Answer:
1.29, 1.202, 1.02
Step-by-step explanation:
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 
The X-intercepts: (1,0),(-7,0)
Axis of symmetry: x= - 3
The vertex: (-3, -8)
The Y-intercept: (0, -7/2)
Concave up or down: concave up
Sketch: a poorly drawn picture on the bottom.
I hope that helps.
-3 11 -1 20
4 -12 -4 -20
-3 -1 -5 0
Bring down the -3,
Multiply -3 * 4
Add 11 and -12
Multiply -1 * 4
Add -1 and -4
Multiply -5 * 4
Add 20 and -20
Answer: The power is reduced by one from the
Dividend. -3x^2 – x - 5
I tried to upload a file also but I don't know if it will show up.
Answer:
Below in bold.
Step-by-step explanation:
x^2 - y^2 = 11
2x^2 + y^2 = 97
From the first equation:
y^2 = x^2 - 11
Substituting in the second equation:
2x^2 + x^2 - 11 = 97
3x^2 = 108
x^2 = 36
x = 6, -6.
Substituting for x in the first equation:
(6)^2 - y^2 = 11
y^2 = 36 - 11 = 25
y = 5, -5.