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:
f(1) = 8
Common ratio: 0.5
Step-by-step explanation:
f(1) means the firs term in a sequence.
In the function f(n), represented by 8, 4, 2, 1, .., the first term is 8.
f(1) = 8
To find the common ratio, divide any term by the term before it.
We can use any two of the given terms in the sequence EXCEPT for 8 because it is the first term and does not have a term before it.
I choose to divide the second term by the first term:
4/8 = 1/2 = 0.5
Firstly attention should pe paid to the exact words given in the problem and then only the answer can be easily deduced.
It is a 2 digit number and that is said in the problem
Now For the tens digit it has been said that it is 8 more than the ones digit. The ones digit cannot be zero and that is already mentioned in the problem.
So if we take the ones digit to be 1, then the tens digit will be= 8 +1
= 9
Now if we take the ones digit to be 2, then the tens digit will be = 2 + 8
= 10
So it cannot be 2 as the tens digit takes the number to a 3 digit number and it is not possible.
So the answer to this problem is 91.
Answer:
no its not
Step-by-step explanation:
from 2 random samples you cant figure it out
