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Alex
2 years ago
4

Make a frequency distribution and find the relative frequencies for the following number set. Round the relative frequency to th

e nearest tenth of a percent. Some of the answers will be used more than once and some may not be used. 10 30 40 50 60 70 10 30 40 60 60 80 20 30 50 60 70 90 20 30 50 60 70 90 Number Frequency Relative Frequency 10 ______ ______% 20 ______ ______% 30 ______ ______% 40 ______ ______% 50 ______ ______% 60 ______ ______% 70 ______ ______% 80 ______ ______% 90 ______ ______%
Mathematics
2 answers:
Lelechka [254]2 years ago
8 0
10 30 40 50 60 70 10 30 40 60 60 80 20 30 50 60 70 90 20 30 50 60 70 90 
There are 24 items.
Relative Frequency:
10 - (2) - 8.33%
20 - (1) - 4.17%
30 - (4) - 16.67%
40 - (2) - 8.33%
50 - (3) - 12.5%
60 - (5) - 20.83%
70 - (3)<span> - 12.5%</span>
80 - (1) - 4.17%
90 - (2) - 8.33%
sladkih [1.3K]2 years ago
3 0

close my man just did the review here are the answers

(10) 2 --8.3%

(20) 4--16.7%

(30) 2--8.3%

(40) 2--8.3%

(50) 3--12.5%

(60) 5--20.8%

(70) 1--4.2%

(80) 3--12.5%

(90) 2--8.3%

Step-by-step explanation:

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A marketing firm wants to estimate how much root beer the average teenager drinks per year. A previous study found a standard de
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Answer:

At least 832 teenargers must be interviewed.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

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Now, find the margin of error M as such

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

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How many teenagers must the firm interview in order to have a margin of error of at most 0.1 liter when constructing a 99% confidence interval

At least n teenargers must be interviewed.

n is found when M = 0.1.

We have that \sigma = 1.12

So

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

0.1 = 2.575*\frac{1.12}{\sqrt{n}}

0.1\sqrt{n} = 2.575*1.12

\sqrt{n} = \frac{2.575*1.12}{0.1}

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Rounding up

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