One scarf=s
One hat=h
2s+h=13
s+2h=14
h=13-2s
(plug that into the other equation)
s+2(13-2s)=14
s+26-4s=14
26-3s=14
-3s=-12
s=4
One hat=$5
One scarf=$4
5+4=9
Price of one hat AND one scarf is $9
9514 1404 393
Answer:
B. Figure B
Step-by-step explanation:
The figure is difficult to read. We assume the height of the pyramid is 9, and the radius of the cylinder is 1.
The pyramid volume is ...
V = 1/3Bh
V = 1/3(4²)(9) = 48 . . . cubic units
__
The cylinder volume is ...
V = πr²h
V = π(1²)(48) = 48π . . . cubic units
The cylinder has π times as much volume as the pyramid. Figure B is larger.
_____
<em>Additional comment</em>
If the diameter (not the radius) of the cylinder is 1 unit, then its volume is 12π cubic units and the pyramid has more volume.
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 
Answer:

Step-by-step explanation:




Answer:
z = 31
Step-by-step explanation:
A triangle adds up to 180 degrees
Therefore this equation is formed: 61 + 3z + z -5 = 180
Simplify:
4z = 124
z = 31