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alexandr402 [8]

3 years ago

13

A small publishing company is planning to publish a new book. The production costs include one time fixed costs and variable cos

ts. There are two production methods it could use. With one method, the one time fixed costs will total $18,673, and the variable costs will be $18.50 per book. With the other method, the one time fixed costs will total $43,969, and the variable costs will be $10 per book. For how many books produced will the costs from the two methods be the same ?

1 answer:

kirill115 [55]3 years ago

8 0

Answer:

2976 books

Step-by-step explanation:

Let books produced be x

Cost of First Method:

18,673 + 18.5x

Cost of Second Method:

43,969 + 10x

We equate both the cost expressions and find the value of x (the number of books for which costs are EQUAL):

18,673 + 18.5x = 43,969 + 10x

18.5x - 10x = 43,969 - 18,673

8.5x = 25,296

x = 25,296/8.5

x = 2,976

So for 2976 books produced, the two method's cost will be same

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A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece

Alex_Xolod [135]

Answer:

The number of seat when n is odd $S_n=\frac{n^2+2n+1}{4}$

The number of seat when n is even $S_n=\frac{n^2+2n}{4}$

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

$S_n=\frac{n}{2}[2a+(n-1)d]$

$=\frac{n}{2}[a+l]$

a = first term of the series.

d= common difference.

n= number of term

l= last term

$n^{th}$ term of a A.P series is

$T_n=a+(n-1)d$

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=1 and d=2

$n=1+(t-1)2$

⇒(t-1)2=n-1

$\Rightarrow t-1=\frac{n-1}{2}$

$\Rightarrow t = \frac{n-1}{2}+1$

$\Rightarrow t = \frac{n-1+2}{2}$

$\Rightarrow t = \frac{n+1}{2}$

Last term l= n,, the number of term $=\frac{ n+1}2$ , First term = 1

Total number of seat

$S_n=\frac{\frac{n+1}{2}}{2}[1+n}]$

$=\frac{{n+1}}{4}[1+n}]$

$=\frac{(1+n)^2}{4}$

$=\frac{n^2+2n+1}{4}$

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=2 and d=2

$n=2+(t-1)2$

⇒(t-1)2=n-2

$\Rightarrow t-1=\frac{n-2}{2}$

$\Rightarrow t = \frac{n-2}{2}+1$

$\Rightarrow t = \frac{n-2+2}{2}$

$\Rightarrow t = \frac{n}{2}$

Last term l= n, the number of term $=\frac n2$ , First term = 2

Total number of seat

$S_n=\frac{\frac{n}{2}}{2}[2+n}]$

$=\frac{{n}}{4}[2+n}]$

$=\frac{n(2+n)}{4}$

$=\frac{n^2+2n}{4}$

4 0

3 years ago

Tom took a trip of 1,300 miles. He traveled by train at 50 miles an hour and the same number of hours by plane at 275 mph. How m

Yuri [45]

<span>Equation:
train distance + plane distance = 1300 miles

50x + 275x = 1300
x(325) = 1300
x = 4 hours
4x2=8
---
The trip took 8 hrs.</span>

3 0

3 years ago

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Why is decrease an math a integer and an operation

sergeinik [125]

Decrease means minus or '-'
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3 years ago

Plz help me brainliest will be reward if right and 25 points

slega [8]

Answer:

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Step-by-step explanation:

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2 years ago

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How many three-digit positive integers have three different digits?

olganol [36]

648 is the three-digit positive integers have three different digits

According to the statement

we have given that there are three positive digit number are formed with three distinct digits.

And we have to find that the how many words are formed with distinct numbers.

So, to solve this type of problem the Combination formula is best.

Because it provides the all possibilities that from how many ways numbers are formed.

So, from a combination formula

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So, 648 is the three-digit positive integers have three different digits.

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2 years ago