E^(3x) = 3247
3x = log 3247
x = [log 3247]/3
29 mg = 0.029 g hope this helps
Answer:
Right 3.5 units
Step-by-step explanation:
Well, the coordinates go (x,y) so just plug it in..
(3.5, 0)
Go across (to the right) 3.5 and up 0, that's it.
Hope this helps :)
Given the function. y = 3x + 5
For, x = -3; y = 3(-3) + 5 = -9 + 5 = -4
For x = 1; y = 3(1) + 5 = 3 + 5 = 8
For x = 4; y = 3(4) + 5 = 12 + 5 = 17
Thus the table representing the function is the table with: -3, 1 and 4 as x-values and -4, 8, 17 as y-values.
For x = 0; y = 3(0) + 5 = 0 + 5 = 5
For y = 0; 0 = 3x + 5; 3x = -5 and x = -5/3
Thus the graph of the function is a straight line passing through points (0, 5) and (-5/3, 0).
To illustrate the fuction as a word statement we say that y is five more than three times x
From the given descriptions, the graph does not represent the graph of y = 3x + 5.
Therefore, the one that does not describe the same situation is the graph.
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 