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alukav5142 [94]
3 years ago
14

Steve is trying to increase his average pace per mile by running hills. The hill on 1st Avenue rises 3 vertical feet for each ho

rizontal foot. The hill on 16th Avenue rises 1 vertical foot for every 3 horizontal feet.
Which hill will be more difficult for Steve to run up? Explain your reasoning.

Mathematics
2 answers:
marissa [1.9K]3 years ago
7 0
1st Avenue would be more difficult because it’s rise and run is for every one foot forward it is 3 feet up. Meanwhile avenue 16th would start at (3,1) and the rise and run would be for every 3 feet it would go up 1 foot.
Liula [17]3 years ago
7 0

Answer:

The first Avenue will be more difficult  for Steve to run up.

Step-by-step explanation:

Before to get started, please take a look  at the attached sketch below.

Take in account that slope is a number that describes the steepness of a line. It  can be computed   through the ratio of the vertical change to the horizontal change.

According to this, the slope of the first avenue is 3/1 and the slope of the 16th avenue is 1/3. Therefore, the first avenue inclination is greater than 16 th avenue. This means that the first avenue requires  more effort  to be run up.

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Answer:

The large sample n = 190.44≅190

The  large  sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

<u>Step-by-step explanation</u>:

Given  population proportion was estimated to be 0.3

p = 0.3

Given maximum of error E = 0.04

we know that maximum error

M.E = \frac{Z_{\alpha } \sqrt{p(1-p)} }{\sqrt{n} }

The 85% confidence level z_{\alpha } = 1.44

\sqrt{n} = \frac{Z_{\alpha } \sqrt{p(1-p)} }{m.E}

\sqrt{n} = \frac{1.44X\sqrt{0.3(1-0.3} }{0.04}

now calculation , we get

√n=13.80

now squaring on both sides n = 190.44

large sample n = 190.44≅190

<u>Conclusion</u>:-

Hence The  large  sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

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