Here's a couple of ideas.
1) Use concept of rational numbers.
Start by assigning 2 variables with a square root, representing each student.
Then multiply the 2 variables together resulting in a whole number. (canceling the square root)
Then say that being by yourself is irrational. But working together is rational.
![boy = \sqrt{3} , girl = \sqrt{3} \\ \\ boy*girl = \sqrt{3}*\sqrt{3} = 3](https://tex.z-dn.net/?f=boy%20%3D%20%5Csqrt%7B3%7D%20%2C%20girl%20%3D%20%5Csqrt%7B3%7D%20%20%5C%5C%20%20%5C%5C%20boy%2Agirl%20%3D%20%5Csqrt%7B3%7D%2A%5Csqrt%7B3%7D%20%3D%203)
2) Set up a linear equation to show how the more they work together the higher their grades will be.
For 8th grade a simple one is all that is needed.
![G = 80 + t](https://tex.z-dn.net/?f=G%20%3D%2080%20%2B%20t)
where t is hours working together.
Provide a graph or table demonstrating the positive slope.
Good luck
Answer:
The estimated value for the population standard deviation is $3,750.
Step-by-step explanation:
The interval within which the annual starting salaries for college graduates with degrees in business administration are generally expected to be is $30,000 and $45,000.
The standard deviation is the measure of dispersion or spread. And so is the range of the data.
Both the measures are used to determine how spread out the data is from the mean.
The relationship between standard deviation and range is not defined but according to the rule of thumb the standard deviation is one-fourth of the range of the data.
The formula of range is:
![Range=Maximum-Minimum](https://tex.z-dn.net/?f=Range%3DMaximum-Minimum)
Compute the range of the data of starting salaries of college graduates with degrees in business administration as follows:
![Range=Maximum-Minimum\\=45000-30000\\=15000](https://tex.z-dn.net/?f=Range%3DMaximum-Minimum%5C%5C%3D45000-30000%5C%5C%3D15000)
The range of the data is, $15,000.
Compute the estimated value of the standard deviation as follows:
![SD=\frac{Range}{4}=\frac{15000}{4}=3750](https://tex.z-dn.net/?f=SD%3D%5Cfrac%7BRange%7D%7B4%7D%3D%5Cfrac%7B15000%7D%7B4%7D%3D3750)
Thus, the estimated value for the population standard deviation is $3,750.
I think it’s 15 I hope it helps
Hi there!
The answer is $140.
Could you please Brainliest me?
Answer:
See the proof below
Step-by-step explanation:
Let's assume that our random variable of interest is Y and we have a set of parameters
in the original network.
And let's assume that we add an additional parameter
and we want to see if the likehood for ![x_1, x_2,.....,x_k, x_{k+1}](https://tex.z-dn.net/?f=x_1%2C%20x_2%2C.....%2Cx_k%2C%20x_%7Bk%2B1%7D)
We don't know the distribution for each parameter
but we can say that the likehood function for the original set of parameters is given by:
![F=L(y| x_1,x_2,....,x_k)](https://tex.z-dn.net/?f=F%3DL%28y%7C%20x_1%2Cx_2%2C....%2Cx_k%29)
And in order to maximize this function we need to take partial derivates respect to each parameter like this:
![\frac{dF}{dx_i} =\frac{dL}{dx_i}, i=1,2,....,k](https://tex.z-dn.net/?f=%5Cfrac%7BdF%7D%7Bdx_i%7D%20%3D%5Cfrac%7BdL%7D%7Bdx_i%7D%2C%20i%3D1%2C2%2C....%2Ck)
We just need to set up the last derivate equal to zero and solve for the parameters who satisfy the condition.
If we add a new parameter the new likehood function would be given by:
![F=L(y| x_1,x_2,....,x_k,x_{k+1})](https://tex.z-dn.net/?f=F%3DL%28y%7C%20x_1%2Cx_2%2C....%2Cx_k%2Cx_%7Bk%2B1%7D%29)
And in order to maximize this function again we need to take partial derivates respect to each parameter like this:
![\frac{dF}{dx_i} =\frac{dL}{dx_i}, i=1,2,....,k,k+1](https://tex.z-dn.net/?f=%5Cfrac%7BdF%7D%7Bdx_i%7D%20%3D%5Cfrac%7BdL%7D%7Bdx_i%7D%2C%20i%3D1%2C2%2C....%2Ck%2Ck%2B1)
We are ssuming that we have the same parameters from 1 to k for the new likehood function. So then the likehood for the data would be unchanged and if we have more info for the likehood function we are maximizing the function since we are adding new parameters in order to estimate the function.
![max L(y| x_1,x_2,....,x_k,x_{k+1}) \geq max L(y| x_1,x_2,....,x_k)](https://tex.z-dn.net/?f=max%20L%28y%7C%20x_1%2Cx_2%2C....%2Cx_k%2Cx_%7Bk%2B1%7D%29%20%5Cgeq%20max%20L%28y%7C%20x_1%2Cx_2%2C....%2Cx_k%29)