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4 years ago

Please help me it’s due tomorrow and I need all work to

Mathematics

2 answers:

allochka39001 [22]4 years ago

3 0

1. 10/48 = 5/24
2. 168/360= 28/60= 7/15
3. 70/15= 14/3
4. 9/4
5. 22/5 x 10/3= 220/15= 44/3
6. 8/3 x 29/4 = 232/12= 116/6= 58/3
Answers are in bold.

jeka944 years ago

3 0

For number 1 is 2/4 what is equal 1/2 so
1/2 × 5/12 = 5/24
2. 12/20×14/18=7/15
3. 10/15×7= 14/3 or 4 2/3
4. 3/4×3= 9/4 or 2 1/4
5. 4 2/5×3 1/3= 8/5 or 1 3/5
6. 2 2/3 ×7 1/4= 7/3 or 2 1/3

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nexus9112 [7]

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b) We need a sample of at least 282 young men.

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

This Zscore is how many standard deviations the value of the measure X 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.

(a) What standard deviation must x have so that 99.7% of allsamples give an x within one-half inch of μ?

To solve this problem, we use the 68-95-99.7 rule. This rule states that:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviations of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we want 99.7% of all samples give X within one-half inch of $\mu$ . So $X - \mu = 0.5$ must have $Z = 3$ and $X - \mu = -0.5$ must have $Z = -3$ .

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

$3 = \frac{0.5}{\sigma}$

$3\sigma = 0.5$

$\sigma = \frac{0.5}{3}$

$\sigma = 0.167$

(b) How large an SRS do you need to reduce the standard deviationof x to the value you found in part (a)?

You know from government data that heights of young men are approximately Normal with standard deviation about 2.8 inches. This means that $\sigma = 2.8$