of 5.5% including total amount in \$3500.When I Add the type the following formula into any Excel cell:=25%*50- which gives the result 12.5.For further examples of calculating a percentage of a number, see the How to Calculate a Percentage of a http://ubuntinho.com/in-excel/excel-division-formula-percentage.html
Total Sales : \$1,000 Customer A : \$100 ; Total contribution : 10% Customer B : \$200 ; Total contribution for A + B = 30% Customer C : \$100 ; To get around this, you can calculate your numbers as percentages first. God Bless you. Thanks!
## Formula For Percentage Of Total In Excel
LTIFR formula is lost time incidents* 200000/total man-hours worked. And now, let's see how you can use the Excel percentage formula on real-life data. Excel always performs calculations on that underlying value, which is a decimal (0.1). In Excel, the underlying value is always stored in decimal form.
1. Example 1.
2. If you need a quick refresher on percentages, I recommend starting with a few excellent videos from The Khan Academy: Percent and decimals Solving percent problems Slightly harder percent problems Displaying
3. If its to much.
Format empty cells: Excel behaves differently when you pre-format empty cells with percentage formatting and then enter numbers. Reply Anonymous says: March 20, 2015 at 12:17 pm it means you've put to many nested if's in your formula. Reply Dean says: June 12, 2016 at 5:01 pm Hello, I would like to make a calculator that I can use to work out a percentage of a running amount, It How To Calculate Percentage Of Marks In Excel In cell B3, divide the second year’s sales (\$598,634.00) by the first year (\$485,000.00), and then subtract 1.
Reply rakesh says: March 15, 2016 at 5:04 am for example my target is 10 and i have complete 8 so how can i define my percentage in Excel Reply Carol How To Calculate Percentage In Excel 2010 Maybe this is one step too far for you at this stage, but it shows you one of the many other powerful features Excel has to offer. 2/7 Completed! Thanks. So pls I need your help on this question...you as a business manager is expected to manage and keep your clients records, client I'D, client name and amount credited are inputs
Train 24/7 on any device. Percentage Formula In Excel Multiple Cells Schließen Weitere Informationen View this message in English Du siehst YouTube auf Deutsch. Thanks Reply Art says: January 6, 2016 at 1:38 pm Hello, Looking for a formula to use for a workbook to track percentage of visits I make to my customers throughout Currently I have it set up to calculate up to the first 6% but need to change it to only calculate everything AFTER the first 6%. =IF(B14>=0.06,"6%",IF(B14>=0.05,"5%",IF(B14>=0.04,"4%",IF(B14>=0.03,"3%",IF(B14>=0.02,"2%",IF(B14>=0.0001,"1%",IF(B14=0,"0%"))))))) Reply Maria Azbel (Ablebits.com
## How To Calculate Percentage In Excel 2010
Schließen Ja, ich möchte sie behalten Rückgängig machen Schließen Dieses Video ist nicht verfügbar. To increase an amount by a percentage, use this formula: = Amount * (1 + %) For example, the formula .comment-author .vcard 0 increases the value in cell A1 by 20%. Formula For Percentage Of Total In Excel There’s nothing special about an array and you don’t have to define it. How To Calculate Percentage In Excel 2013 Suppose there are 5 sales transactions and only percentages are given (say 8%, 13%, 21%, 17%, 35%) then what is overall percentage.
The LTIFR for 2015 is 1.30 and 2016 is 4.35 until month of June. his comment is here Example of what I am looking for; Column C has zip code 015 from rows 3 through 43. Reply Bryan says: February 8, 2016 at 2:37 pm Hi, I'm trying to calculate the percentage of a range of figures greater than a given value, in this case a percentage. When I do the .40/12 in the formula I come up with 1,434. Excel Percentage Difference Formula
Your selected filters are: Where do you use Office? Reply rafiu says: April 13, 2016 at 6:33 pm can i have percentage formula for 3years calculating for net profit Reply Sony says: April 18, 2016 at 11:42 am I want First, download the source files for free: Excel percentages worksheets. this contact form Thanks in Advance Best Rahul Thakur Reply MATHEW says: August 19, 2016 at 6:24 a Try Microsoft Edge, a fast and secure browser that's designed for Windows 10 Get started Sign
How do you do that? Excel Formula To Calculate Percentage Difference Between Two Numbers In our example, if A2 is your current expenditures and B2 is the percentage you want to increase or reduce that amount by, here are the formulas you'd enter in cell I have units sold and no of returns.
## Let's say that you type 10 into cell A2, and you then apply the percentage number format.
That’s just what Excel calls the range of cells you plug into a formula.Excel has two functions for percentage ranking. Column A of per person cost. DATES : 31/12/13 = 31/12/13 : 12.36%, >= 01/04/2015 : 14% PL. How To Add Percentage In Excel results are either less than or equals to 100.
Are you looking for something different? Again, Excel always uses the underlying value to perform calculations. Buy now Welcome to the Office Blogs. http://ubuntinho.com/in-excel/how-to-calculate-total-revenue-in-excel.html Anmelden Transkript Übersetzen 590.533 Aufrufe 405 Dieses Video gefällt dir?
IE: Total merits/total demerits= what%. good job. So I did =C14*D14 (That's where it starts since I'm using a template.) I also want to add a column for Profit, And the total below cost price and the total Also, If anyone can please lead me to straight forward info on how to, Add the asterisk to all my item numbers that I made up so I could create bar
Reply Rahul Thakur says: August 10, 2016 at 8:43 am Hi Svetlana, You are doing a good job. Reply Svetlana Cheusheva says: June 29, 2015 at 3:13 pm Hi Vicki, You probably meant -14%? All Rights Reserved. I need to find out how many of participants chose number 1, 2, 3, 4,5 to put in table 2.
The total is at the end of the table in a certain cell A very common scenario is when you have a total in a single cell at the end of BusinessDevHomeITNon-profitOn-premisesPartnerPublic sectorSchoolSmall BusinessClear allApply filtersWhat do you want to see? Can the formula be set to match the total number of cells in the column? Calculating percentage change between rows In case you have one column of numbers, say column C that lists weekly or monthly sales, you can calculate percentage change using this formula: .comment-author
That duplicates the header, including the formatting.Step 7In C3, type % Change, replacing the text that’s already there.All percent changes calculated 7.
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## Number 9 worksheets
The number 9 has joined the Number Parade! The number 9 worksheets will introduce your child to the formation of the number 9, illustrate the amount the number 9 represents, and show the word for the number 9. Each of the four number worksheets also includes an activity suggestion for you to try with your child to help strengthen his knowledge of the number.
The first worksheet teaches your child the formation of the number 9 in a bright, bold illustration. Your child can become familiar with this number by tracing the number using the pointer finger on his dominant hand. Point briefly to the upper right part of the number, indicating where your child should begin “writing” with his finger. This number is written in one stroke, so remind your child to stop briefly after he writes the initial circle and before he changes direction to draw the downward line, but not to lift his finger until he has completed the number. Direct him to use his finger again as he touches each picture one time when he counts to 9. Touching each picture one time will reinforce the idea of “one to one correspondence” which means each number is counted only once. The word nine is the only number word between one and ten that begins with the letter N, so it is relatively easy for children to recognize.
The number scramble worksheet that comes next gives children practice identifying the number 9. Often the number 6 is confused with the number 9 because both numbers consist of a circle and line. The number 8, with two circles, can also be misleading. So remind your child to look for the circle at the top of the number, followed by a straight line. Your child can touch each number 9 along the path, moving his finger up, down, or across until he reaches the colored square with the 9 inside. After discovering the path, your child can use a pencil to connect each number 9.
The third worksheet provides counting practice. Counting groups of nine can be challenging, especially when the illustrations are smaller in size. Suggest to your child that he mark each picture with his pencil as he counts to keep track of those he has counted. This will reinforce one-to-one correspondence and promote accurate counting.
