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Klio2033 [76]
3 years ago
15

1. Lucy has a jewelry business. To make a certain necklace, she spends $3.38 on materials and $5.57 on labor. She then sells the

necklace for $15.99. Lucy’s profit on the necklace is equal to the sale price minus the cost of materials and labor.
(a) How much profit does Lucy earn when she sells a necklace? Show your work.
(b) Write a fraction of the profit per sale price of the necklace. Record your answer as a fraction using whole numbers.
(c) What decimal of every dollar of the sale price of the necklace is profit using the fraction from Part (b)? Show your work. Round your answer to the nearest hundredth.
(d) Lucy collected $223.86 selling necklaces at a craft fair. How many necklaces did Lucy sell? Show your work.
(e) How much profit did she earn from the number of necklaces she sold? Show your work.
Mathematics
2 answers:
VARVARA [1.3K]3 years ago
7 0
A. Profit is calculated by subtracting the total cost from total revenue.

    TR = $15.99 and TC = $3.38 + $5.57

           P = $15.99 - ($3.38 + $5.57)
   <em>          P = $7.04</em>

b. The fraction of the profit and sale price is,

    F = Profit / sale price = 7.04 / 15.99

Simplifying,

   <em>F = 704/1599</em>

 c. If the fraction in b is expressed in decimal places to the nearest hundredths is<em> 0.44</em>.

d. To determine the number of necklaces sold, divide the total amount collected by the price of each.

     n = $223.86 / $15.99 = 14

Answer:<em> 14</em>

e. The profit is calculated by multiplying the answers from a and d.
   
     TP = 14($7.04)
    <em>  TP = $98.56 </em>
son4ous [18]3 years ago
5 0
<span>(a) $7.04 (b) 704/1599 (c) 0.44 (d) 14 (e) $98.56 (a) How much profit does Lucy earn when she sells a necklace? Since the problem states that her profit is the sale price minus the cost of materials and labor, we have the following equation. P = $15.99 - $3.38 - $5.57 P = $12.61 - $5.57 P = $7.04 So her profit is $7.04 per necklace. (b) Write a fraction of the profit per sale price of the necklace. Record your answer as a fraction using whole numbers. The raw fraction is 7.04/15.99, to get rid of the decimal point, multiply top and bottom by 100, getting 704/1599. The prime factors of 704 are 2,2,2,2,2,2,11 and the prime factors of 1599 are 3,13,41. Since neither 704, nor 1599 share any common factors, the fraction can't be reduced and the final answer is 704/1599. (c) What decimal of every dollar of the sale price of the necklace is profit using the fraction from Part (b)? Pardon the lack of formatting 704/1599 = 0.4403 Rounded to the nearest hundredth gives 0.44 (d) Lucy collected $223.86 selling necklaces at a craft fair. How many necklaces did Lucy sell? Show your work. This question is asking for total cost. So divide the $223.86 by the sale price of $15.99. $223.86 / $15.99 = 14 14 necklaces were sold. (e) How much profit did she earn from the number of necklaces she sold? Show your work. Since we know from (a) that she has $7.04 profit per necklace, just multiply the amount of profit by the number sold. So $7.04 * 14 = $98.56</span>
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Step-by-step explanation:

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With the information above, we can solve the question.

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Step-by-step explanation:

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Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So \mu = 53, \sigma = 11

So:

Pr(53-11 \leq X \leq 53+11) = 0.6827

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Pr(42 \leq X \leq 64) = 0.6827

Pr(31 \leq X \leq 75) = 0.9545

Pr(20 \leq X \leq 86) = 0.9973

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have Pr(31 \leq X \leq 75) = 0.9545 = 0.9545 subtracted by Pr(42 \leq X \leq 64) = 0.6827 is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

P = {0.9545 - 0.6827}{2} = 0.1359

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

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