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dedylja [7]
3 years ago
8

2÷(−2)=?2, divided by, left parenthesis, minus, 2, right parenthesis, equals, question mark

Mathematics
2 answers:
dolphi86 [110]3 years ago
5 0

Answer:

2÷(−2) = -1

Step-by-step explanation:

charle [14.2K]3 years ago
5 0

Answer:

-1 <3

Step-by-step explanation:

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For each given p, let ???? have a binomial distribution with parameters p and ????. Suppose that ???? is itself binomially distr
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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 X_i with the following distribution:

X_i Bin (1,p) = Be(p) 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:

Z = \sum_{i=1}^N X_i

From the info given we know that N \sim Bin (M,q)

We need to proof that Z \sim Bin (M, pq) by the definition of binomial random variable then we need to show that:

E(Z) = Mpq

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The deduction is based on the definition of independent random variables, we can do this:

E(Z) = E(N) E(X) = Mq (p)= Mpq

And for the variance of Z we can do this:

Var(Z)_ = E(N) Var(X) + Var (N) [E(X)]^2

Var(Z) =Mpq [p(1-p)] + Mq(1-q) p^2

And if we take common factor Mpq we got:

Var(Z) =Mpq [(1-p) + (1-q)p]= Mpq[1-p +p-pq]= Mpq[1-pq]

And as we can see then we can conclude that   Z \sim Bin (M, pq)

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I am not sure if my answer is true.
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