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AVprozaik [17]
3 years ago
12

Makenna surveyed 48 of her female classmates to determine the number of stores each one visited before purchasing a dress for th

e spring dance. She wants to represent her results in a box plot.
How many values in her data set will be above the third quartile for her box plot?
Physics
2 answers:
Maslowich3 years ago
6 0

48/4 = 12

the answer is 12


Aleonysh [2.5K]3 years ago
5 0

Answer: 12 values

Explanation:

In statistics, a Boxplot is a graph that is a good indicator of how the values in the data are spread out. It is useful when we need to compare distributions between many groups or datasets. It is composed of Outliers, which display low minimums and high maximums. Also, it shows The Interquartile Range by a square composed of Q1 (quartile 1), Q2 (quartile 2) and Median.

Quartiles are values that divide data into quarters, where Q1 represents the lowest 25% of numbers, Q2 represents the next lowest 25% of numbers, Q3 represents the second-highest 25% of numbers above the median and finally, Q4 represents the highest 25% of numbers.

In this case,  her data set is composed of 48 values obtained by the survey Makenna did in her class. Then, numbers above the third quartile for her box plot will be Q4 values, hence, the highest 25% of numbers are:

                                              48 x 25% = 12

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By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.

<h3>What are the limits?</h3>

First, we need to find the limits.

We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.

So if 58 in is the 100%, the 26% and 43% of that are:

  • 26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
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But we know that the CG is found to be 45.5% MAC, then it measures:

(45.5%/100%)*58in = 0.455*58in = 26.39 in

We need to compare it with the largest limit, so we get:

26.39 in - 24.94 in = 1.45 in

This means that the CG is 1.45 inches out of limits.

If you want to learn more about percentages, you can read:

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2 years ago
Easy physics question help.!!!
stellarik [79]

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Explanation:

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2 years ago
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At time=0s, the object is at the 21.0-meter position along the roadway. Where is the object at time = 10 s?
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3 years ago
A particular planet has a moment of inertia of 8.26 × 1036 kg⋅m2 and a mass of 6.54 × 1022 kg. Based on these values, what is th
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Answer: Correct answer is B = 1.776933×10^{7}

Explanation:

Given :

Moment of inertia I = 8.26×10^{36} kgm^{2}

Mass of planet m = 6.54×10^{22} kg

Also, Planet is solid sphere so that, Moment of inertia is I = \frac{2}{5} mR^{2} =0.4mR^{2}

Where R is radius of planet

Putting into calculation

We get,

I = \frac{2}{5} mR^{2}

8.26×10^{36} =  0.4×6.54×10^{22}×R^{2}

8.26×10^{14} = 2.616 R^{2}

3.15749235×10^{14} = R^{2}

R =  1.776933×10^{7}

Thus, Correct answer is B = 1.776933×10^{7}

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Answer: They travel away from the focus of the earthquake in all directions.

Explanation:

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