The fourth worksheet offers a fun cut and paste activity in addition to counting practice. Ask your child to identify the plate that has the number 9 on it. Then allow him to count out 9 of his favorite foods from the pictures on the bottom of the page and paste them on or above the number 9 on the plate. If your child is also familiar with the number 6, he can cut out 6 food items and paste them on the other plate on the worksheet. Remind your child to avoid covering the number when he pastes the food on the plate.
To reinforce the number 9 at home, show your child a tic-tac-toe board. Ask him to count the spaces by writing the numbers from 1 through 9 in each of the boxes. Then have some fun by showing him how to play tic-tac-toe. After the game is over, you can each count how many squares are covered. Then count how many blank squares there are, and explain to your child how those two numbers add up to 9.
You can also demonstrate for your child how the number 9 is made up of 3 groups of 3. To illustrate this point, show your child 1 triangle and let him count the 3 points or sides on it. The add 2 more triangles for a total of 3 triangles and ask your child to count all of the sides or points he sees. You can also show your child a picture of 3 tricycles and ask him to count the total number of wheels on all 3 tricycles.
After your child has completed all of the number 9 worksheets, don’t forget to print the number 9 tracing numbers worksheet. This worksheet will give your child important practice tracing and writing the number 9. This can be a challenging number to write because it consists of both a circle and a straight line, which means there are many direction changes to be made without lifting the pencil. Some children like to think of this number as a balloon on a stick.
Coming next week: The number 10!
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## Intermediate Algebra for College Students (7th Edition)
$x=-19$
For easier work, it is better to get rid of the fractions. This can be achieved by multiplying the LCD of $24$ on both sides of the equation to obtain: $24(5+\frac{x-2}{3}) = 24(\frac{x+3}{8}) \\24(5) + \frac{24(x-2)}{3} = \frac{24(x+3)}{8} \\120+8(x-2) = 3(x+3) \\120+8x-16 = 3x+9 \\8x+104=3x+9$ Subtract $3x$ and $104$ on both sides to obtain: $8x-3x=9-104 \\5x=-95 \\\frac{5x}{5} = \frac{-95}{5} \\x = -19$ Check: $\begin{array}{ccc} &5+\frac{-19-2}{3} &= &\frac{-19+3}{8} \\&5+\frac{-21}{3} &= &\frac{-16}{8} \\&5+(-7) &= &-2 \\&-2 &= &-2\end{array}$
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# Every Group of Order 24 Has a Normal Subgroup of Order 4 or 8
## Problem 568
Prove that every group of order $24$ has a normal subgroup of order $4$ or $8$.
## Proof.
Let $G$ be a group of order $24$.
Note that $24=2^3\cdot 3$.
Let $P$ be a Sylow $2$-subgroup of $G$. Then $|P|=8$.
Consider the action of the group $G$ on the left cosets $G/P$ by left multiplication.
This induces a permutation representation homomorphism
$\phi: G\to S_{G/P},$ where $S_{G/P}$ is a group of bijective maps (permutations) on $G/P$.
This homomorphism is defined by
$\phi(g)(aP)=gaP$ for $g\in G$ and $aP\in G/P$.
Then by the first isomorphism theorem, we see that
$G/\ker(\phi) \cong \im(\phi) < S_{G/P}.$ This implies that the order of $G/\ker(\phi)$ divides the order of $S_{G/P}$. Note that as $|G/P|=3$, we have $|S_{G/P}|=|S_3|=6$. Thus, we must have $4\mid |\ker{\phi}|$.
Also note that $\ker(\phi) < P$. To see this let $x\in \ker(\phi)$. Then we have $xP=\phi(x)(P)=\id(P)=P.$ Here $\id$ is the identity map from $G/P$ to itself. Hence $x\in P$. It follows that $|\ker(\phi)|$ divides $|P|=8$.
Combining these restrictions, we see that $|\ker(\phi)|=4, 8$.
Being the kernel of a homomorphism, $\ker(\phi)$ is a normal subgroup of $G$.
Hence the group $G$ of order $24$ has a normal subgroup of order $4$ or $8$.
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• Every Sylow 11-Subgroup of a Group of Order 231 is Contained in the Center $Z(G)$ Let $G$ be a finite group of order $231=3\cdot 7 \cdot 11$. Prove that every Sylow $11$-subgroup of $G$ is contained in the center $Z(G)$. Hint. Prove that there is a unique Sylow $11$-subgroup of $G$, and consider the action of $G$ on the Sylow $11$-subgroup by […]
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##### Every Group of Order 12 Has a Normal Subgroup of Order 3 or 4
Let $G$ be a group of order $12$. Prove that $G$ has a normal subgroup of order $3$ or $4$....
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Metamath Proof Explorer
Theorem plysubcl
Description: The difference of two polynomials is a polynomial. (Contributed by Mario Carneiro, 24-Jul-2014)
Ref Expression
Assertion plysubcl ${⊢}\left({F}\in \mathrm{Poly}\left({S}\right)\wedge {G}\in \mathrm{Poly}\left({S}\right)\right)\to {F}{-}_{f}{G}\in \mathrm{Poly}\left(ℂ\right)$
Proof
Step Hyp Ref Expression
1 plyssc ${⊢}\mathrm{Poly}\left({S}\right)\subseteq \mathrm{Poly}\left(ℂ\right)$
2 simpl ${⊢}\left({F}\in \mathrm{Poly}\left({S}\right)\wedge {G}\in \mathrm{Poly}\left({S}\right)\right)\to {F}\in \mathrm{Poly}\left({S}\right)$
3 1 2 sseldi ${⊢}\left({F}\in \mathrm{Poly}\left({S}\right)\wedge {G}\in \mathrm{Poly}\left({S}\right)\right)\to {F}\in \mathrm{Poly}\left(ℂ\right)$
4 simpr ${⊢}\left({F}\in \mathrm{Poly}\left({S}\right)\wedge {G}\in \mathrm{Poly}\left({S}\right)\right)\to {G}\in \mathrm{Poly}\left({S}\right)$
5 1 4 sseldi ${⊢}\left({F}\in \mathrm{Poly}\left({S}\right)\wedge {G}\in \mathrm{Poly}\left({S}\right)\right)\to {G}\in \mathrm{Poly}\left(ℂ\right)$
6 addcl ${⊢}\left({x}\in ℂ\wedge {y}\in ℂ\right)\to {x}+{y}\in ℂ$
7 6 adantl ${⊢}\left(\left({F}\in \mathrm{Poly}\left({S}\right)\wedge {G}\in \mathrm{Poly}\left({S}\right)\right)\wedge \left({x}\in ℂ\wedge {y}\in ℂ\right)\right)\to {x}+{y}\in ℂ$
8 mulcl ${⊢}\left({x}\in ℂ\wedge {y}\in ℂ\right)\to {x}{y}\in ℂ$
9 8 adantl ${⊢}\left(\left({F}\in \mathrm{Poly}\left({S}\right)\wedge {G}\in \mathrm{Poly}\left({S}\right)\right)\wedge \left({x}\in ℂ\wedge {y}\in ℂ\right)\right)\to {x}{y}\in ℂ$
10 neg1cn ${⊢}-1\in ℂ$
11 10 a1i ${⊢}\left({F}\in \mathrm{Poly}\left({S}\right)\wedge {G}\in \mathrm{Poly}\left({S}\right)\right)\to -1\in ℂ$
12 3 5 7 9 11 plysub ${⊢}\left({F}\in \mathrm{Poly}\left({S}\right)\wedge {G}\in \mathrm{Poly}\left({S}\right)\right)\to {F}{-}_{f}{G}\in \mathrm{Poly}\left(ℂ\right)$
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Cody
# Problem 44484. Separate even from odd numbers in a vector - with a loop
Solution 2089924
Submitted on 12 Jan 2020 by anan jeraise
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
filetext = fileread('oddEven.m'); assert(isempty(strfind(filetext, 'regexp')),'regexp hacks are forbidden')
2 Pass
filetext = fileread('oddEven.m'); loopUsed = ~isempty(strfind(filetext, 'while')) || ~isempty(strfind(filetext, 'for')); assert(loopUsed, 'Must use at least one loop')
3 Pass
v = []; w_correct = []; assert(isequal(oddEven(v),w_correct))
w = []
4 Pass
v = [2; 7; 0; 3; 2]; w_correct = [7; 3; 2; 0; 2]; assert(isequal(oddEven(v),w_correct))
e = 2 o = 7 e = 2 0 o = 7 3 e = 2 0 2 w = 7 3 2 0 2
5 Pass
v = [1, 0, 2, 9, 3, 8, 8, 4]; w_correct = [1, 9, 3, 0, 2, 8, 8, 4]; assert(isequal(oddEven(v),w_correct))
o = 1 e = 0 e = 0 2 o = 1 9 o = 1 9 3 e = 0 2 8 e = 0 2 8 8 e = 0 2 8 8 4 w = 1 9 3 0 2 8 8 4
6 Pass
odd = 2 * randi([-4, 4], 1, randi([4,10])) - 1; even = 2 * randi([-4, 4], 1, randi([4,10])); v = [even, odd]; w_correct = [odd, even]; assert(isequal(oddEven(v),w_correct))
e = 8 e = 8 2 e = 8 2 4 e = 8 2 4 -2 e = 8 2 4 -2 -6 e = 8 2 4 -2 -6 6 e = 8 2 4 -2 -6 6 6 o = 1 o = 1 -9 o = 1 -9 1 o = 1 -9 1 -1 o = 1 -9 1 -1 7 o = 1 -9 1 -1 7 7 o = 1 -9 1 -1 7 7 -3 o = 1 -9 1 -1 7 7 -3 -5 w = 1 -9 1 -1 7 7 -3 -5 8 2 4 -2 -6 6 6
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# Mean Deviation
In statistics and mathematics, the deviation is a measure that is used to find the difference between the observed value and the expected value of a variable. In simple words, the deviation is the distance from the center point. Similarly, the mean deviation is used to calculate how far the values fall from the middle of the data set. In this article, let us discuss the definition, formula, and examples in detail.
