Answer:
ist the question
Step-by-step explanation:
write question first
The area of a circle is represented by the equation:

We know that the diameter is two times the radius: 
So if we know that the diameter of the circle is 24m, we can divide this by two in order to get the radius:
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So then we can plug this radius in to the equation for the area of a circle:


We are told to use 3.14 as pi, and not pi itself, so let's plug in 3.14 for pi:


Now we know that the area of this circle is 452.16 square meters.
Asimply measure its length. What else could you measure? After all, length is the only feature a segment has. You’ve got your short, your medium, and your long segments.
Step-by-step explanation:
length
Answer:
3(7 + 4)2 − 24 ÷ 6 = 62
Step-by-step explanation:
3(7 + 4)2 − 24 ÷ 6 is the given expression.
Now, by the rule of BODMAS, where B = Bracket, O= of, D = divide,
M = multiplication, A = addition and S = subtraction
we try and solve the following expression in the same order.
Solving the bracket first, we get
3<u>(7 + 4)</u>2 − 24 ÷ 6 = 3(<u>11</u>)2 − 24 ÷ 6 =<u> 66</u> − 24 ÷ 6
Next, we solve divide,
66 − <u>24 ÷ 6</u> = 66 - <u>4</u>
Next, solving the subtraction, 66 - 4 = 62
Hence, 3(7 + 4)2 − 24 ÷ 6 = 62
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 