Answer:
(0,2)
Step-by-step explanation:
2:4 means one part is 2/(2+4)=1/3 of AB and the other part is 2/3 of AB
Add 1/3 of the distance from -2 to 4. (1/3)(4+2)=2. -2+2=0 The x coordinate is 0
Subtract 1/3 of the distance from 6 to -6, (1/3)6+6)=4 6-4=2 The y coordinate is 2
The point is (0,2)
If you would like to know which expression represents the amount of time Joe exercises, you can calculate this using the following steps:
Keith ... x hours a week
Troy ... three hours less than Keith: x - 3
Joe ... two times more than Troy: 2 * (x - 3)
The correct result would be D) 2 * (x - 3).
A(c - b) = d
ac - ab = d
ac = d + ab
c = (d + ab)/a
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:
I dont understand that sorry
