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bazaltina [42]
3 years ago
10

For each given p, let Z have a binomial distribution with parameters p and N. Suppose that N is itself binomially distributed wi

th parameters q and M. For- mulate Z as a random sum and show that Z has a binomial distribution with parameters pq and M.
Mathematics
1 answer:
8090 [49]3 years ago
7 0

Explanation:

Lets interpret Z with M trials. First we have M trials, each trial can be a success or not. The number of success is called N. Each trial that is a success becomes a trial, and if it is a success it becomes a success for Z. Thus, in order for a trial to be successful, it needs first to be successful for the random variable N (and it is with probability q), and given that, it should be a success among the N trials of the original definition of Z (with probability p).

This gives us that each trial has probability pq of being successful. Note that this probability is pq independently of the results of the other trials, because the results of the trials of both N and the original definition of Z are independent. This shows us that Z is the total amount of success within M independent trials of an experiment with pq probability of success in each one. Therefore, Z has Binomial distribution with parameters pq and M.

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Is -128/5 a terminating or repeating decimal??
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terminating

Step-by-step explanation:

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3 years ago
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Solve: 3n - 6 = -8(6 +5n)<br> and explain how to solve it
hjlf

Answer:

Step-by-step explanation:

First use distributive property: a(b + c) = a*b + a*c

3n - 6 = -8(6 + 5n)

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Now add 40n to both sides

3n - 6 + 40n = -48 - 40n + 40n

43n - 6 = -48

Now add 6 to both sides

43n - 6 + 6 = -48 + 6

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4 0
3 years ago
Read 2 more answers
Question about a rational expression
sesenic [268]
-9=6/v

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If you multiply by 6, 6/v x 6 = 36/v, that’s why it don’t work.

Answer:
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v=-6/9
v=-2/3
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2 years ago
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3 years ago
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