Answer:
The expected value of X is
and the variance of X is 
The expected value of Y is
and the variance of Y is 
Step-by-step explanation:
(a) Let X be a discrete random variable with set of possible values D and probability mass function p(x). The expected value, denoted by E(X) or
, is

The probability mass function
of X is given by

Since the bus driver is equally likely to drive any of the 4 buses, the probability mass function
of Y is given by

The expected value of X is
![E(X)=\sum_{x\in [28,32,42,44]} x\cdot p_{X}(x)](https://tex.z-dn.net/?f=E%28X%29%3D%5Csum_%7Bx%5Cin%20%5B28%2C32%2C42%2C44%5D%7D%20x%5Ccdot%20p_%7BX%7D%28x%29)

The expected value of Y is
![E(Y)=\sum_{x\in [28,32,42,44]} x\cdot p_{Y}(x)](https://tex.z-dn.net/?f=E%28Y%29%3D%5Csum_%7Bx%5Cin%20%5B28%2C32%2C42%2C44%5D%7D%20x%5Ccdot%20p_%7BY%7D%28x%29)

(b) Let X have probability mass function p(x) and expected value E(X). Then the variance of X, denoted by V(X), is
![V(X)=\sum_{x\in D} (x-\mu)^2\cdot p(x)=E(X^2)-[E(X)]^2](https://tex.z-dn.net/?f=V%28X%29%3D%5Csum_%7Bx%5Cin%20D%7D%20%28x-%5Cmu%29%5E2%5Ccdot%20p%28x%29%3DE%28X%5E2%29-%5BE%28X%29%5D%5E2)
The variance of X is
![E(X^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{X}(x)](https://tex.z-dn.net/?f=E%28X%5E2%29%3D%5Csum_%7Bx%5Cin%20%5B28%2C32%2C42%2C44%5D%7D%20x%5E2%5Ccdot%20p_%7BX%7D%28x%29)


The variance of Y is
![E(Y^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{Y}(x)](https://tex.z-dn.net/?f=E%28Y%5E2%29%3D%5Csum_%7Bx%5Cin%20%5B28%2C32%2C42%2C44%5D%7D%20x%5E2%5Ccdot%20p_%7BY%7D%28x%29)


If the points are collinear, then that means they lie on the same line when connected to each other. In order to prove this, the distances between the points must sum up to the length of the line.
AB + BC = AC
7 + 4 ? 9
11 ≠ 9
Since they do not sum up, therefore, the points are not collinear.
I do believe the answers in a row are A: x, B: 40, C: 12, D:y
If 6 cans of soup cost $1.50
9 cans cost .......................x?
x = 9 * 1.50 / 6
x = 2.25
Answer
9 cans cost $2.25