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
<span>216,000,000
</span><span>2.2 × 10 to the 3rd power = 2.2 * 10^3 = 2,200
</span>3.1 × 10 to the 7th power = 3.1 * 10^7 = 31,000,000
<span>310,000
Increasing order:
</span>2,200 , 310,000 , 216,000,000 , 31,000,000
So increasing order with original numbers:
2.2 × 10 to the 3rd power, 310,000, <span>216,000,000, 3.1 × 10 to the 7th power</span>
Answer:
40
Step-by-step explanation:
-12*F
Temperature in morning=-4*F
Temperature has dropped by=8*F
Temperature now=-4-8= -12*F
Well, you have to find all numbers that go into 24. 2, 3, 4, 6, 8, and 12. So, if I didn't miss any it should be 6 ways to divide.
