If x=6 it is
9. 5
10. 18
11. 23
12. 5.5
13.68
if x=3 it is
2
15
8
11
41
Here are your points (-9,-8), (-6,-2), (-2,-6)
Answer:
See the proof below.
Step-by-step explanation:
Assuming this complete question: "For each given p, let Z have a binomial distribution with parameters p and N. Suppose that N is itself binomially distributed with parameters q and M. Formulate Z as a random sum and show that Z has a binomial distribution with parameters pq and M."
Solution to the problem
For this case we can assume that we have N independent variables 
with the following distribution:
bernoulli on this case with probability of success p, and all the N variables are independent distributed. We can define the random variable Z like this:
From the info given we know that 
We need to proof that 
by the definition of binomial random variable then we need to show that:
The deduction is based on the definition of independent random variables, we can do this:
And for the variance of Z we can do this:
![Var(Z)_ = E(N) Var(X) + Var (N) [E(X)]^2](https://tex.z-dn.net/?f=%20Var%28Z%29_%20%3D%20E%28N%29%20Var%28X%29%20%2B%20Var%20%28N%29%20%5BE%28X%29%5D%5E2%20)
![Var(Z) =Mpq [p(1-p)] + Mq(1-q) p^2](https://tex.z-dn.net/?f=%20Var%28Z%29%20%3DMpq%20%5Bp%281-p%29%5D%20%2B%20Mq%281-q%29%20p%5E2)
And if we take common factor 
we got:
![Var(Z) =Mpq [(1-p) + (1-q)p]= Mpq[1-p +p-pq]= Mpq[1-pq]](https://tex.z-dn.net/?f=%20Var%28Z%29%20%3DMpq%20%5B%281-p%29%20%2B%20%281-q%29p%5D%3D%20Mpq%5B1-p%20%2Bp-pq%5D%3D%20Mpq%5B1-pq%5D)
And as we can see then we can conclude that
D) (-2,3)
-2 is the x-value
3 is the y-value
If you look at the graph on the x-axis (horizontal line) the numbers are increasing and decreasing by 1.
Since the point of intersection (big blue dot) is on the left side of 0, it falls directly at the -2 mark
As for the y-axis (vertical lines) the numbers also increase and decrease by 1. The point of intersection falls above 0 at +3
Therefore your point of intersection is (-2,3)
We can tell it’s not a permutation because it’s picking two people. Order doesn’t matter.
The number of total possible groups of men is 9 C 2 = 9!/((9-2)!(2!)) = 36
The number of total possible groups of women is 11 C 2 = 11!/((11-2)!(2!)) = 55
55 * 36 = 1980 different groups in the committee
