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frez [133]
2 years ago
6

Together, Emily and Charlotte have a total of 60 strawberries. Emily sats 4 times the amount that Charlotte eats. How many straw

berries does Charlotte eat?
Answer choices:

A. 6

B. 8

C. 10

D. 12

E. 14

F. 15
Mathematics
1 answer:
ehidna [41]2 years ago
7 0

Answer:

12

Step-by-step explanation:

Since Emily eats four times the amount that Charlotte eats, then Emily will eat: = (4 × x) = 4x. Therefore, Charlotte eats 12 strawberries

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Arada [10]

Answer:

Ron's ERA has a z-score of -2.03.

Karla's ERA has a z-score of -1.86.

Due to the lower z-score, Ron had a better yean than Karla relative to their peers.

Step-by-step explanation:

Z-score:

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 question:

Since the lower the ERA, the better the pitcher, whoever's ERA has the lower z-score had the better year relative to their peers.

Ron

ERA of 3.06, so X = 3.06

For the males, the mean ERA was 5.086 and the standard deviation was 0.998. This means that \mu = 5.086, \sigma = 0.998

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

Z = \frac{3.06 - 5.086}{0.998}

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Ron's ERA has a z-score of -2.03.

Karla

ERA of 3.28, so X = 3.28

For  the females, the mean ERA was 4.316 and the standard deviation was 0.558. This means that \mu = 4.316, \sigma = 0.558

So

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

Z = \frac{3.28 - 4.316}{0.558}

Z = -1.86

Karla's ERA has a z-score of -1.86.

Which player had the better year relative to their peers, Ron or Karla?

Due to the lower z-score, Ron had a better yean than Karla relative to their peers.

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We substitute our given values to calculate the sample size:

n\geq (\frac{z\sigma}{ME})^2\\\\\geq (\frac{1.96\times 12}{5})^2\\\\\geq 22.13\approx23

Hence, the smallest desired sample size is 23

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