## Mean Deviation Definition
The mean deviation is defined as a statistical measure that is used to calculate the average deviation from the mean value of the given data set. The mean deviation of the data values can be easily calculated using the below procedure.
Step 1: Find the mean value for the given data values
Step 2: Now, subtract the mean value from each of the data values given (Note: Ignore the minus symbol)
Step 3: Now, find the mean of those values obtained in step 2.
## Mean Deviation Formula
The formula to calculate the mean deviation for the given data set is given below.
Mean Deviation = [Σ |X – µ|]/N
Here,
Σ represents the addition of values
X represents each value in the data set
µ represents the mean of the data set
N represents the number of data values
| | represents the absolute value, which ignores the “-” symbol
## Mean Deviation for Frequency Distribution
To present the data in the more compressed form we group it and mention the frequency distribution of each such group. These groups are known as class intervals.
Grouping of data is possible in two ways:
1. Discrete Frequency Distribution
2. Continuous Frequency Distribution
In the upcoming discussion, we will be discussing mean absolute deviation in a discrete frequency distribution.
Let us first know what is actually meant by the discrete distribution of frequency.
### Mean Deviation for Discrete Distribution Frequency
As the name itself suggests, by discrete we mean distinct or non-continuous. In such a distribution the frequency (number of observations) given in the set of data is discrete in nature.
If the data set consists of values x1,x2, x3………xn each occurring with a frequency of f1, f2… fn respectively then such a representation of data is known as the discrete distribution of frequency.
To calculate the mean deviation for grouped data and particularly for discrete distribution data the following steps are followed:
Step I: The measure of central tendency about which mean deviation is to be found out is calculated. Let this measure be a.
If this measure is mean then it is calculated as,
where $$N=\sum_{i=1}^{n}\;f_{i}$$
If the measure is median then the given set of data is arranged in ascending order and then the cumulative frequency is calculated then the observations whose cumulative frequency is equal to or just greater than N/2 is taken as the median for the given discrete distribution of frequency and it is seen that this value lies in the middle of the frequency distribution.
Step II: Calculate the absolute deviation of each observation from the measure of central tendency calculated in step (I)
StepIII: The mean absolute deviation around the measure of central tendency is then calculated by using the formula
If the central tendency is mean then,
In case of median
Let us look into the following examples for a better understanding.
### Mean Deviation Examples
Example 1:
Determine the mean deviation for the data values 5, 3,7, 8, 4, 9.
Solution:
Given data values are 5, 3, 7, 8, 4, 9.
We know that the procedure to calculate the mean deviation.
First, find the mean for the given data:
Mean, µ = ( 5+3+7+8+4+9)/6
µ = 36/6
µ = 6
Therefore, the mean value is 6.
Now, subtract each mean from the data value, and ignore the minus symbol if any
(Ignore”-”)
5 – 6 = 1
3 – 6 = 3
7 – 6 = 1
8 – 6 = 2
4 – 6 = 2
9 – 6 = 3
Now, the obtained data set is 1, 3, 1, 2, 2, 3.
Finally, find the mean value for the obtained data set
Therefore, the mean deviation is
= (1+3 + 1+ 2+ 2+3) /6
= 12/6
= 2
Hence, the mean deviation for 5, 3,7, 8, 4, 9 is 2.
Example 2:
In a foreign language class, there are 4 languages, and the frequencies of students learning the language and the frequency of lectures per week are given as:
Language Sanskrit Spanish French English No. of students(xi) 6 5 9 12 Frequency of lectures(fi) 5 7 4 9
Calculate the mean deviation about the mean for the given data.
Solution: The following table gives us a tabular representation of data and the calculations
## Frequently Asked Questions on Mean Deviation
### What does the mean deviation tell us?
The mean deviation gives information about how far the data values are spread out from the mean value.
### Mention the procedure to find the mean deviation.
The procedure to find the mean deviation are:
Step 1: Calculate the mean value for the given data
Step 2: Subtract the mean from each data value (Distance)
Step 3: Finally, find the mean for the distance
### What are the advantages of using the mean deviation?
The advantages of using mean deviation are:
It is based on all the data values provided, and hence it will give a better measure of dispersion.
It is easy to understand and calculate.
### What is the mean deviation about a median?
The mean deviation about the median is similar to the mean deviation about mean. Instead of calculating the mean for the given set of data values, find the median value by arranging the data values in the ascending order and then find the middle value. After finding the median, now subtract the median from each data value, and finally, take the average
### What are the three different ways to find the mean deviation?
The mean deviation can be calculated using
Individual Series
Discrete series
Continuous series
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# Thread: Problems in GATE preparation and discussion and their solutions
1. ## Problems in GATE preparation and discussion and their solutions
Question :- The cost function for the product in a firm is given by 5q2, where q is the amount of production. The firm can sell the product at a market price of rs 50 per unit. The number of units to be produced to maximize the profit is
A) 5
B) 10
C) 15
D) 20
2. ## Re: Problems in GATE preparation and discussion and their solutions
Well, i think answer must be 5. As profit = selling price - production cost. So in all 4 cases, profit will be as follows :
1.) Profit = (5 x 50) - (5 x 5 x 5) = 125
2.) Profit = (10 x 50) - (5 x 10 x 10) = 0
3.) Profit = (15 x 50) - (5 x 15 x 15) = - 375
4.) Profit = (20 x 50) - (5 x 20 x 20) = 0
And i have done mech engineering 7 years before so may be i am wrong in above given approach as trying such question after a long time...
3. ## Re: Problems in GATE preparation and discussion and their solutions
Basically u r checking it by the options, but by your approach I got an idea how to solve it.
Assume no. of units as 'x'
so, profit = 50x - 5x2
so differentiate it to maximize.
---------- Post added at 05:05 PM ---------- Previous post was at 05:01 PM ----------
Everybody is invited for their doubts and problems in civil engineering, either numeric or theoretical..!!
4. ## Re: Problems in GATE preparation and discussion and their solutions
you are absolutely right that by differentiation we can get answer directly (50 = 10x)...anyways.....that's the beauty of objective questions....
