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mars1129 [50]
3 years ago
9

A small publishing company is planning to publish a new book. The production costs will include one-time fixed costs (such as ed

iting) and variable costs (such as printing). The one-time fixed costs will total 49982. The variable costs will be $8.50 per book. The publisher will sell the finished product to bookstores at a price of 25.25 per book. How many books must the publisher produce and sell so that the production costs will equal the money from sales
Business
1 answer:
den301095 [7]3 years ago
7 0

Answer:

Break-even point in units= 2,984 units

Explanation:

Giving the following information:

The one-time fixed costs will total 49982. The variable costs will be $8.50 per book. The publisher will sell the finished product to bookstores for 25.25 per book

<u>To calculate the break-even point in units, we need to use the following formula:</u>

Break-even point in units= fixed costs/ contribution margin per unit

Break-even point in units= 49,982/ (25.25 - 8.5)

Break-even point in units= 2,984 units

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Car X can come with any of these 5 additional features: sunroof, stereo, tinted windows, leather seats and cruise control.
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This question is incomplete! here is the complete question with answer!

Car X can come with any of these 5 additional features: sunroof, stereo, tinted windows, leather seats and cruise control.

Quantity A : Number of different combinations possible

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A The quantity in Column A is greater

B The quantity in Column B is greater

C The two quantities are equal.

D The relationship cannot be determined from the information given.

Answer:

A The quantity in Column A is greater

Explanation:

Lets first analyze the number of different combinations possible!

To find out the combinations we use the following formula:

The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter.

nCr=\frac{n!}{r!(n-r)!}

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Car X has 2 options for each feature: Included or Not included

If car X has no feature at all then:

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If car X has 1 feature included then:

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If car X has 2 features included then:

5C2=\frac{5!}{2!(5-2)!}=\frac{5!}{2!*3!}=10

If car X has 3 features included then:

5C3=\frac{5!}{3!(5-3)!}=\frac{5!}{3!*2!}=10

If car X has 4 features included then:

5C4=\frac{5!}{4!(5-4)!}=\frac{5!}{4!*1!}=5

If car X has all of the 5 features included then:

5C5=\frac{5!}{5!(5-5)!}=\frac{5!}{5!*0!}=1

Now sum all the possible combinations: 1+5+10+10+5+1=32

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