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BartSMP [9]
3 years ago
9

Please help and show work! :)

Mathematics
1 answer:
Sedaia [141]3 years ago
7 0

Step-by-step explanation:

how do you show work cuz I screenshot it but it won't let me write over the thing and put the answer so can I just type it out

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A large sample of tires from cabs driven within a city have an average tire tread depth of 0.25cm at the end of the winter. If t
frutty [35]

Answer:

0% probability cab tires depths would be shallower than 0.25cm.

Step-by-step explanation:

Problems of normally distributed samples can be 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 = 2.2, \sigma = 0.33

What is the probability cab tires depths would be shallower than 0.25cm.

This probability is the pvalue of Z when X = 0.25. So

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

Z = \frac{0.25 - 2.2}{0.33}

Z = -5.9

Z = -5.9 has a pvalue of 0.

0% probability cab tires depths would be shallower than 0.25cm.

7 0
3 years ago
A university surveyed recent graduates of the English department for their starting salaries. Four hundred graduates returned th
NeX [460]

Answer:

Step-by-step explanation:

We want to determine a 95% confidence interval for the mean salary of all graduates from the English department.

Number of sample, n = 400

Mean, u = $25,000

Standard deviation, s = $2,500

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z × standard deviation/√n

It becomes

25000 ± 1.96 × 2500/√400

= 25000 ± 1.96 × 125

= 25000 ± 245

The lower end of the confidence interval is 25000 - 245 =24755

The upper end of the confidence interval is 25000 + 245 = 25245

Therefore, with 95% confidence interval, the mean salary of all graduates from the English department is between $24755 and $25245

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4 years ago
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Answer:

multiply

Step-by-step explanation:88 16 then 12

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3 years ago
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Answer:

the third one...............

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