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Setler [38]
2 years ago
12

How will $105 be shared between Jack, Ken and Aaron; if Jack gets double of what Ken gets, and Ken gets double of what Aaron get

s?
Mathematics
1 answer:
igomit [66]2 years ago
8 0

1) Jack : Ken = 2 : 1
2) Ken : Aaron = 2 : 1

Jack : Ken : Aaron
4 : 2 : 1

4 + 2 + 1 =7
$105 / 7 = $15

Jack = 4 x $15 = $60
Ken = 2 x $15 = $30
Aaron = 1 x $15 = $15
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A​ half-century ago, the mean height of women in a particular country in their 20s was 64.7 inches. Assume that the heights of​
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Answer:

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Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of​ today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of​ today's women in their 20's have been taken.

<u><em /></u>

<u><em>Let </em></u>\bar X<u><em> = sample mean heights</em></u>

The z-score probability distribution for sample mean is given by;

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where, \mu = population mean height of women = 64.7 inches

            \sigma = standard deviation = 2.07 inches

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the sample of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches is given by = P(\bar X \geq 65.86 inches)

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<em>The above probability is calculated by looking at the value of x = 2.57 in the z table which has an area of 0.99492.</em>

<em />

Therefore, 99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.

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