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frozen [14]
3 years ago
13

This map shows the location of trees in a yard.

Mathematics
2 answers:
satela [25.4K]3 years ago
8 0
The answer here is c. You get this by following the path of 3 to the right and up 8
Luba_88 [7]3 years ago
3 0
It's c (3,8) because you have to run first then jump
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1 of 5<br>What is the nth term rule of the linear sequence below?<br>2,5, 8, 11, 14, ...<br>w<br>Tn​
Alika [10]

Answer: 3n-1

The difference between each of the numbers is 3.  This means that we have 3n.

If we assume that for the first number n=1, then 3*1=3, but the value we have is 2. This means that we must subtract 1 to get from 3 to 2.

6 0
3 years ago
what is this number? 1,010,020,030,040,050,060,070,080,090,010,020,030,040,050,060,070,080,090,010,020,030,040,050,060,070,080,0
Andrei [34K]

Answer:

that isn't a number

Step-by-step explanation:

3 0
2 years ago
True or False: A relation passing the horizontal line test is a function.
rosijanka [135]

Answer:

True.

Step-by-step explanation:

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7 0
3 years ago
Can anyone decompose it into multipliers?​
Alex787 [66]

Answer:

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
For each given p, let ???? have a binomial distribution with parameters p and ????. Suppose that ???? is itself binomially distr
pshichka [43]

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

Var (Z) = Mpq(1-pq)

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)

8 0
3 years ago
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