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MArishka [77]
3 years ago
14

Karen runs a print shop that makes posters for large companies. It is a very competitive business. The market price is currently

$1.00 per poster. She has fixed costs of $250.00. Her variable costs are $1,500 for the first thousand posters, $1,200 for the second thousand, and then $800 for each additional thousand posters.What is her AFC per poster if she prints 1,000 posters? What is her AVC and ATC?
Mathematics
1 answer:
kumpel [21]3 years ago
7 0
AFC = FC / Quantity printed 

<span>So given she prints 1,000 posters: AFC = 250.00/1000 = $0.25 </span>
<span>Given she prints 2,000 posters: AFC = 250.00/2000 = $0.125 </span>
<span>Given she prints 10,000 posters: AFC = 250.00/2000 = $0.025 </span>

<span>ATC = TC / Quantity printed </span>

<span>where TC = FC + Variable C * Quantity printed </span>

<span>If she prints 1000: TC = 250 + 2000*1000 = 2,000,250 </span>
<span>ATC = 2,000,250/1000 = 2000.25 </span>

<span>If she prints 2000: TC = 250 + 1600*2000 = 3,200,250 </span>
<span>ATC = 3,200,250/2000 = 1600.125 </span>

<span>If she prints 10000: TC = 250 + 1600*2000 + 1000*8000 ($1000 for each additional poster after 2000) = 11,200,250 </span>
<span>ATC = 11,200,250/10000 = 1120.025</span>
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Answer:

80 tickets

Step-by-step explanation:

Given the profit, y, modeled by the equation, y = x^2 – 40x – 3,200, where x is the number of tickets sold, we are to find the total number of  tickets, x, that need to be sold for the drama club to break  even. To do that we will simply substitute y = 0 into the given the equation and calculate the value of x;

y = x^2 – 40x – 3,200,

0 = x^2 – 40x – 3,200,

x^2 – 40x – 3,200 = 0

x^2 – 80x  + 40x – 3,200 = 0

x(x-80)+40(x-80) = 0

(x+40)(x-80) = 0

x = -40 and x = 80

x cannot be negative

Hence the total number of  tickets, x, that need to be sold for the drama club to break  even is 80 tickets

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3 years ago
Find the 10th term of an arithmetic sequence if the common difference is 32 and the first term is 2
zaharov [31]

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290

Step-by-step explanation:

If the first term is 2 and the common difference is 32, then the formula for a(n) is:

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Need help fast don't understand both A and B show work please
Shkiper50 [21]
Problem A
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2^n - 1 <<<<< Answer

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5 0
3 years ago
While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem
Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = <em><u>population proportion</u></em>.

So, Null Hypothesis, H_0 : p = 0.013      {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, H_A : p > 0.013      {means that this is an unusually high number of faulty modems}

The test statistics that would be used here <u>One-sample z-test</u> for proportions;

                             T.S. =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~  N(0,1)

where, \hat p = sample proportion faulty modems= \frac{10}{367} = 0.027

           n = sample of modems = 367

So, <u><em>the test statistics</em></u>  =  \frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }

                                     =  2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. <u>Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.</u>

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0
3 years ago
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