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Zigmanuir [339]
3 years ago
8

Some one please help me with this 3 part assignment. I will give you as many points as you want. Just please do it correctly. Th

ank you.
David and his dad were going camping. When they went into the garage to get their tent they noticed it was ripped. David’s dad said he could fix the tent. He would buy some new material to cover the one rectangular side that ripped.

A) How much material does he need to cover one rectangular side of the tent?
Justify your answer using equations/formulas, models, and/or words to explain your mathematical reasoning.
Answer:




B) If David’s dad wanted to re-cover the whole tent, how much material would he need? Justify your answer using equations/formulas, models, and/or words to explain your mathematical reasoning.
Answer:




C) If David found a roll of tent material in the garage that was 65 ft2, would he have enough material to cover the entire tent? Explain your reasoning.
Answer:

Mathematics
1 answer:
dsp733 years ago
8 0
The complete question in the attached figure
A) How much material does he need to cover one rectangular side of the tent?
to cover the rectangular side ABCD, calculate the area
Area=CD*AC=9*6=54 ft²
the answer part A is 54 ft²

B) <span>If David’s dad wanted to re-cover the whole tent, how much material would he need? 

the area total=2*[Area ABCD]+2*[Area ACE]

</span>Area ABCD=9*6=54 ft²
Area ACE=(1/2)*7*6=21 ft²
the area total=2*[54]+2*[21]=150 ft²

the answer part B is 150 ft²

C) <span>If David found a roll of tent material in the garage that was 65 ft2, would he have enough material to cover the entire tent? Explain your reasoning.

to cover the entire tent is required 150 </span>ft²
150 ft²> 65 ft²------> David not have enough material to cover the entire tent
<span>
the answer part C is 
</span>David not have enough material to cover the entire tent

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snow_tiger [21]

Answer:

First question: LCL = 522, UCL = 1000.5

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Step-by-step explanation:

First question:

Lower bound:

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Upper bound:

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Second question:

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

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The margin of error is:

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

The project manager believes that p will turn out to be approximately 0.11.

This means that \pi = 0.11

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.

The project manager wants to estimate the proportion to within 0.03

This means that the sample size needed is given by n, and n is found when M = 0.03. So

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

0.03 = 1.96\sqrt{\frac{0.11*0.89}{n}}

0.03\sqrt{n} = 1.96\sqrt{0.11*0.89}

\sqrt{n} = \frac{1.96\sqrt{0.11*0.89}}{0.03}

(\sqrt{n})^2 = (\frac{1.96\sqrt{0.11*0.89}}{0.03})^2

n = 417.9

Rounding up

A sample size no smaller than 418 is needed.

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