It’s kinda blurry can u take another picture?
Solution :
a). ![$\text{Var} (X|Y) =E ((X-E(X|Y))^2 |Y)$](https://tex.z-dn.net/?f=%24%5Ctext%7BVar%7D%20%28X%7CY%29%20%3DE%20%28%28X-E%28X%7CY%29%29%5E2%20%7CY%29%24)
Now, if X = Y, then :
![P(X|Y)=\left\{\begin{matrix} 1,& \text{if } x=y \\ 0, & \text{otherwise }\end{matrix}\right.](https://tex.z-dn.net/?f=P%28X%7CY%29%3D%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%201%2C%26%20%5Ctext%7Bif%20%7D%20x%3Dy%20%5C%5C%200%2C%20%26%20%5Ctext%7Botherwise%20%7D%5Cend%7Bmatrix%7D%5Cright.)
Then, E[X|Y] = x = y
So, ![$\text{Var} (X|Y) =E((X-X)^2 |Y)$](https://tex.z-dn.net/?f=%24%5Ctext%7BVar%7D%20%28X%7CY%29%20%3DE%28%28X-X%29%5E2%20%7CY%29%24)
![$=E(0|Y)$](https://tex.z-dn.net/?f=%24%3DE%280%7CY%29%24)
= 0
Therefore, this statement is TRUE.
b). If X = Y , then Var (X) = Var (Y)
And as Var (X|Y) = 0, so Var (X|Y) ≠ Var (X), except when all the elements of Y are same.
So this statement is FALSE.
c). As defined earlier,
![$\text{Var} (X|Y) =E ((X-E(X|Y))^2 |Y=y)$](https://tex.z-dn.net/?f=%24%5Ctext%7BVar%7D%20%28X%7CY%29%20%3DE%20%28%28X-E%28X%7CY%29%29%5E2%20%7CY%3Dy%29%24)
So, this statement is also TRUE.
d). The statement is TRUE because
.
e). FALSE
Because, ![$\text{Var} (X|Y) =E ((X-E(X|Y=y))^2 |Y=y)$](https://tex.z-dn.net/?f=%24%5Ctext%7BVar%7D%20%28X%7CY%29%20%3DE%20%28%28X-E%28X%7CY%3Dy%29%29%5E2%20%7CY%3Dy%29%24)
We have that
cos A=0.25
so
A=arc cos (0.25)-------> using a calculator----> A=75.5225°
Round to the nearest hundredth-----> A=75.52²
the answer is
the option <span>75.52°</span>
Answer: use pemdas to complete this
Step-by-step explanation:
distribute the -4.6 to all of the ( ) and then combine like numbers :) lmk if you need more help