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larisa [96]
3 years ago
6

All movies until 4:00 pm require a discount ticket, and all movies after 4:00 PM required a regular price ticket. each theater a

t the movie Plex seats 98 people.what is the least number of showings the theater could have shown after 4:00 PM last Saturday? explain how you found your answer.

Mathematics
2 answers:
steposvetlana [31]3 years ago
8 0
Not enough info, its part c of a question that has more to go off of
alexdok [17]3 years ago
3 0

Answer:

13 showings.

Step-by-step explanation:

Using the information about the question given in the comments, we know that there are 1456 tickets sold the weekend in question.  We also know that they sold 6 times as many regular price tickets as they did discount tickets.

Let r represent regular price tickets and d represent discount tickets.  Since there were 6 times more regular price than discount, we get

r = 6d

Together, there were 1456 tickets sold:

r + d = 1456

Substitute our value from the first equation into the second:

6d + d = 1456

Combine like terms:

7d = 1456

Divide both sides by 7:

7d/7 = 1456/7

d = 208

There were 208 discount tickets sold.  This means there were 6(208) = 1248 regular price tickets sold.

The regular price tickets are sold after 4:00.  There are 98 seats in each showing; this means they needed

1248/98 = 12.7 ≈ 13 showings.

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The equation 18 + p = 34 represents the sum of Reese's and Ana's ages, where p represents Reece's age. How old is Reece?
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3 years ago
Please help! I don't understand how to solve this problem
Ilia_Sergeevich [38]

Using the z-distribution, a sample of 142,282 should be taken, which is not practical as it is too large of a sample.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm z\frac{\sigma}{\sqrt{n}}

The margin of error is:

M = z\frac{\sigma}{\sqrt{n}}

In which:

  • \overline{x} is the sample mean.
  • z is the critical value.
  • n is the sample size.
  • \sigma is the standard deviation for the population.

Assuming an uniform distribution, the standard deviation is given by:

S = \sqrt{\frac{(4 - 0)^2}{12}} = 1.1547

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96.

The sample size is found solving for n when the margin of error is of M = 0.006, hence:

M = z\frac{\sigma}{\sqrt{n}}

0.006 = 1.96\frac{1.1547}{\sqrt{n}}

0.006\sqrt{n} = 1.96 \times 1.1547

\sqrt{n} = \frac{1.96 \times 1.1547}{0.006}

(\sqrt{n})^2 = \left(\frac{1.96 \times 1.1547}{0.006}\right)^2

n =  142,282.

A sample of 142,282 should be taken, which is not practical as it is too large of a sample.

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

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Answer:

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Step-by-step explanation:

7 0
3 years ago
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