Answers: 7x^2y^2 -3x+6y
I am not sure how to explain it but here you go
Answer:
The answer is B
Step-by-step explanation:
The Original Price is $40
Explanation:
50 1 20
—- = — Which means ——
100 2 40
Answer:

We can find the second moment given by:

And we can calculate the variance with this formula:
![Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246](https://tex.z-dn.net/?f=%20Var%28X%29%20%3DE%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%207.496%20-%282.5%29%5E2%20%3D%201.246)
And the deviation is:

Step-by-step explanation:
For this case we have the following probability distribution given:
X 0 1 2 3 4 5
P(X) 0.031 0.156 0.313 0.313 0.156 0.031
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
We can verify that:

And 
So then we have a probability distribution
We can calculate the expected value with the following formula:

We can find the second moment given by:

And we can calculate the variance with this formula:
![Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246](https://tex.z-dn.net/?f=%20Var%28X%29%20%3DE%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%207.496%20-%282.5%29%5E2%20%3D%201.246)
And the deviation is:

Answer:-1/3
Step-by-step explanation: I really don’t know