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Thepotemich [5.8K]
3 years ago
8

Car X and car Y traveled the same 80-mile route. If car X took 2 hours and car Y traveled at an average speed that was 50 percen

t faster than the average speed of car X, how many hours did it take car Y to travel the route?
Mathematics
1 answer:
KiRa [710]3 years ago
7 0

Answer:

Step-by-step explanation:

Both cars, X and Y travelled 80 miles each.

Speed = distance/time

If car X took 2 hours to travel 80 miles, it means that the speed at which car X travelled is

80/2 = 40 mph

If car Y traveled at an average speed that was 50 percent faster than the average speed of car X, it means that the speed at which car X travelled is

40 + 0.5 × 40 = 60 mph

Time = distance/speed

Therefore, time taken by car Y to travel 80 miles is

80/60 = 1.33 hours

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Answer:

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Answer:

a) 0.5.

b) 0.8413

c) 0.8413

d) 0.6826

e) 0.9332

f) 1

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 = 6, \sigma = 0.2

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

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

Z = \frac{6-6}{0.2}

Z = 0

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1 - 0.5 = 0.5.

(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

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

Z = \frac{6.2-6}{0.2}

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Z = 1 has a pvalue of 0.8413

(c) P(x ≤ 6.2) =

In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X  5.8.

X = 6.2

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

Z = \frac{6.2-6}{0.2}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 5.8

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

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Z = -1

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This is 1 subtracted by the pvalue of Z when X = 5.7.

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

Z = \frac{5.8-6}{0.2}

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This is 1 subtracted by the pvalue of Z when X = 5.

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

Z = \frac{5-6}{0.2}

Z = -5

Z = -5 has a pvalue of 0.

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