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o-na [289]
3 years ago
6

A movie theater has a seating capacity of 271. The theater charges $5.00 for children, $7.00 for students,

Mathematics
1 answer:
nirvana33 [79]3 years ago
6 0

Answer:

Children attended = 154

Students attended = 40

Adults attended = 77

Step-by-step explanation:

Seating capacity = 271

Charges for each children ticket = $5

Charges for each student ticket = $7

Charges for each adult ticket = $12

Let number of children tickets = 2x

Ticket price for children = \$ 2x \times 5

As per question statement, there were half as many adults as there are children,

so number of adult tickets = x

Ticket price for adults = \$ x \times 12

Now, number of students =

(271-x-2x)\\\Rightarrow (271-3x)

Ticket price for students = \$ (271-3x) \times 7

Total ticket sales = $1974

\Rightarrow 12x+2x\times 5 +(271-3x)\times 7 = 1974\\\Rightarrow 22x-21x +(271 \times 7) = 1974\\\Rightarrow x = 1974 - 1897\\\Rightarrow x = 77

So, number of adults attended = 77

number of children attended = 77 \times 2 = 154

number of students attended = 271 - 77 - 154 = 40

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antiseptic1488 [7]

Answer:

<h2>C.</h2>

Step-by-step explanation:

Look at the pictures.

x - sausage

y - bacon

A. 20 pounds of sausage and 90 pound of bacon

x = 20 → y = 100 > 90

B. 40 pound of sausage and 40 pound of bacon

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C. 60 pound of sausage and 80 pound of bacon

x = 60 → y = 50 < 80

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3 years ago
The mean percentange of a population of people eating out at least once a week is 57℅
Sidana [21]

Answer:

<u>The correct answer is B. between 56.45% and 57.55% </u>

Complete statement and question:

The mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.50%. Assume that a sample size of 40 people was surveyed from the population a significant number of times. In which interval will 68% of the sample means occur?

between 55.89% and 58.11%

between 56.45% and 57.55%

between 56.54% and 57.46%

between 56.07% and 57.93%

Source: brainly.com/question/1068489

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Mean percentage of a population of people eating out at least once a week  = 57%

Standard deviation = 3.5%

Sample size = 40

Confidence level = 68%

2. In which interval will 68% of the sample means occur?

For answering this question, we should find out the standard deviation of the sample, using this formula:

Standard deviation of the sample = Standard deviation of the population/√Sample size

Standard deviation of the sample = 3.5/√40

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57 - 1 * 0.55 = 57 - 0.55 = 56.45

57 + 1  * 0.55 = 57 + 0.55 = 57.55

<u>The correct answer is B. between 56.45% and 57.55% </u>

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