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Lunna [17]
2 years ago
12

Based on past musical productions, a theater predicts selling 400−8p number of tickets when each ticket is sold at p dollars. Us

ing the table from the last question answer the following questions:
1. For which ticket prices will the theater earn no revenue? Explain how you know.


2. At what ticket prices should the theater sell the tickets if it must earn at least $3,200 in revenue to break even (to not lose money) on the musical production? Explain how you know.
Mathematics
1 answer:
Dvinal [7]2 years ago
7 0

To earn at least $3,200 in revenue, the ticket prices should be $10 or $40.

<h3>What is an Equation</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Revenue = number of tickets sold * price per ticket = (400 - 8p)p

Revenue = 400p - 8p²

For no revenue:

400p - 64p² =  0

p = $0 or p = $50

For revenue of atleast 3200:

400p - 8p² =  3200

p = $10 or p = $40

To earn at least $3,200 in revenue, the ticket prices should be $10 or $40.

Find out more on Equation at: brainly.com/question/1214333

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A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are
Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

P = P_{1} + P_{2} + P_{3}

P_{1} is the probability that all three of them are 13-watt. So:

P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277

P_{2} is the probability that all three of them are 18-watt. So:

P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415

P_{3} is the probability that all three of them are 23-watt. So:

P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173

P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245

There is a 12.45% probability that one bulb of each type is selected.

3 0
3 years ago
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Oksana_A [137]

Answer:

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Step-by-step explanation:

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2 years ago
2+2 idk what is the answer and im just confused of what it is
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<h3>Answer:</h3>

2+2 = 4

<h3>Very Simple!</h3>

▶Add 2 to 2 , you may count it with your fingers.

<h2>Additionally Information:</h2>

  • (-) This sign means subtraction i.e. you must take out!

  • (×) This sign means multiplication i.e.you must multiply !

  • (÷) This sign means division i.e. you must divide!

This are for beginners ✔

<h2>Hope you understand</h2>
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