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B. A variable is something we do not know and we use a letter to represent it.
Answer:
![E(W) = 4 [np_x] + 6 [m p_y]](https://tex.z-dn.net/?f=%20E%28W%29%20%3D%204%20%5Bnp_x%5D%20%2B%206%20%5Bm%20p_y%5D)
![Var(W) = 16[np_x (1-p_x)] +36 [mp_y (1-p_y)]](https://tex.z-dn.net/?f=%20Var%28W%29%20%3D%2016%5Bnp_x%20%281-p_x%29%5D%20%2B36%20%5Bmp_y%20%281-p_y%29%5D)
Step-by-step explanation:
For this case we have the following distributions given:
![X \sim Binom (n, p_x)](https://tex.z-dn.net/?f=%20X%20%5Csim%20Binom%20%28n%2C%20p_x%29)
The expected value for x is ![E(X) = n p_x](https://tex.z-dn.net/?f=%20E%28X%29%20%3D%20n%20p_x)
And the variance ![Var(X) = np_x (1-p_x)](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20np_x%20%281-p_x%29)
![Y \sim Binom (m, p_y)](https://tex.z-dn.net/?f=%20Y%20%5Csim%20Binom%20%28m%2C%20p_y%29)
The expected value for x is ![E(Y) = n p_y](https://tex.z-dn.net/?f=%20E%28Y%29%20%3D%20n%20p_y)
And the variance ![Var(Y) = mp_y (1-p_y)](https://tex.z-dn.net/?f=%20Var%28Y%29%20%3D%20mp_y%20%281-p_y%29)
We define a new random variable:
![W = 4X +6Y](https://tex.z-dn.net/?f=%20W%20%3D%204X%20%2B6Y)
We need to find the expectd value and the variance for W.
For the expected value we have this:
![E(W) = E(4X+6Y) = E(4X) + E(6Y) = 4E(X) + 6E(Y)](https://tex.z-dn.net/?f=%20E%28W%29%20%3D%20E%284X%2B6Y%29%20%3D%20E%284X%29%20%2B%20E%286Y%29%20%3D%204E%28X%29%20%2B%206E%28Y%29)
And if we replace the parameters we got:
![E(W) = 4 [np_x] + 6 [m p_y]](https://tex.z-dn.net/?f=%20E%28W%29%20%3D%204%20%5Bnp_x%5D%20%2B%206%20%5Bm%20p_y%5D)
Now foe tha variance of W we know that X and Y are indpenendet variable so then ![Cov(X,Y) =0](https://tex.z-dn.net/?f=%20Cov%28X%2CY%29%20%3D0)
![Var(W)= Var(4X +6Y) = Var(4X) +Var(6Y) + 2 Cov(X,Y)](https://tex.z-dn.net/?f=%20Var%28W%29%3D%20Var%284X%20%2B6Y%29%20%3D%20Var%284X%29%20%2BVar%286Y%29%20%2B%202%20Cov%28X%2CY%29)
For this case we can use the property that is a is a constant and X a random variable then
, and we got this:
![Var(W) = 16 Var(X) + 36 Var(Y)](https://tex.z-dn.net/?f=%20Var%28W%29%20%3D%2016%20Var%28X%29%20%2B%2036%20Var%28Y%29)
And if we replace we got:
![Var(W) = 16[np_x (1-p_x)] +36 [mp_y (1-p_y)]](https://tex.z-dn.net/?f=%20Var%28W%29%20%3D%2016%5Bnp_x%20%281-p_x%29%5D%20%2B36%20%5Bmp_y%20%281-p_y%29%5D)
We know that the perimeter of a rectangle is 2(width)+2(length). We can multiply the length by 2 to find out how much remaining fencing we have, then divide that number by 2 to find the width.
15.5*2=31
170.5-31=139.5
139.5/2=69.75
A possible width is 69.75 feet.
Answer:
what u need help with
Step-by-step explanation: