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Len [333]
3 years ago
7

The graph shows the height of a hiker above sea level.

Mathematics
2 answers:
Afina-wow [57]3 years ago
5 0

My answer for this question is B


The answer is B

dimaraw [331]3 years ago
3 0

The <u>correct answer</u> is:


B) The variables are height and time. For the first part of the graph, the height is increasing slowly, which means the hiker is walking up a gentle slope. Flat parts of the graph show where the elevation does not change, which means the trail is flat here. The steep part at the end of the graph shows that the hiker is descending a steep incline.


Explanation:


The variables are marked on the graph. Time is marked along the x-axis, which means it is the independent variable. Height is marked along the y-axis, which means it is the dependent variable.


The first part of the graph rises slowly. This means the elevation does not change much over the time; this would be consistent with a gentle slope being climbed.


The flat areas are where the elevation does not change. This would be consistent with the hiker resting.


The steep decrease at the end shows that the elevation goes down quickly. This is consistent with the hiker climbing down a steep slope.

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Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

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So, 98% confidence interval for the average age, \mu is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.3263) = 0.98

P( -2.3263 \times {\frac{\sigma}{\sqrt{n} } < {\bar X - \mu} < 2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

P( \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } < \mu < \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

98% confidence interval for \mu = [ \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } , \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ]

                                                  = [ 25 - 2.3263 \times {\frac{1.8}{\sqrt{36} } , 25 + 2.3263 \times {\frac{1.8}{\sqrt{36} } ]

                                                  = [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0
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