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Marina CMI [18]
3 years ago
9

The cost c (in dollars) for the care and maintenance of a horse and carriage is c= 15x 2000, where x is the number of rides? how

many rides are needed to break even?
Mathematics
1 answer:
Semmy [17]3 years ago
3 0
133.33
Is the answer for
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Evaluate the expression : common log of square root 9/25.
tekilochka [14]

You want log √(9/25).  Recognizing that √9 = 3 and that √25 = 5, we get

log 3/5, which by rules of logs comes out to log 3 - log 5.

To four decimal places:

log 3 - log 5 = 0.4771 - 0.6990, or -0.2218.

4 0
3 years ago
1/4 Divided by 16 3/4 (Detailed, worked out answer please)
scZoUnD [109]
14 would probably be the answer idk
8 0
3 years ago
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Need help plz its on the screenshot
user100 [1]
The answers are a and c
4 0
3 years ago
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−3x − 8y = 20 −5x + y = 19 solved by substitution
beks73 [17]

Delete as this is the wrong answer. Made a typo in the answer.

Answer:

x = \frac{-212}{3}

y = 24

Step-by-step explanation:

-3x - 8y = 20  .... (1)

-5x + y = 19   .... (2)

What we can do is take the (2) equation

-5x + y = 19

and solve for y by adding 5x to both sides

y = 24

now that we have y we substitute it in for y in equation (1)

-3x - 8(24) = 20

-3x - 192 = 20

we want to solve for x so we need to get its self.

add 192 to both sides

-3x = 212

now we divide both sides by -3. It's going to be a fraction.

x = \frac{-212}{3}

7 0
3 years ago
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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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