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2 years ago

13

mark transferred songs from his computer onto his portable music player. He transferred 2 6/7 songs in 1 2/3 minutes. How many s

ongs did he transfer per minute?

1 answer:

Rainbow [258]2 years ago

7 0

(2 6/7) / (1 2/3) = x/1
cross multiply...and turn mixed numbers to improper fractions
5/3x = 20/7
x = 20/7 * 3/5
x = 60/35 = 1 5/7 songs per minute

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Step-by-step explanation:

Let p be the proportion of people who leave one space after a period.

Given: Sample size : n= 525

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Formula to find the confidence interval :

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$0.838\pm (1.645)\sqrt{\dfrac{0.838(1-0.838)}{525}}\\\\=0.838\pm (1.645)\sqrt{0.00025858285}\\\\=0.838\pm (1.645)(0.01608)\\\\= 0.838\pm0.0265\\\\=(0.838-0.0265,\ 0.838+0.0265)\\\\=(0.8115,\ 0.8645)$

Hence, a 90% confidence interval for the proportion of people who leave one space after a period: (0.8115, 0.8645)

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Answer:

Due to the higher z-score, David has the higher standardized score

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.

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Whoever had the higher z-score.

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$Z = \frac{52 - 70}{11}$

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

$Z = \frac{52 - 64}{6}$

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Due to the higher z-score, David has the higher standardized score

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