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andrew-mc [135]
3 years ago
11

After writing part of his novel, Thomas is now writing 16 pages per week. After 4 weeks, he has written 85 pages. Assume the rel

ationship is linear. Find and interpret the rate of change and the initial value.
Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
7 0

The initial value is 21 and rate of change is 16 pages per week

<em><u>Solution:</u></em>

Given that After writing part of his novel, Thomas is now writing 16 pages per week

After 4 weeks, he has written 85 pages.

Given that assume the relationship to be linear

Linear relationships can be expressed either in a graphical format or as a mathematical equation of the form y = mx + c

y = mx + c

where "y" is the number of pages written after 4 weeks

x = 4 weeks and m = 16 pages

Therefore,

85 = 16(4) + c

85 = 64 + c

c = 85 - 64

c = 21

Therefore, initial value is 21 and rate of change is 16 pages per week

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The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a st
Alinara [238K]

Answer:

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

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}

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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so \mu = 0.5, \sigma = 0.05.

What is the probability that a line width is greater than 0.62 micrometer?

That is P(X > 0.62)

So

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

Z = \frac{0.62 - 0.5}{0.05}

Z = 2.4

Z = 2.4 has a pvalue of 0.99180.

This means that P(X \leq 0.62) = 0.99180.

We also have that

P(X \leq 0.62) + P(X > 0.62) = 1

P(X > 0.62) = 1 - 0.99180 = 0.0082

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

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