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LekaFEV [45]
3 years ago
12

HELP ME OUT PLEASE I AHVE 5 MORE QUESTIONS

Mathematics
1 answer:
ziro4ka [17]3 years ago
4 0
The answer is
14
87
14
2
87
14

In a row
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2 PLEASE HELP! WILL GIVE BRAINLIEST!
Vladimir79 [104]

Answer:

Area

Step-by-step explanation:

If you are paving the parking lot, you would need every sqaure inch of the parking lot, not only the outside (which is what the perimeter is)

5 0
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Suppose your boss wants you to obtain a sample to estimate a population mean. Based on previous analyses, you estimate that 49 i
VashaNatasha [74]

Answer:

the minimum sample size n = 11.03

Step-by-step explanation:

Given that:

approximate value of the population standard deviation \sigma = 49

level of significance ∝ = 0.01

population mean = 38

the minimum sample size n = ?

The minimum sample size required can be determined by calculating the margin of error which can be re[resented by the equation ;

Margin of error = Z_{ \alpha /2}} \times \dfrac{\sigma}{\sqrt{n}}

38 = \dfrac{2.576 \times 49}{\sqrt{n}}

\sqrt{n} = \dfrac{2.576 \times 49}{38}

\sqrt{n} = \dfrac{126.224}{38}

\sqrt{n} = 3.321684211

n= (3.321684211)^2

n ≅ 11.03

Thus; the minimum sample size n = 11.03

5 0
3 years ago
11 relaxing after work. the 2010 general social survey asked the question: "after an average work day, about how many hours do y
Bogdan [553]
A confidence interval tells us how many percents we are confident about the range of a parameter. In this problem, <span>a 95% confidence interval for the mean number of hours spent relaxing or pursuing activities they enjoy was (1.38, 1.92). That means we're 95% confident that the Americans spend from 1.38 hours to 1.92 hours per day on average relaxing or pursuing activities they enjoy. In other words, 95% of the samples of the same size would have a mean number of hours relaxing or pursuing activities they enjoy between 1.38 to 1.92.</span>
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3 years ago
Using the segment addition postulate, which is true?
gulaghasi [49]
The answer is BC + CD = BD
7 0
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Divide 269.50 by 14 and the answer is the average cost per visit
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