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
Answer:
g^5h^2
Step-by-step explanation:
12g^5h^4, g^5h^2
This is one way of doing it. Break down every number and every variable into a product of the simplest factors. Then see how many of each factor appear in both monomials.
12g^5h^4 = 2 * 2 * 3 * g * g * g * g * g * h * h * h * h
g^5h^2 = g * g * g * g * g * h * h
So far you see every single prime factor of each monomial.
Now I will mark the ones that are present in both. Those are the common factors.
12g^5h^4 = 2 * 2 * 3 * g * g * g * g * g * h * h * h * h
g^5h^2 = g * g * g * g * g * h * h
The greatest common factor is the product of all the factors that appear in both monomials.
GCF = g * g * g * g * g * h * h = g^5h^2
6 + 3 or you could do 4.5 + 4.5. but the first one is probably what they're looking for.
