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vesna_86 [32]
2 years ago
14

Can someone help please

Mathematics
1 answer:
myrzilka [38]2 years ago
3 0
<h3>Answer:   15x^(7/3) - 8x^(7/4) + x + 9000</h3>

=========================================================

Explanation:

If you know the cost function C(x), to find the marginal cost, we apply the derivative.

Marginal cost = derivative of cost function

Marginal cost = C ' (x)

Since we're given the marginal cost, we'll apply the antiderivative (aka integral) to figure out what C(x) is. This reverses the process described above.

\text{Cost} = \text{antiderivative of marginal cost}\\\\\displaystyle C(x) = \int \left(35x^{4/3} - 14x^{3/4} + 1\right)dx\\\\

C(x) = \frac{1}{1+4/3}*35x^{4/3+1} - \frac{1}{1+3/4}*14x^{3/4+1} + x + D\\\\C(x) = \frac{1}{7/3}*35x^{7/3} - \frac{1}{7/4}*14x^{7/4} + x + D\\\\C(x) = \frac{3}{7}*35x^{7/3} - \frac{4}{7}*14x^{7/4} + x + D\\\\C(x) = 15x^{7/3} - 8x^{7/4} + x + D\\\\

D represents a fixed constant. I would have used C as the constant of integration, but it's already taken by the cost function C(x).

To determine the value of D, we plug in x = 0 and C(x) = 9000. This is because we're told the fixed costs are $9000. This means that when x = 0 units are made, you still have $9000 in costs to pay. This is the initial value. You'll find that all of this leads to D = 9000 because everything else zeros out.

Therefore, we go from this

C(x) = 15x^{7/3} - 8x^{7/4} + x + D\\\\

to this

C(x) = 15x^{7/3} - 8x^{7/4} + x + 9000\\\\

which is the final answer.

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Answer:

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c. 252

Step-by-step explanation:

First take into account the concept of combination and permutation:

In the permutation the order is important and it is signed as follows:

P (n, r) = n! / (n - r)!

In the combination the order is NOT important and is signed as follows:

C (n, r) = n! / r! (n - r)!

Now, to start with part a, which corresponds to a combination because the order here is not important. Thus

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C (10, 2) = 10! / 2! * (10-2)! = 10! / (2! * 8!) = 45

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Part b, would also be a combination, defined as follows

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C (10, 0) = 10! / 0! * (10-0)! = 10! / (0! * 10!) = 1

C (10, 1) = 10! / 1! * (10-1)! = 10! / (1! * 9!) = 10

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C (10, 3) = 10! / 3! * (10-3)! = 10! / (2! * 7!) = 120

The sum of all these scenarios would give us the number of possible total scenarios:

1 + 10 + 45 + 120 = 176 possible total scenarios.

part c, also corresponds to a combination, and to be equal it must be divided by two since the coin is thrown 10 times, it would be 10/2 = 5, that is our r = 5

Knowing this, the combination formula is applied:

C (10, 5) = 10! / 5! * (10-5)! = 10! / (2! * 5!) = 252

252 possible scenarios to be the same amount of heads and tails.

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