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)


Answer:
13.0
Step-by-step explanation:
sin(65°) × 14.3 would give u 12.9 then u round it to the 13
Step-by-step explanation:
maybe 11 I'm not a 100% just trying to help because that's how many full squares it has
Y=-3x+2 bc the line is going down so it’s negative and the y intercept is 2
Answer:
f(-4)= -2
Step-by-step explanation:
f(x)=x^2+5x+2 find(-4)
replace x= -4 in function
f(-4) = (-4)^2+ 5*-4+2
= 16 -20 +2
= 18-20
= -2