Answer:
Step-by-step explanation:
sports store sells three different packs of tennis balls. Brand A costs $2.50 for two balls, brand B sells three balls for $3.90, and brand C has a sale of four balls for $5.12. Which inequality about the unit price per ball is correct
Brand A:
$2.50 for 2
Brand B :
$3.90 for 3
Brand C:
$5.12 for 4
We calculate the unit price :
Brand A:
$2.50 / 2 = $1.25
Brand B:
$3.90 / 3 = $1.30
Brand C:
$5.12 / 4 = $1.2
So Sahra is 3 times as old as her daugter. That means that the sum of their age is the daughters age times 4.
We can prove that by saying the daughters age is x. Sahra's age must be 3x. Their sum must be x+3x = 4x.
So to get the daughters age we divide the sum by 4, which is 8.
Answer:Graph A
Step-by-step explanation: -3 is your y intercept and your slope is up 2 over 1.
<h3>
Answer: 5</h3>
=================================================
Work Shown:
x^2 - 5x + 1 = 0
x^2 + 1 - 5x = 0
x^2 + 1 = 5x
(x^2 + 1)/x = 5 .... where x is nonzero
(x^2)/x + (1/x) = 5
x + (1/x) = 5
---------------
An alternative method involves solving the original equation using the quadratic formula. After you get the two roots x = p and x = q, you should be able to find that p + 1/p = 5 and also q + 1/q = 5 as well.
In this case,
p = (5 + sqrt(21))/2
q = (5 - sqrt(21))/2
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