5. ## Re: Problems in GATE preparation and discussion and their solutions
Choose the grammatically INCORRECT sentence:
The committee initially asked for a funding of Fifty Lakh rupees, but later settled for a lesser sum.
This country’s expenditure is not less than that of Bangladesh.
They gave us the money back less the service charges of Three Hundred rupees.
This country’s expenditure on educational reforms is very less
6. ## Re: Problems in GATE preparation and discussion and their solutions
Plz give the answer of this question....
7. ## Re: Problems in GATE preparation and discussion and their solutions
The answer to this problem is D.
I think the mistake is in 'very less' but I am not sure.
And if u know the reason then explain.
8. ## Re: Problems in GATE preparation and discussion and their solutions
May be "The answer to this question is A"
"funding" is incorrect ,fund should be mentioned here instead of funding.
9. ## Re: Problems in GATE preparation and discussion and their solutions
Answer is definitely D, as I checked it from the answer keys also. And all the four statements are focussing mainly on 'less'. So considering that we have to check keeping it in context.
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# A function is given by a table of values, a graph, a formula, or a verbal description. Determine whether it is one-to-one.$g(x) = 1 - \sin x$
## function is not one-to-one.
### Discussion
You must be signed in to discuss.
##### Catherine R.
Missouri State University
##### Kristen K.
University of Michigan - Ann Arbor
##### Samuel H.
University of Nottingham
Lectures
Join Bootcamp
### Video Transcript
we're deciding whether or not this function is 1 to 1. And to be 1 to 1 means that every X value has only one y value. And every y value has only one X value. So one way to figure it out is to look at the graph. So imagine this graph. First of all, imagine a sign graph, Michael. Sign of X would look something like this and then imagine that you flipped it upside down and got y equals negative sine of X. So that would look something like this. And then imagine that you shifted it up one. And that's where we get G of X. So that would look something like this. Okay, well, does every x value have only one y value and every y value have only one X value? Clearly not because if we pass a horizontal line through, it passes through multiple points. That means there are multiple X values with the same y value. So this is not 1 to 1
Oregon State University
##### Catherine R.
Missouri State University
##### Kristen K.
University of Michigan - Ann Arbor
##### Samuel H.
University of Nottingham
Lectures
Join Bootcamp
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, 14.05.2020batopusong81
# What is the sum of 0.2 + 300% + 1 4⁄5 ?
1. multiply the variables.
2. add the constants and multiply them by the variable.
3. multiply the constants.
answer: x² - 8x + 7
h=8+r
vcone=1/3 π
v(r)= 1/3 π
v=1/3 8π
Sum = 5
Step-by-step explanation:
Convert everything to decimal
300% = 3
1 4/5 = 1.80
0.2 + 3 + 1.80 = 5
Sum = 5
### Other questions on the subject: Math
Math, 28.10.2019, elaineeee
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Try to search it at you tube...Read More
Math, 28.10.2019, danigirl12
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##### i have a test tomorrow and i cant remember how to do this!!!
label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
the square root of a number increased by twice the number is 8. find the number
Apr 29th, 2015
do you have any options?
if so i can answer accordingly
Apr 29th, 2015
√x + 2x = 8
√x= 8 - 2x
x = (8 -2x)^2
x = 64 - 32x + 4x ^2
4x^2 - 33x + 64 = 0 (a=4 , b= -33,c =64)
by solving
x=( -b +/- √(b^2 - 4ac)) / 2a
numbers are,
x1=33/8√ 65 / 8.....................................
x2=33/8− √ 65 / 8...........................................
Apr 29th, 2015
i think the question is square of a number instead of square root of a number
if the question is
the square ( instead of square root) of a number increased by twice the number is 8.
then ,
x^2 + 2x = 8
x^2 + 2x - 8 =0
(x + 4)(x -2) = 0
(x + 4)= 0 or (x -2) = 0
x= -4 or x = 2
numbers are 2, -4.......
Apr 29th, 2015
...
Apr 29th, 2015
...
Apr 29th, 2015
Oct 24th, 2017
check_circle
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Cody
# Problem 563. How to add?
Solution 1211093
Submitted on 13 Jun 2017 by Said BOUREZG
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
X='98765432109876543210987654321098765432109876543210987654321'; Y='98765432109876543210987654321098765432109876543210987654321'; Z='197530864219753086421975308642197530864219753086421975308642'; assert(isequal(dda(X,Y),Z))
ans = Columns 1 through 29 1 9 7 5 3 0 8 6 4 2 1 9 7 5 3 0 8 6 4 2 1 9 7 5 3 0 8 6 4 Columns 30 through 58 2 1 9 7 5 3 0 8 6 4 2 1 9 7 5 3 0 8 6 4 2 1 9 7 5 3 0 8 6 Columns 59 through 60 4 2 ans = '197530864219753086421975308642197530864219753086421975308642'
2 Pass
X='6546468768680988454345'; Y='5757557542432424209808098908085353545657657'; Z='5757557542432424209814645376854034534112002'; assert(isequal(dda(X,Y),Z))
ans = '5757557542432424209814645376854034534112002'
3 Pass
X='122'; Y='323'; Z='445'; assert(isequal(dda(X,Y),Z))
ans = '445'
4 Pass
X='767678686868667868635435353545'; Y='465464643244242424249787979'; Z='768144151511912111059685141524'; assert(isequal(dda(X,Y),Z))
ans = '768144151511912111059685141524'
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# In a parallelogram PQRS if angle …
## CBSE, JEE, NEET, NDA
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In a parallelogram PQRS if angle p=3x-5angleQ=2X+50 find x
Sia 🤖 1 month, 1 week ago
Similar as (Q=2x+15):
{tex}\angle P+\angle Q=180^{\circ}{/tex} (Angles on the same side of a transversal are supplementary)
{tex}\Rightarrow \ 3 x-5+2 x+15=180^{\circ}{/tex}
5x + 10 = 180o
{tex}\Rightarrow{/tex} 5x + 170o {tex}\Rightarrow{/tex} x = 34o
Priyanka Kumari 1 month, 1 week ago
3x-5+2x+50=180 (angle on the same side of a transversal are supplementary) 5x+45=180 5x=135 X= 27
Related Questions:
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# Metric-Dual Vector Space
1. Sep 17, 2014
### ChrisVer
a metric is also used to raise/lower indices.
$g_{\nu \mu } x^{\mu} = x_{\nu}$
$g^{ \nu \mu} x_{\mu} = x^{\mu}$
In general a metric [with lower indices] is a map from $V_{(1)} \times V_{(2)} \rightarrow \mathbb{R}$
whereas the upper indices are the map from $V^{*}_{(3)} \times V^{*}_{(4)} \rightarrow \mathbb{R}$
I used the subscript to denote the vectors later I'm going to take, and the * means the dual space.
In this case then you have:
$g(x_1, x_2) = s \in \mathbb{R}$
$g(x_{3}, x_{4}) = s \in \mathbb{R}$ , with $x_{i} \in V_{(i)}^{(*)}$
Which means that $g_{\mu \nu} x_{1}^{\mu} x_{2}^{\nu} = x_{1 \nu} x_{2}^{\nu}$
and also $g^{\mu \nu} x_{3\mu} x_{4 \nu} = x^{\nu}_3 x_{4 \nu}$
Now if these are equal then it means that $g(x_{1})=g(x_4)$ or in other words the metric maps a vector in a vector space to its dual.
But isn't the dual space basis given by derivatives? So if $x^{\mu}$ is a vector, then $x_{\mu}$ should be written in terms of derivatives?
2. Sep 17, 2014
### Staff: Mentor
No; the dual space of a vector space is the space of linear maps from the vector space into the reals. So the metric maps a vector to its dual, i.e., it maps a vector into a map from vectors into reals, i.e., the covector $x_{\nu} = g_{\mu \nu} x^{\mu}$ is just the linear map that takes the vector $x^{\mu}$ into its squared length. So the inverse metric just takes a map from vectors into reals, back into a vector; i.e., the vector $x^{\mu} = g^{\mu \nu} x_{\nu}$ is the one that gets mapped to its squared length by the linear map (covector) $x_{\nu}$.
