Answer:
c 6 d 4 I hope it will help you please follow me
Answer:
A right 4 down 5
Step-by-step explanation:
counting the grid spaces from any point for example R and going to the point R' (R prime) will give you the translation
Answer:
-5
Step-by-step explanation:
2 - 1 = 1
5 - 10 = -5
-5 divided by 1 = -5
Answer:
The answer is: 
Step-by-step explanation:
angles 1 and 5 are equal to each other since 
so 
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 