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AfilCa [17]
3 years ago
5

Put scales on the axes that make sense.explain why you chose your scale.

Mathematics
1 answer:
mel-nik [20]3 years ago
7 0

The scale of a graph is the measure of each division on the axis

A graph scale can alter the information displayed

The question appears incomplete. However, the effect of scales on the can be shown in a situation, where the relationship between the variables in non linear, for example in the accelerating motion, the relationship between the displacement and time is not linear, such that the scale affects the shape of the graph as follows;

Scale 1: 2 Unit Time Intervals Scale:

When the time is given in two units intervals, we have;

\begin{array}{|c|cc|} \mathbf{Time } &&\mathbf{Position}\\0&&0\\2&&0\\4&&0\\6&&0\\8&&2\\10&&4\\12&&5\end{array}

Please find attached the graph for the above table

Scale 2: 0.5 Unit Time Interval Scale:

When the time is given in 0.5 units intervals, we have;

\begin{array}{|c|cc|} \mathbf{Time } &&\mathbf{Position}\\0&&0\\0.5&&-0.375\\1&&-0.5\\1.5&&-0.375\\2&&0\\2.5&&0.375\\3&&0.5\\3.5&&0.375\\4&&0\\4.5&&-0.375\\5&&-0.5\\5.5&&-0.375\\6&&0\\6.5&&0.5\\7&&1\\7.5&&1.5\\8&&2\\8.5&&2.5\\9&&3\\9.5&&3.5\\10&&4\\10.5&&4.4375\\11&&4.75\\11.5&&4.9375\\12&&5\end{array}

Please find attached the graph of the same motion with half time unit intervals used in the calculation

Therefore, the choice of scale depends on the relationships between the variables

Learn more about the estimating the scales on the axis of a graph here:

brainly.com/question/1988417

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Jennifer's height is 63.7 inches.

Step-by-step explanation:

Let <em>X</em> = heights of adult women in the United States.

The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 65 inches and standard deviation <em>σ</em> = 2.4 inches.

To compute the probability of a normal random variable we first need to convert the raw score to a standardized score or <em>z</em>-score.

The standardized score of a raw score <em>X</em> is:

z=\frac{X-\mu}{\sigma}

These standardized scores follows a normal distribution with mean 0 and variance 1.

It is provided that Jennifer is taller than 70% of the population of U.S. women.

Let Jennifer's height be denoted by <em>x</em>.

Then according to the information given:

    P (X > x) = 0.70

1 - P (X < x) = 0.70

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⇒  P (Z < z) = 0.30

The <em>z</em>-score related to the probability above is:

<em>z</em> = -0.5244

*Use a <em>z</em>-table.

Compute the value of <em>x</em> as follows:

          z=\frac{X-\mu}{\sigma}

-0.5244=\frac{x-65}{2.4}

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Thus, Jennifer's height is 63.7 inches.

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