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Ne4ueva [31]
3 years ago
12

HEEEEEELLLLPPPPPP IVE FALLEN AND I CANT GET UP HELP ME WITH THIS PROBLEM !! ASAP AND GET BRAINLIEST ​

Mathematics
1 answer:
BartSMP [9]3 years ago
4 0
The answer is 28.3.
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Solve the formula for converting temperature from degrees Celsius to
Harlamova29_29 [7]

Answer:

C. C=9/5F+32

Step-by-step explanation:

divided by 5 then multiply by 9 then add 32

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Compute the exact value of the function for the given x-value without using a calculator. You must show your work for full credi
yKpoI14uk [10]
F(2) = 5(2)
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A car travels 2.83 km in the x-direction, then turns left 65.3 ◦ to the original direction and travels an additional distance of
umka21 [38]

Answer:

  4.24 km

Step-by-step explanation:

The x-component of the displacement after the turn is ...

  d2·cos(θ) = (3.38 km)cos(65.3°) ≈ 1.41239 km

Adding this to the displacement before the turn, we have ...

  x-component of displacement = 2.83 km + 1.41 km = 4.24 km

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3 years ago
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Akimi4 [234]
Idk you kno I am trying to get points right
3 0
3 years ago
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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