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telo118 [61]
3 years ago
15

Use rounding to estimate $6.79 - $3.95 to the nearest half dollar ​

Mathematics
1 answer:
Nataliya [291]3 years ago
8 0

Answer:

$3.00

Step-by-step explanation:

$6.79-$3.95= 2.84

because 84 is closer to 100 then to 50, round to .00 instead of .50

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The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation St
AysviL [449]

Answer:

a) 0.0668 = 6.68% probability that a domestic airfare is $550 or more

b) 0.1093 = 10.93% probability than a domestic airfare is $250 or less

c) 0.6313 = 63.13% probability that a domestic airfare is between $300 and $500

d) The cost for the 3% highest domestic airfares are $592 and higher.

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 = 385, \sigma = 110

a. What is the probability that a domestic airfare is $550 or more (to 4 decimals)?

This is 1 subtracted by the pvalue of Z when X = 550. So

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

Z = \frac{550 - 385}{110}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% probability that a domestic airfare is $550 or more

b. What is the probability than a domestic airfare is $250 or less (to 4 decimals)?

This is the pvalue of Z when X = 250. So

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

Z = \frac{250 - 385}{110}

Z = -1.23

Z = -1.23 has a pvalue of 0.1093

0.1093 = 10.93% probability than a domestic airfare is $250 or less

c. What if the probability that a domestic airfare is between $300 and $500 (to 4 decimals)?

This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 300. So

X = 500

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

Z = \frac{500 - 385}{110}

Z = 1.045

Z = 1.045 has a pvalue of 0.8519

X = 300

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

Z = \frac{300 - 385}{110}

Z = -0.77

Z = -0.77 has a pvalue of 0.2206

0.8519 - 0.2206 = 0.6313

0.6313 = 63.13% probability that a domestic airfare is between $300 and $500

d. What is the cost for the 3% highest domestic airfares?

At least the 97th percentile, so at least X when Z has a pvalue of 0.97. So X when Z = 1.88.

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

1.88 = \frac{X - 385}{110}

X - 385 = 1.88*110

X = 592

The cost for the 3% highest domestic airfares are $592 and higher.

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