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Arturiano [62]
2 years ago
6

olivia is making scarves. each scarf will have 5 rectangles, and 2/5 of the rectangles will be purple. how many purple rectangle

s does she need for 3 scarves?
Mathematics
1 answer:
Ipatiy [6.2K]2 years ago
5 0
We have to find how many purple rectangles does Olivia need for 3 scarves. We know that each scarf will have 5 rectangles and 2/5 of the rectangles will be purple: 2/5 * 5 = 2. It means that each scarf will have 2 purple rectangles. And for 3 scarves : 3 * 2 = 6. Answer: She needs <span>6 purple rectangles.</span>
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Answer:

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Step-by-step explanation:

To find the number of black ribbons one would have subtract the amount of red ribbons from the total number of ribbons. Then one would have to express it as a proportion. Part over whole equals percent over 100. The number of black ribbons over total ribbons equals "b" over 100.

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Step-by-step explanation:

7y +7x +5 = > answer

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There are two college entrance exams that are often taken by students, Exam A and Exam B. The composite score on Exam A is appro
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Answer:

B.The score on Exam A is better, because the percentile for the Exam A score is higher.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Two exams. The exam that you did score better is the one in which you had a higher zscore.

The composite score on Exam A is approximately normally distributed with mean 20.1 and standard deviation 5.1.

This means that \mu = 20.1, \sigma = 5.1.

You scored 24 on Exam A. So

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

Z = \frac{24 - 20.1}{5.1}

Z = 0.76

The composite score on Exam B is approximately normally distributed with mean 1031 and standard deviation 215.

This means that \mu = 1031, \sigma = 215.

You scored 1167 on Exam B, s:

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Z = 0.632

You had a better Z-score on exam A, so you did better on that exam.

The correct answer is:

B.The score on Exam A is better, because the percentile for the Exam A score is higher.

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