You want log √(9/25). Recognizing that √9 = 3 and that √25 = 5, we get
log 3/5, which by rules of logs comes out to log 3 - log 5.
To four decimal places:
log 3 - log 5 = 0.4771 - 0.6990, or -0.2218.
14 would probably be the answer idk
Delete as this is the wrong answer. Made a typo in the answer.
Answer:
x = 
y = 24
Step-by-step explanation:
-3x - 8y = 20 .... (1)
-5x + y = 19 .... (2)
What we can do is take the (2) equation
-5x + y = 19
and solve for y by adding 5x to both sides
y = 24
now that we have y we substitute it in for y in equation (1)
-3x - 8(24) = 20
-3x - 192 = 20
we want to solve for x so we need to get its self.
add 192 to both sides
-3x = 212
now we divide both sides by -3. It's going to be a fraction.
x = 
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 