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Carolyn makes and sells necklaces. The material to make each necklace costs
$3.20. Last week she made 12 necklaces and wants to make a total profit of $54.

Using the equation 12(x+3.20)=54, Carolyn determined that she will make a profit of $7.70 on each necklace. Which error did Carolyn make?
A. The solution is the amount she should sell each necklace for, and she will make a profit of $4.50 on each necklace.
B. She should have used the equation 12(x+3.20)=54 and found that she will make a profit of $1.30 on each necklace.
C. She should have used the equation x=54-12(3.20) and found that she will make a profit of $12.40 on each necklace.
D. The solution is the amount that it costs to make each necklace, and she should add $3.20 to find that she will make a profit of $10.90 on each necklace.

Respuesta :

wegnerkolmp2741o

Answer:

A. The solution is the amount she should sell each necklace for, and she will make a profit of $4.50 on each necklace.

Step-by-step explanation:

Let x be the cost of each necklace.  She needs to subtract the cost of the materials from each necklace to get the profit.

x-3.20 is the profit on each necklace.

She sells 12 necklaces

12* (x-3.20) = 54

Divide each side by 12

12/12* (x-3.20) = 54/12

x-3.20 = 4.50

Add 3.20 to each side

x-3.20 +3.20 = 4.50 +3.20

x = 7.70

She sells the necklaces for 7.70 and makes a 4.50 profit on each necklace

imlittlestar

Answer:

The correct answer is option A

The solution is the amount she should sell each necklace for, and she will make a profit of $4.50 on each necklace.

Step-by-step explanation:

It is given that,

The material to make each necklace costs  $3.20. Last week she made 12 necklaces and wants to make a total profit of $54.

To find the selling price of necklace

Let x be the profit on each necklace.

The we can write,

12(x - 3.20)=54

x - 3.20 = 54/12 = 4.5

x = 4.5 + 3.20 = 7.7

From this we get,

the amount she should sell each necklace  =  $7.7

And she will make a profit of $4.50 on each necklace.

The correct answer is Option A.

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