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Ira Lisetskai [31]
2 years ago
8

A teacher promised a movie day to the class that did better, on average, on their test. The box plot shows the results of the te

st:
box plot labeled 2nd period with min at 77, Q1 at 78, median at 89, Q3 at 92, max at 100. Box plot labeled 4th period with min at 72, Q1 at 82.5, median at 89, Q3 at 96, max at 97.5

Which class should get the reward, and why?
Mathematics
2 answers:
Pavel [41]2 years ago
8 0

Answer:

The 4th period class should get the reward.

Step-by-step explanation:

  • The max score in 2nd period is 100% but the average students got below 95% in 2nd period class.
  • Though the medians are the same, but the first and third quartiles are higher in 4th period than that of 2nd period,
  • So the students did better on average in 4th period class than in the 2nd period class.

So, relatively 4th period class did better in their test and should get the reward


masha68 [24]2 years ago
4 0
I say it is C. Basically I just eliminate the potential answered down. A could not be the one since not all student in 2nd period got 100% and average students got below 95%. B cannot be it since both box plot have the same median. I do not think D is it because of how the answer is told. "The 4th period class should get the reward. Their lowest score is an outlier, and should be thrown out," it sound childish and makeing a joke to put "and should be thrown out." I may be wrong but that is my opinion. The relatively the best and reasonable answer is C.
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Step-by-step explanation: I got a 100% on the quiz

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Answer:

Option A

Option D

Option E

Step-by-step explanation:

we know that

If the height and base of triangle B are proportional to the height and base of triangle A

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Remember that

If two triangles are similar then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

so

\frac{h_A}{h_B} =\frac{b_A}{b_B}

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b_A and b_B are the base of triangle A and triangle B

In his problem we have

h_A=2.5\ cm\\b_A=1.6\ cm

substitute

\frac{2.5}{h_B} =\frac{1.6}{b_B}

Rewrite

\frac{2.5}{1.6} =\frac{h_B}{b_B}

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<u><em>Verify all the options</em></u>

A) we have

h_B=2.75\ cm\\b_B=1.76\ cm

Find the ratio of the height to the base of triangle B and compare the result with the ratio of height to the base of triangle A (the value is 1.5625)

substitute the values in the proportion

\frac{2.75}{1.76}=1.5625

The ratios are the same

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B) we have

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substitute the values in the proportion

\frac{9.25}{9.16}=1.0098

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These values could not be the height and base of triangle B

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h_B=3.2\ cm\\b_B=5\ cm

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substitute the values in the proportion

\frac{3.2}{5}=0.64

The ratios are not the same

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These values could not be the height and base of triangle B

D) we have

h_B=1.25\ cm\\b_B=0.8\ cm

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substitute the values in the proportion

\frac{1.25}{0.8}=1.5625

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E) we have

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For (0, 160)

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(-2/5)(-8/9)(1/3)(2/7)= 32/945

4 0
2 years ago
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