7.
(2b^2+7b^2+b)+(2b^2-4b-12)
(9b^2+b)+(2b^2-4b-12)
9b^2+b+2b^2-4b-12
11b^2+b-4b-12
11b^2-3b-12
8.
(7g^3+4g-1)+(2g^2-6g+2)
7g^3+4g-1+2g^2-6g+2
7g^3-2g-1+2g^2+2
7g^3-2g+1+2g^2
7g^3+2g^2-2g+1
Hope this helps!
Answer:-0.0625
Step-by-step explanation:
Answer:
Area =
square units
Step-by-step explanation:
<u>Key skills needed: Area of a parallelogram</u>
1) This shape is a parallelogram, and the way to find the area of a parallelogram is:

A = area
b = base (the side at the bottom
h = the height ( the length that runs from one side to another and makes a right angle)
2) The base (b) would be 3x. The height (h) would also be 3x
3) This means A =
3 times 3 is 9 --> and x times x is 
Therefore the area is --> 
<em>Hope you understood and have a nice day!! :D</em>
Answer:
Fermented cucumbers
Step-by-step explanation:
A pickled cucumber is a cucumber that has been pickled in a brine, vinegar, or other solution and left to ferment for a period of time, by either immersing the cucumbers in an acidic solution or through souring by lacto-fermentation. Pickled cucumbers are often part of mixed pickles.
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 