Step-by-step explanation:
Cost per pound, c = 25cents
number of cans per pound, n = 100/4 = 25cans
total cans collected, t = 330cans
payment for total cans collected,p
where p = (t/n)×c = (330/25)×25 = 330cents
11 liters of 25% orange juice, 4 liters of 10% orange juice:
1' let x to be 25% orange juice and y be 10% orange juice
2' .25x+.10y=(15).21
3' x+y=15
4' 2.5x+y=31.5 divide all sides by 10
5' -x-y=-15 multiply all sides by -1
6' 1.5x=16.5
7' x=11
8' y=15-11=4
so 11 and 4 are the answers.
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 
Oh how I miss the days when math was as simple as 2+2=4.
Educated guess here: By putting parentheses.