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AlladinOne [14]
3 years ago
6

10 points HELP ASAP PLEASE AND THANK YOU!!

Mathematics
2 answers:
BARSIC [14]3 years ago
8 0

x=3

multiply 2 to both sides to cancel denominator

subtract 8x from both sides and then divide by 2

iogann1982 [59]3 years ago
6 0

\dfrac{6+8x}{2}=5x\\\\\dfrac{6}{2}+\dfrac{8x}{2}=5x\\\\3+4x=5x\qquad|\text{subtract 4x from both sides}\\\\3=x\to x=3

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B. 10.1%

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3 years ago
Assuming that the heights of college women are normally distributed with mean 64 inches and standard deviation 1.5 inches, what
professor190 [17]

Answer:

15.74% of women are between 65.5 inches and 68.5 inches.

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.

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\mu = 64, \sigma = 1.5

What percentage of women are between 65.5 inches and 68.5 inches?

This percentage is the pvalue of Z when X = 68.5 subtracted by the pvalue of Z when X = 65.5.

X = 68.5

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

Z = \frac{68.5 - 64}{1.5}

Z = 3

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Z = \frac{65.5 - 64}{1.5}

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Z = 1 has a pvalue of 0.8413

So 0.9987 - 0.8413 = 0.1574 = 15.74% of women are between 65.5 inches and 68.5 inches.

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3 years ago
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Answer:

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