3. Sep 17, 2014
### Fredrik
Staff Emeritus
g is a map from $V\times V$ into $\mathbb R$. If $(e_\mu)_{\mu=0}^3$ is an ordered basis for V, then the 16 numbers $g_{\mu\nu}$ are called the components of g, with respect to that ordered basis. If we let $M$ denote the matrix with $g_{\mu\nu}$ on row $\mu$, column $\nu$, then $g^{\mu\nu}$ denotes the number on row $\mu$, column $\nu$ of $M^{-1}$.
Since g is a map from $V\times V$ into $\mathbb R$, it's not a map from V into V*, but you can use g to define such a map. First you note that for all x in V, the map $y\mapsto g(x,y)$ with domain V is an element of V*. It's convenient to denote this map by $g(x,\cdot)$. The map $x\mapsto g(x,\cdot)$ with domain V, is a map from V into V*.
You're using the symbol g for a lot of different things. First you use it for a metric on V, then for a metric on V* (specifically the one with components $g^{\mu\nu}$). In that final equality, I don't even know what it means. I just see that the g on the left is not the same as the g on the right (since the inputs are in different spaces).
I also don't understand what you're doing. Are you trying to determine what $x_3$ and $x_4$ would have to be in order to ensure that with this choice of metric on V*, we have $g(x_1,x_2)=g(x_3,x_4)$?
The ordered basis of V* that's dual to $(e_\mu)_{\mu=0}^3$ is the 4-tuple $(e^\mu)_{\mu=0}^3$ defined by $e^\mu(e_\nu)=\delta^\mu_\nu$ for all $\mu,\nu$.
In differential geometry, V will be the tangent space at some point p, and V* will be its dual. We can use a coordinate system x (with p in its domain) to define an ordered basis for V. For each $\mu$, we define
$$e_\mu=\frac{\partial}{\partial x^\mu}\!\bigg|_p.$$ Let me know if you don't know the definition of the right-hand side. The dual basis is defined as in the preceding paragraph. For reasons that I don't have time to go into now, the dual basis vectors can be written $\mathrm{d}x^\mu|_p$.
4. Sep 17, 2014
### dextercioby
No, you're mixing concepts. X's are coordinates on the space-time manifold. A 4-tuple of real values uniquely identify a point on the manifold (an event, or a point where one observer resides). That's why they only have 1 type of indices, namey contravariant. I hate it when people put the mu downstairs. As for the derivatives, well they form a basis of the tangent space to any point on the space-time manifold and it's easy to see that once you use the proper definition of a tangent vector.
5. Sep 17, 2014
### ChrisVer
In fact, I have a problem with this thing.
I don't understand the difference between $x^{\mu}$ and $e^{\mu}$ I think they are the same things...$x \in V$ is just a vector, so it can be written in the $e_{\mu}$ basis. And same for an $x^* \in V^*$ which can be expressed with the basis of V* $e_{*}^{\mu}$.
(http://www.itkp.uni-bonn.de/~metsch/GRC2014/grca1a.pdf)
I can't understand though how a vector can be expanded in terms of derivative basis vectors.
6. Sep 17, 2014
### DrGreg
If $\textbf{v}\in V$ you can write$$\textbf{v} = v^\mu \textbf{e}_\mu$$$v^0, v^1$, etc, are (contravariant) coordinates (i.e. numbers) and $\textbf{e}_0, \textbf{e}_1$, etc, are vectors (i.e. elements of V). Similarly if $\textbf{v}^*\in V^*$ you can write$$\textbf{v}^* = v_\mu \textbf{e}^\mu$$$v_0, v_1$, etc, are (covariant) coordinates (i.e. numbers) and $\textbf{e}^0, \textbf{e}^1$, etc, are covectors (i.e. elements of V*).
7. Sep 17, 2014
### ChrisVer
and then in order for $e^{\mu}_{*}(e_{\nu}) = \delta^{\mu}_{\nu}$ you would have to write:
$\textbf{v}^* = v_\mu \partial^{\mu}$?
8. Sep 17, 2014
### Fredrik
Staff Emeritus
You mean the same type of things? For example, if one of them is a real-valued function, then so is the other. If one of them is a 4-tuple of real numbers, then so is the other. $x$ and $e_\mu$ are the same type of things in that sense, but $x^\mu$ is a component of $x$, so it's a real number regardless of what type of object $x$ and $e_\mu$ are.
You need to know the definition of the tangent space at p to understand that. The simplest definition says that it's the vector space spanned by the set $\big\{\frac{\partial}{\partial x^\mu}\big|_p\big\}_{\mu=0}^3$. This definition makes your question trivial. A tangent vector is by definition a linear combination of the four partial derivative functionals.
There are two issues with this definition. It's ugly and it's hard to see why these things are called "tangent vectors". The ugliness issue is addressed by an equivalent definition that's coordinate independent. It goes like this: Let M be a smooth manifold. Let F be the set of smooth functions from M into $\mathbb R$. The tangent space at p is defined as the set of all linear $v:F\to\mathbb R$ such that $v(fg)=v(f)g(p)+f(p)v(g)$ for all $f,g\in F$.
The tangent space at p is denoted by $T_pM$. If we take the coordinate-independent definition of $T_pM$ as our starting point, we will have to state and prove a theorem that says that if x is a coordinate system with p in its domain, then $\big\{\frac{\partial}{\partial x^\mu}\big|_p\big\}_{\mu=0}^3$ is a basis for $T_pM$.
The other issue, that it's hard to see why these things are called tangent vectors, is best addressed by a definition of $T_pM$ that's not equivalent to the one above. This definition makes it obvious why the vectors are called tangent vectors, but is much harder to work with. So we would end up proving a theorem that says that the tangent space defined this way is isomorphic to the tangent space defined as above. And then we will end up working with the tangent space defined above anyway.
If you want to study that alternative definition of $T_pM$, check out Isham's book on differential geometry.
9. Sep 17, 2014
### Fredrik
Staff Emeritus
If $v\in V$ and $\omega\in V^*$, you can write $v=v^\mu e_\mu$ and $\omega=\omega_\mu e^\mu$. If $V=T_pM$, and $x:U\to\mathbb R^n$ is a coordinate system such that $p\in U$, then $e_\mu=\frac{\partial}{\partial x^\mu}\!\big|_p$ and $e^\mu=\mathrm dx^\mu\big|_p$. The explanation for the "d" notation is a bit complicated. You need to study 1-forms to understand it.
10. Sep 17, 2014
### ChrisVer
I meant that if $x^{\mu}$ is a vector,then so is $e^{\mu}$. The only difference is the name, and that e is some basis (like $e_{x,y,z}$ is a basis for the 3dimensional euclidean space).
I guess the tangent space is easier to see in a 2dim surface which is the mapping $x: U_2 \rightarrow S$ where $U$ is the 2d (u,v) parameter space and $S$ is the surface, where then the tangent space is spanned by the derivatives of x wrt the parameters of your surphase... So $d \textbf{x}(u,v) = \textbf{x}_u du + \textbf{x}_v dv$ right? so $\textbf{x}_{u,v}$ are the vectors which span the tangent plane of the surface. Now on some point p, then you have to take the derivatives at the corresponding point $\textbf{x}_{u,v} (u_0,v_0)$ if p is in $(u_0,v_0)$.
Now going to a 4dimensional case, I have $\textbf{X}(x^{\nu})$,_where now $\textbf{X}: U_{4} \rightarrow M$ maps the $x^{\mu} \in U_4$ coords to some manifold M, and thus in the same way:
$d \textbf{X} (x^{\mu})= \partial_{\mu} \textbf{X} dx^{\mu}$
and thus the spanning vectors are the $\partial_{\mu} \textbf{X}$
Last edited: Sep 17, 2014
11. Sep 17, 2014
### Staff: Mentor
It isn't. It's a component of a vector. If $\vec{e}_{\mu}$ (note that the index is lower here, not upper) is a vector--a basis vector--then the vector $\vec{x}$ can be written in component form as $x^{\mu} \vec{e}_{\mu}$. In other words, the vector itself is a linear combination of the basis vectors; the components of that vector are just the coefficients in the linear combination.
