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Bond [772]
3 years ago
15

Convert 17.42 m to customary units. A.57'-17/8" B. 36-10 1/2" C. 442 1/2" D. 367/8" E. None of these answers is reasonable.

Mathematics
1 answer:
mars1129 [50]3 years ago
5 0

Answer:

Option E - None of these answers is reasonable.

Step-by-step explanation:

To find : Convert 17.42 m to customary units ?

Solution :

The customary units is defined as the measure length and distances in the customary system are inches, feet, yards, and miles.

The options belong to feet and inches.

We have to convert meter into inches, feet.

Meter into feet,

1 \text{ feet} = 0.3048 \text{ meter}

1 \text{ meter} = \frac{1}{0.3048}\text{ feet}

17.42 \text{ meter} = \frac{17.42}{0.3048}\text{ feet}

17.42 \text{ meter} = \frac{174200}{3048}\text{ feet}

17.42 \text{ meter} =57 \frac{464}{3048}\text{ feet}

17.42 \text{ meter} =57 \frac{58}{381}\text{ feet}

Now, Feet into inches

1 \text{ feet} = 12\text{ inches}

\frac{58}{381} \text{ feet} = 12\times \frac{58}{381}\text{ inches}

\frac{58}{381} \text{ feet} =\frac{232}{381}\text{ inches}

i.e.  17.42 \text{ meter} =57\text{ feet }\frac{232}{381}\text{ inches}

or 17.42 \text{ meter} =57'\frac{232}{381}''

None of these answers is reasonable.

Therefore, Option E is correct.

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Answer:

a) 56.91% probability that the customer will have to wait between 5 and 10 minutes.

b) 65.49% probability that the client will have to wait less than 6 minutes of more than 9 minutes.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

\mu = 9.2, \sigma = 2.6

(a) Between 5 and 10 minutes

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 5. So

X = 10

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

Z = \frac{10 - 9.2}{2.6}

Z = 0.31

Z = 0.31 has a pvalue of 0.6217

X = 5

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

Z = \frac{5 - 9.2}{2.6}

Z = -1.62

Z = -1.62 has a pvalue of 0.0526

0.6217 - 0.0526 = 0.5691

56.91% probability that the customer will have to wait between 5 and 10 minutes.

(b) Less than 6 minutes or more than 9 minutes

Less than 6

pvalue of Z when X = 6

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

Z = \frac{6 - 9.2}{2.6}

Z = -1.23

Z = -1.23 has a pvalue of 0.1230

12.30% probability that the client will have to wait less than 6 minutes

More than 9

1 subtracted by the pvalue of Z when X = 9.

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

Z = \frac{9 - 9.2}{2.6}

Z = -0.08

Z = -0.08 has a pvalue of 0.4681

1 - 0.4681 = 0.5319

53.19% probability that the client will have to wait more than 9 minutes

Less than 6 or more than 9

12.30 + 53.19 = 65.49% probability that the client will have to wait less than 6 minutes of more than 9 minutes.

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