Answer: 0.41
Step-by-step explanation: 0.59 is 59%. To find the percentage of chance that it will not rain, you subtract 1 (which is 100%) minus 0.59. You get an answer of 0.41 (or 41% if you prefer.)
If the question meant f(x)=2^(x) -3 then the answer should be 1 over 4 or 1/4
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
9514 1404 393
Answer:
- x ≤ 4
- x > 10
- x ≤ -7
Step-by-step explanation:
We're guessing you want to solve for x in each case. You do this in basically the same way you would solve an equation.
1. 3x +2 ≤ 14
3x ≤ 12 . . . . . subtract 2
x ≤ 4 . . . . . . . divide by 3
__
2. -5 +2x > 15
2x > 20 . . . . . . add 5
x > 10 . . . . . . . . divide by 2
__
3. -2x +4 ≥ 18
4 ≥ 18 +2x . . . . . add 2x
-14 ≥ 2x . . . . . . . subtract 18
-7 ≥ x . . . . . . . . . divide by 2
_____
<em>Additional comment</em>
The statement above that the same methods for solving apply to both equations and inequalities has an exception. The exception is that some operations reverse the order of numbers, so make the inequality symbol reverse. The usual operations we're concerned with are <em>multiplication and division by a negative number</em>: -2 < -1; 2 > 1, for example. There are other such operations, but they tend to be used more rarely for inequalities.
You will note that we avoided division by -2 in the solution of the third inequality by adding 2x to both sides, effectively giving the variable term a positive coefficient. You will notice that also changes its relation to the inequality symbol, just as if we had left the term where it was and reversed the symbol: -2x ≥ 14 ⇔ -14 ≥ 2x ⇔ x ≤ -7 ⇔ -7 ≥ x