I agree that this notation for components and vectors can be confusing (the index $\mu$ labels both components and basis vectors). But that's the standard notation.
Now you're confusing the notation: is $x$ supposed to be a vector? Vectors aren't mappings; they're elements of a vector space. Covectors are mappings, but they're mappings of vectors into numbers, not mappings of pairs (or n-tuples) of numbers into points in a surface. The latter type of mapping is a coordinate chart, though properly speaking it should be given the other way around: a coordinate chart is a mapping of points in a manifold into n-tuples of real numbers, where n is the dimension of the manifold.
Perhaps what you're trying to say here is that the *components* of a vector $x$ can be viewed as n-tuples of numbers, and therefore, if we choose some particular point in the surface as the origin, we can view the components of a vector $x$ as the n-tuple that the coordinate chart maps a given point to, where the point is the one at the tip of the vector when its tail is at the origin. However, you should be aware that this doesn't work in a curved manifold, only in a flat one.
Strictly speaking, vectors do not "live" in the manifold itself; they live in the tangent space at a particular point (there is a different tangent space at each point); but in a flat manifold, you can get away with identifying the tangent space at any given point with the manifold itself, described in a chart where the given point is the origin. This is a bad habit to get into, though, because it doesn't generalize to curved manifolds. It's better to keep vectors, points, and coordinate n-tuples all separate in your mind, and to keep in mind also that all of these objects "live" in the tangent space at a given point in the manifold, not in the manifold itself.
No, not "derivatives of x". First of all, that just repeats the above confusion: is $x$ supposed to be a vector? A point? An n-tuple of coordinates? Second, the derivatives that span the tangent space are not derivatives "of" anything in particular; they are directional derivative operators along particular orthogonal directions--more precisely, they are directional derivatives along curves in particular orthogonal directions, evaluated at the given point. The reason this works is that there is a bijective correspondence between tangent vectors at a given point, and directional derivative operators along curves at that point: the operator corresponding to a given tangent vector is just the directional derivative along the curve that the vector is tangent to. So the derivatives that span the tangent space are just the directional derivatives corresponding to the basis vectors.
Last edited: Sep 18, 2014
12. Sep 18, 2014
### Fredrik
Staff Emeritus
$x^\mu$ would be an awful notation for a vector. It typically denotes a component of a vector $x=x^\mu e_\mu$, or a component of a coordinate system. If $x$ and $e_\mu$ are 4-tuples of real numbers, then $x^\mu$ is a real number. But if $x:U\to\mathbb R^n$ is a coordinate system on a manifold and $p\in U$, we would write $x(p)=(x^1(p),x^1(p),x^2(p),x^n(p))$. (The default choice for indices is to let them go from 1 to n, but in relativity we let them go from 0 to 3 instead of 1 to 4). So we would also use the notation $x^\mu$ for the map that takes a point p in U to it's $\mu$th coordinate in the coordinate system x.
In this case it would make sense to define the tangent space at $(a,b)\in U_2$ as derivatives of curves in S through the point (a,b). The maps $u\mapsto \mathbf x(u,b)$ and $v\mapsto\mathbf x(a,v)$ are curves in S through $\mathbf x(a,b)$. Their derivatives at the points in their domains that are mapped to $\mathbf x(a,b)$ are $\mathbf x_u(a,b)$ and $\mathbf x_v(a,b)$, where $\mathbf x_u$ and $\mathbf x_v$ denote the partial derivatives of $\mathbf x$. $\mathbf x_u(a,b)$ and $\mathbf x_v(a,b)$ are elements of $\mathbb R^3$. It makes sense to think of the plane spanned by these vectors as the tangent plane of the surface $\mathbf x$ at the point $\mathbf x(a,b)$.
But this is not how it's done in differential geometry. If $C:\mathbb R\to M$ is a curve in a manifold M, you can't just compute $C'(0)$ and call it a tangent vector, because there's no addition operation on M. But if x is a coordinate system, then $x\circ C$ is a curve in $\mathbb R^n$. The components of the tangent vector of C at the point C(0), in the coordinate system x, are $(x^\mu\circ C)'(0)$. The tangent vector can be written as $(x^\mu\circ C)'(0)e_\mu$, but the $e_\mu$ in this equality are not the standard basis vectors for $\mathbb R^n$. They're the partial derivative functionals $\frac{\partial}{\partial x^\mu}\!\big|_{C(0)}$.
13. Sep 20, 2014
### stevendaryl
Staff Emeritus
Well, there is a convention--which I don't particularly like, although lots of people follow it--that a variable with a superscript is interpreted as a vector, and a variable with a subscript is interpreted as a covector. So when someone following this convention writes $v^\mu$, they mean what you would write as: $v^\mu e_\mu$.
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# Exploring an interview question from the Oxford computer science department
I saw a fun set of problems from this tweet yesterday:
We used one of the problems for our project this morning:
We stared the project today by having the boys read the problem and talking about some potential strategies for how to approach it:
They started by looking at some simple examples. These examples helped them get a feel for how the problem worked, but didn’t quite point them toward the solution.
From the last discussion they began to believe that they game would always end the same way no matter how you played it. So they decided to see what would happen if they ran through the version of the game presented in the problem.
Their way of running through the game was fascinating to me
Now we talked about why there was only one way to end. They found their way to the idea that the even and odd numbers of white beans was the key to the problem.
Finally I showed them one way of thinking about how the odds and even number of white and black beans changes each turn.
I really like this problem. It is a super fun problem for kids to explore.
# Sharing Kendra Lockman’s Desmos activity with my son
I saw this tweet from Kendra Lockman yesterday:
It looked like a fun activity to try, so we spent 20 minutes this morning going through it. It was nice to year what my son son thinking about fractions throughout the activity. The first 4 videos below show his work and the last is some quick thoughts from him on the morning:
Part 1:
Part 2:
Part 3:
Part 4:
Here’s his summary of the activity:
# A strange homework problem
One of my older son’s homework problems asked him to find 3 digit multiples of 7 whose digit sums were also multiples of 7. I was puzzled by this problem had it on my mind most of the day today.
I hoped that talking through it would help me understand what the math idea was behind the problem. Sadly no, but we still had a good talk.
Here’s the problem and the work my son did:
So – still quite puzzled about the problem – I decided to see if there was anything quirky that came up looking at a divisibility rule for 7 with 3 digit numbers. This gave us a nice opportunity to talk about modular arithmetic:
Finally, since I wasn’t making any progress seeing the point of the original problem, I had him talk about other divisibility rules that he knew:
So, a nice conversation, but I’m actually baffled. I’ll have to ask the author of the problem what he was trying to get at – I feel like I’m missing the point.
# Talking through 3 AMC 8 problems with my son
My son was working on the 1993 AMC 8 yesterday and had trouble with a few problems. Today we sat down and talked through those problem.
The first was problem 17:
Here’s what my son had to say:
Next was problem 19:
Next was problem 24:
I like these old contest problems. They lead to really nice conversations!
# A mistake that led to a great conversation
My older son had a homework problem that asked him to find the area of the region bounded by the two equations:
(i) $| 2x + 3y | \leq 6$, and
(ii) $| x - 2y | \leq 4$
Mathematica’s picture of that shape is here:
He told me that he used Pick’s theorem and found that the area of the shape was 13 square units.
There’s just one small problem – you can’t use Pick’s theorem to find the area of this shape since the corners of the shape are not lattice points of the grid.
What to do . . . .
I wrote a quick little program that picked 100 million random points in the 10×10 square centered at (0,0) and tested whether or not they were part of the shape. That program found that 13.71% of the points were part of the shape – that was enough to convince him that the area might be larger than 13 square units.
Next I had him re-read Pick’s theorem to see what went wrong. He saw pretty quickly that the shape didn’t meet the condition of having the corners lie on lattice points.
