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3 years ago

one section of an auditorium has 12 rows of seats each row has 13 seats what is the total number of seats in that section

Mathematics

2 answers:

Svet_ta [14]3 years ago

8 0

12 x 13 =156
There is 156 seats in that section

max2010maxim [7]3 years ago

5 0

The answer is 156 hope this helpes

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Gabby says that 8 is a common multiple of 16 and 24. Mark says that 8 is a common factor of 16 and 24. Who is correct?

SVEN [57.7K]

Answer:

gabby is correct

Step-by-step explanation:

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2 years ago

2+6i/2-4i = a + bi What is the value of a? What is the value of b?

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2 years ago

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fgiga [73]

Lets p = salesman's commission as percent
15,200 * p = 912
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answer
6%

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3 years ago

A researcher wants to estimate the percentage of all adults that have used the Internet to seek pre-purchase information in the

Lubov Fominskaja [6]

Answer:

The required sample size for the new study is 801.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

25% of all adults had used the Internet for such a purpose

This means that $\pi = 0.25$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

What is the required sample size for the new study?

This is n for which M = 0.03. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.96\sqrt{\frac{0.25*0.75)}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.25*0.75}$

$\sqrt{n} = \frac{1.96\sqrt{0.25*0.75}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.25*0.75}}{0.03})^2$

$n = 800.3$

Rounding up:

The required sample size for the new study is 801.

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