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WINSTONCH [101]
3 years ago
13

(NO LINKS) A space shuttle travels at 2.6 x 10,000 feet per second. An hour is 3.6 x 1,000 seconds. This expression can be used

to find the number of feet the space shuttle travels in an hour. How many feet does the shuttle travel in an hour?

Mathematics
1 answer:
pogonyaev3 years ago
5 0

Answer:

1) 7.22 ft in a hr & 2) Choice C.  y = 0.50x + 2

Step-by-step explanation:

1) Space shuttle

2.6 x 10,000 = 26,000ft

3.6 x 1,000 = 3,600 seconds = 1 hr

26,000/3,600 = 7.22 ft in a hr

2) Taxi company

y = 0.50x + 2

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The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.895 g and a standard deviation
kvasek [131]

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3.67% probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.

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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

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

In this question, we have that:

\mu = 0.895, \sigma = 0.292, n = 37, s = \frac{0.292}{\sqrt{37}} = 0.048

Find the probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.

This is the pvalue of Z when X = 0.809.

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

By the Central Limit Theorem

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

Z = \frac{0.809 - 0.895}{0.048}

Z = -1.79

Z = -1.79 has a pvalue of 0.0367

3.67% probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.

5 0
3 years ago
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