I really wanted to try to find a way to make Pick’s theorem work with this shape. I had him determine the y-coordinate for the far right corner. The value was y = -2/7.
After finding that value, we had a good talk about scaling. To make the new grid larger we had to *divide* the x and y coordinates in the equations by 7. Here’s Mathematica’s picture of the new shape and grid (note that the x and y values run from -25 to 25 in this picture):
With this shape we are able to use Pick’s theorem to calculate the area.  We counted 40 grid points on the boundary without too much difficulty.  Counting the ones in the middle was a little bit more of a pain, so we wrote a short program to perform that calculation for us. Note that we have to change the “less than or equal to” from the original equations to “strictly less than” since we want to be inside the shape:
So, we have 653 lattice points in the inside and 40 on the boundary. Pick’s theorem tells us that the area is equal to the number of lattice points on the interior plus half the number of lattice points in the boundary minus 1. That’s 672 units.   In the picture above, 1 unit is equal to 1/49 of a unit in the original picture, so the original area is 96 / 7 or 13 5/7.  Close to what he found originally, but not equal!
Along the way we also talked about alternate ways to find the area – the easiest being dividing the shape into two triangles with a vertical line through the middle.
I’m really excited about the discussion that we had tonight. Funny how many important ideas in math can come up from a problem about absolute value and inequalities.
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# Lessons from a great geometry homework problem
My older son had a terrific homework problem in his enrichment math class. I wanted to walk through the problem again today so that my younger son could see it and also to highlight some of the lessons in the problem.
To start the project we revisted a fun geometry problem that will make a surprise appearance at the end of the homework problem:
Next I introduced the homework problem. My older son is familiar with this problem, but my younger son is seeing it for the first time. In this video my older son highlights the main ideas that we need to solve the problem (well . . . see the next video for the one we forgot!):
Here’s the one extra piece that we missed from the last video:
Next with the triangles labelled properly, we worked to see how we can use the Pythagorean theorem to help solve for the values of the two unknowns. I used this section of today’s project to give my younger son a little algebra practice:
Now comes the task of simplifying the two complicated equations. Hopefully that will help us make some progress towards solving them.
After the simplifying in the last video we are now ready to take a crack at solving for the radius of the smaller circles. Solving the equation involves solving a quadratic and that gave us a chance to talk about factoring.
Finally, we went back to the picture from the homework problem. We hadn’t solved for x in the project, but now we can use the pictures to help us find x’s value. We see an 8-15-17 triangle and also a 3-4-5 triangle. We also see the 5-5-8 triangle from the beginning of the project!
So, a fun project connecting a neat geometry problem that we’ve studied before with a new homework problem.
# Sharing Jim Propp’s base 3/2 essay with kids part 3
I’ve been sharing Jim Propp’s most recent essay (as of Sept. 2017) with the boys this week. The essay and our first two projects are here:
His essay is here:
Jim Propp’s How do you write one hundred in base 3/2?
Sharing Jim Propp’s base 3/2 essay with kids – Part 1
Sharing Jim Propp’s base 3/2 essay with kids – Part 2
Today I had the kids read the post. Here are some of the ideas that they thought were interesting:
With the camera off I started talking about some of the ideas from math that I knew that weren’t “useful” originally but became useful later. So, I turned the camera on and talked about it live. I would not be surprised if several of the things I said in this part are not historically accurate – this wasn’t prepared and was off the top of my head.
Next I asked each of the boys to design their own Engel machine. My older son went first and we counted to 10 using his machine.
My younger son went next. He had a different design and we used it to count to 10, too:
I’ve really enjoyed sharing Propp’s essay with the boys. It is a great way for kids to explore different bases and also a great introduction to some fun advanced ideas from math. I also love that kids can play with the “machines” using blocks – that seems to really keep them engaged.
# Sharing Jim Propp’s base 3/2 essay with kids part 2
I’m going through Jim Propp’s piece on base 3/2 with my kids this week.
His essay is here:
Jim Propp’s How do you write one hundred in base 3/2?
And the our first project using that essay is here:
Sharing Jim Propp’s base 3/2 essay with kids – Part 1
Originally I wanted to have the kids read the essay and give some of their thoughts for part 2, but I changed my mind on the approach this morning. Instead I asked each of them to answer the question in the title of Propp’s essay -> How do you write 100 in base 3/2?
Propp points out in his essay that his approach to base 3/2 via chip firing / Engel machines / exploding dots is not what mathematicians would normally consider to be base 3/2. The boys are not aware of that statement, though, since they have not read the essay yet.
Here’s how my younger son approached writing 100 in base 3/2. The first video is an introduction to the problem and, from knowing how to write numbers like 100 (in base 10) in other integer bases.
I think the first 3 minutes of this video are interesting because you get to hear his ideas about why this approach seems like a good idea. The remainder of this video plus the next two videos are a long march down the road to discovering why this approach doesn’t work in the version of base 3/2 we are studying:
So, after finding that the path we were walking down led to a dead end, we started over. This time my son decided to try to write 100 as 10×10. This approach does work!
Next I introduced the problem to my older son. He also started by trying to solve the problem the same way that you would for integer bases, though his technique was slightly different. He realized fairly quickly (by the end of the video, I mean) that this approach didn’t work:
My older son needed to find a new approach, and he ended up finding an idea different from my younger son’s idea to find 100 in base 3/2. His idea was to use chip firing:
I thought that today’s project would be a quick reminder of how base 3/2 works (at least the version we are studying). That thought was way off base and was completely influenced by me knowing the answer! Instead we found – by accident – a great example of how to explore a challenging problem in math. Sometimes the first few things you try don’t work, and you have to keep trying new things.
Definitely a fun morning!
# Sharing Jim Propp’s base 3/2 essay with kids
Jim Propp’s essay on base 3/2 is fantastic:
Jim Propp’s How do you write one hundred in base 3/2?
and here are links to our two prior base 3/2 projects:
Fun with James Tanton’s base 1.5
Revisiting James Tanton’s base 3/2 exercise
I’m hoping to have time to spend at least 3 days playing around with Propp’s latest blog post. Today we had 20 min free unexpectedly in the morning and I used that time to introduce two of the ideas. They haven’t read the post, yet, but instead I started by having them watch Propp’s short video about the binary Engel machine:
After watching that video I had the boys recreate the idea with snap cubes on our white board. Here’s that work plus a few of their thoughts on the connection with binary:
Next I challenged the boys to draw the base 3/2 version of the machine. After they did that we counted to 10 in base 3/2 and talked about what we saw:
I was happy that the boys were able to understand the idea behind the base 3/2 Engel machine. With the work from today giving them a nice introduction to some of the ideas in Propp’s essay, I think they are ready to try reading the essay tomorrow. It’ll be interesting to see what ideas catch their eye. Hopefully we can do another short project on whatever those ideas are tomorrow morning.
# Sharing Kelsey Houston-Edwards’s Axiom of Choice video with kids
Kelsey Houston-Edwards has a new video out about the Axiom of Choice:
The video is amazing (as usual) and I wanted to be able to share it with the boys. This one is a bit hard than usual – the topic is pretty advanced to begin with and is also pretty far outside of my own knowledge – but we gave it a shot.
Here’s what the boys thought after seeing the video:
Next we reviewed how Houston-Edwards divided the numbers from 0 to 1 into buckets. The boys didn’t quite have the details right, but that actually made talking through the idea pretty easy – I learned from their explanation what points needed to be re-emphasized.
Now we talked through the really challenging part of the video -> creating the set with no size. Given the challenge of explaining this idea to kids, I’m pretty happy with how the conversation went here. Also, I only finally understood the argument myself while I was explaining it to them!
