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enyata [817]
4 years ago
7

A cyclist must travel 800 km. how many days will the trip take if the cyclist travels 8 h/day at an average speed of 16 km/h?

Mathematics
1 answer:
Mrac [35]4 years ago
7 0
First, you need to find out how much the cyclist will travel a single day. Since the cyclist travels 16 km/h and for 8 h/day, you can easily multiply 16 x 8 = which would be 128 km/day. 

To find how long it will take the cyclist to travel 800km, just divide 800 by the amount they will be going in one day. 800/128= 6.25. That would convert to 6 days and 6 hours. 
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Simplify the following expression.<br> 5 [13 + 10 = (3 + 2)] + 9 x 2
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Step-by-step explanation:

5 [13 + 10 = (3 + 2)] + 9 x 2

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The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 30,393 miles, with a standard
Pie

Answer:

52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

Step-by-step explanation:

To solve this question, we have 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 30393, \sigma = 2876, n = 37, s = \frac{2876}{\sqrt{37}} = 472.81

What is the probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

This probability is the pvalue of Z when X = 30393 + 339 = 30732 subtracted by the pvalue of Z when X = 30393 - 339 = 30054. So

X = 30732

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

By the Central Limit Theorem

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

Z = \frac{30732 - 30393}{472.81}

Z = 0.72

Z = 0.72 has a pvalue of 0.7642.

X = 30054

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

Z = \frac{30054 - 30393}{472.81}

Z = -0.72

Z = -0.72 has a pvalue of 0.2358

0.7642 - 0.2358 = 0.5284

52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

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