Answer:
it is Ahmed shoevked more your calculations are wrong
For any distribution, the sum of the probabilities of all possible outcomes must be 1. In this case, we have to have
We're told that 
, and we're given other probabilities, so we have
The expected number of calls would be
![E[X]=\displaystyle\sum_xx\,P(X=x)](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Csum_xx%5C%2CP%28X%3Dx%29)
![E[X]=0\,P(X=0)+1\,P(X=1)+\cdots+4\,P(X=4)](https://tex.z-dn.net/?f=E%5BX%5D%3D0%5C%2CP%28X%3D0%29%2B1%5C%2CP%28X%3D1%29%2B%5Ccdots%2B4%5C%2CP%28X%3D4%29)
![E[X]=1.4](https://tex.z-dn.net/?f=E%5BX%5D%3D1.4)
Answer:
22
Step-by-step explanation:
since q is given as r + 10 we can add 10 to 12 in the missing place and find the value
First let's solve for the rate, "b", by setting up a ratio with the two points given:
16.875/7.5=(ar^3)/(ar)
2.25=r^2
r=1.5
Now we need to solve for the initial value a using either point given...
7.5=a(1.5)^1
7.5=1.5a
a=5
So now we have solved for both variables and have a complete equation:
y=5(1.5)^x
