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Tpy6a [65]
2 years ago
13

Juliet is selling photographs as part of a project for her entrepreneurship class. She sells the first 20 photographs for 15 eac

h for the rest of the project. After her expenses, Juliet earns a profit of 80% of the revenues from her sales. What is the least number of photographs she must sell for the rest of the project to earn a profit of at least $400?A) 18B) 20C) 24D) 32
Mathematics
1 answer:
garik1379 [7]2 years ago
7 0

Answer:

The correct option would be B) 20

Step-by-step explanation:

Consider the complete question is,

"Juliet is selling photographs as part of a project for her entrepreneurship class. She sells the first 20 photographs for 10 each, because the first 20 photographs sold quickly, she raised the price of the photograph to 15 each for the rest of the project. After her expenses, Juliet earns a profit of 80% of the revenues from her sales. What is the least number of photographs she must sell for the rest of the project to earn a profit of at least $400"

Now,  80% of sales = profit

\implies \text{Sales}=\frac{100\times Profit}{80}

If, Profit =  $400

\text{Sales}=\frac{100\times 400}{80}=\$ 500

∵ 20 photographs are sold for 10 each,

Since, revenue in first selling = 20 × 10 = $ 200,

Let she must sold x photographs for the rest of the project,

∵ price of each photograph in second selling = $ 15,

So, revenue in second selling = x × 15 = $ 15x,

Thus, total sales = 200 + 15x

⇒ 200 + 15x = 500

⇒ 15x = 500 -  200

⇒ 15x = 300

⇒ x = 20

Therefore, she must sell 20 photographs for the rest of her project.

i.e. OPTION B) is correct.

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Answer:

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Step-by-step explanation:

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Now we need to find the y-coordinate of the center of mass.

\bar{y}=\frac{1}{M}\int\int y*\rho dydx

\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx

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(\bar{x},\bar{y})=(5,\frac{85}{28})

I hope it helps you!

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