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Sedbober [7]
3 years ago
7

A distance of 150 km was covered by a motorcyclist traveling at an average speed of 75 km/h, by a bus at 60 km/h, a truck at 50

km/h, and a bicyclist at 20 km/h. How much time did each require to travel the entire distance? Explain why the speed and the time needed to travel 150 km are inversely proportional quantities. Find the time in hours needed to travel for motorcyclist,bus,truck and bicylclist
Mathematics
1 answer:
Maurinko [17]3 years ago
7 0
Motorcyclist: 2 hours
Bus: 2.5 hours
Truck: 3 hours
Bicyclist: 7.5 hours
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The scores on the GMAT entrance exam at an MBA program in the Central Valley of California are normally distributed with a mean
Kaylis [27]

Answer:

58.32% probability that a randomly selected application will report a GMAT score of less than 600

93.51%  probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 591, \sigma = 42

What is the probability that a randomly selected application will report a GMAT score of less than 600?

This is the pvalue of Z when X = 600. So

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

Z = \frac{600 - 591}{42}

Z = 0.21

Z = 0.21 has a pvalue of 0.5832

58.32% probability that a randomly selected application will report a GMAT score of less than 600

What is the probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600?

Now we have n = 50, s = \frac{42}{\sqrt{50}} = 5.94

This is the pvalue of Z when X = 600. So

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

Z = \frac{600 - 591}{5.94}

Z = 1.515

Z = 1.515 has a pvalue of 0.9351

93.51%  probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

What is the probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600?

Now we have n = 50, s = \frac{42}{\sqrt{100}} = 4.2

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

Z = \frac{600 - 591}{4.2}

Z = 2.14

Z = 2.14 has a pvalue of 0.9838

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

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Svet_ta [14]

Answer:

x ≤ 2

Step-by-step explanation:

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$17.80 in 1 hr

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The main cost will be 142.69.


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