Now we backed away from the complexity of the Axiom of Choice and reviewed two other slightly easier ideas that came up in our discussion. Here we discuss why $\sqrt{2}$ is irrational:
Finally, we wrapped up by discussing why the rational numbers are countable:
Although kids will have a hard time understanding all of the ideas that Kelsey Houston-Edwards brings up in her Axiom of Choice video, I think it is fun to see which ideas grab their attention. The idea that you can have a set that doesn’t have a size is pretty amazing. I was pretty happy with how things went today – exploring the ultra complex idea first and then backing off to discuss slightly easier ideas involving infinity. Definitely a fun set of ideas to plant in the minds of younger kids 🙂
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# What is Specific Heat Capacity? | Definition, Units, Types – Thermometry and Calorimetry
Specific Heat Definition:
The amount of heat required to raise the temperature of unit mass of the substance through 1°C is called its specific heat.
We are giving a detailed and clear sheet on all Physics Notes that are very useful to understand the Basic Physics Concepts.
## What is Specific Heat Capacity? | Definition, Units, Types – Thermometry and Calorimetry
Specific Heat Symbol:
It is denoted by c or s.
Specific Heat Unit:
Its SI unit is ‘joule/kilogram-°C’ (J/kg-°C) or Jkg-1K-1
Dimensional Formula of Specific Heat:
The dimensional formula is [L2T2θ-1].
Specific Heat of Water:
The specific heat of water is 4200 J kg-1°C-1 or 1 cal g-1 C-1 which is high as compared with most other substances.
Specific Heat Capacity Formula:
The amount of heat energy required to change the temperature of any substance is given by
Q = mcΔt
where,
m = mass of the substance,
c = specific heat of the substance and
Δt = change in temperature
### Gases have two types of specific heat
1. The specific heat capacity at constant volume (Cv).
2. The specific heat capacity at constant pressure (Cp).
Specific heat at constant pressure (Cp) is greater than specific heat at constant volume (Cv), i.e. Cp>Cv.
For molar specific heats, Cp – Cv = R
where, R = gas constant and this relation is called Mayer’s formula.
The ratio of two principal sepecific heats of a gas is represented by γ, i.e.
γ = $$\frac{C _{p}}{C_{V}}$$
The value of γ depends on atomicity of the gas.
Thermometry and Calorimetry:
The thermometer is a device used to check the temperature of an object. This branch of measurement of the temperature of a substance is called thermometry. It is measured in degrees or Fahrenheit, usually.
Calorimetry also means the measurement of heat but in joules. It states the amount of heat lost by the body is the amount of heat gained by its surrounding.
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https://www.mathphysicsbook.com/mathematics/the-divergence-currents-and-tensor-densities/current-forms-and-densities/
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# Current forms and densities
We previously defined the current vector (AKA flux) $${j\equiv\rho u}$$, where $${\rho}$$ is the density of the physical quantity $${Q}$$ and $${u}$$ is a velocity field, and then combined them into the four-current $${J\equiv(\rho,j^{\mu})}$$. There are a number quantities that can be defined around this concept:
QuantityDefinitionMeaning
Current vector$${j\equiv\rho u}$$The vector whose length is the amount of $${Q}$$ per unit time crossing a unit area perpendicular to $${j}$$
Current form
\begin{aligned}\zeta & \equiv i_{j}\mathrm{d}V\\
& =\left\langle j,\hat{n}\right\rangle \mathrm{d}S
\end{aligned}
The $${(n-1)}$$-form which gives the amount of $${Q}$$ per unit time crossing the area defined by the argument vectors
Current density
\begin{aligned}\mathfrak{j} & \equiv\sqrt{g}\,j\\
\Rightarrow\zeta & =\mathfrak{j}^{1}\mathrm{d}^{2}x
\end{aligned}
The vector whose coordinate length is the amount of $${Q}$$ per unit time crossing a unit coordinate area perpendicular to $${j}$$
Current
\begin{aligned}I & \equiv\int_{S}\zeta\\
& =\int_{S}\left\langle j,\hat{n}\right\rangle \mathrm{d}S\\
& =\int_{S}\mathfrak{j}^{1}\mathrm{d}^{2}x
\end{aligned}
The amount of $${Q}$$ per unit time crossing $${S}$$
Four-current$${J\equiv(\rho,j^{\mu})}$$Current vector on the spacetime manifold
Notes: $${\rho}$$ is the density of the physical quantity $${Q}$$, $${u}$$ is a velocity field, $${\hat{n}}$$ is the unit normal to a surface $${S}$$, and $${\mathrm{d}^{3}x}$$ are coordinates with $${x^{1}}$$ constant on $${S}$$ and normal to it. The four-current can be generalized to other Lorentzian manifolds, and can also be turned into a form $${\xi\equiv i_{J}\mathrm{d}V}$$ or a density $${\mathfrak{J}\equiv\sqrt{g}\,J}$$.
Δ Note that the terms flux and current (as well as flux density and current density) are not used consistently in the literature.
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Calorie Basics
Calorie Basics — The Physics involves More Newton than Anyone Else!
We all need energy and when we talk about the energy we need to turn the pedals of our bicycles we generally talk in terms of calories and here are some quick Calorie Basics.
Calorie Basics — What is a Calorie?
The official definition of a calorie is: the heat energy required to raise one gram of water, 1° C. A gram is about the weight of a paperclip and so is not much water and that makes the calorie a small unit of energy. In fact, I once calculated the calories it takes to move me up a flight of stairs and was stunned at the huge number, but when a colleague informed me that in the context I was performing the calculation I need to think in larger units.
Calorie Basics — Calories vs. Kilo-Calories: The Larger Units.
The opening definition of a calorie is a small amount of energy; therefore, the calorie we normally talk about is a kilo-calorie, that is 1,000 calories. So take heart, and that 2,300 calorie budget you are on is actually 2,300,000 calories!
Calorie Basics — Measuring the Calories in Food
It was da-bomb! The old technique involved burning the food in question in a water filled sealed container, the process used electricity and when the food was consumed they would measure the temperature of the water. However, that technique is no longer used due to the expense of it and the genuine concern it overstates the extractable calories from the food.
Instead, they now typically use tables and base the caloric content off the macro-nutrient content. That is, it is generally accepted protein and carbohydrates provide 4 calories per gram and fats provide about 9 calories per gram. There are other nutrients in the mix, but that is the big 3 (alcohol comes in at 7 calories per gram).
Calorie Basics — Digging Deeper What is Energy?
Energy is simply the “stuff” we need to do work and work at its easiest to understand level is changing the motion of an object (making a still object move, bring a moving object to a stop, making an object turn, etc). In order to move your legs, to turn the crank, to move the chain, to spin the rear wheel you need energy. That energy is derived from the calories you eat and drink!
Now, often times when we talk of biking we often hear the word power used. Power is related to but is NOT energy and is actually the rate at which work is performed. Two actors can accomplish the same work using different power, but the one who used the greater power did the work in less time. The energy output is the same the time is different.
Calorie Basics — Biological Facts
Now we have some of the basic physics out of the way we can talk of calories as they relate to biking and weight loss or gain. I have seen a number of people argue that the weight loss equation of `calories in < calories out` is not true, that if you eat smart you will lose weight. Rubbish. Eating smart is always needed, but if in eating smart you take in more calories than you expend you will gain weight.
Our bodies are built to consume, store, and expend energy. Since most of human existence has been in feast-famine mode our bodies are always glad to take on more energy and reluctant to give it up. Who knows, the dry season may come early this year and we will have to go an extra month with little food. This has been the case until very recent in our history. I have read stories of my ancestors in Door County having very little to eat over the winter (and sharing what little they had with others). Now a days, we hear much about obesity.
Calorie Basics — 3,500
It is generally accepted as fact that one pound of human body weight is equivalent to 3,500 calories. If you eat 3,500 more calories than you burn you will gain a pound, if you expend 3,500 calories more than you eat you will lose that pound. Over the course of a year that translates to 10 calories per day. That amounts to +/- 1 peanut M&M per day.
Calorie Basics — Biking
Bicyclists soon learn about the need to add calories judiciously while riding. The evil bane of processed sugars and even soda pop (usually defizzed) is used by professional riders. The reason is because sugars can be quickly broken down and delivered to the muscles in useful form to provide the energy to make the crank turn round n’ round! Sure a hunk prime-rib fat contains a lot more calories but we all know how hard it is to digest that fat and my body is busy trying to set a course record.
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