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Likurg_2 [28]
3 years ago
14

3 hrs 20 minutes- 1 hr 48 minutes

Mathematics
1 answer:
Nana76 [90]3 years ago
5 0

You first convert everything to minutes

3hrs= 60x3=180mins

1hr= 60x1=60mins

then you add 20 to 180

200

and you also add 48 to 60

104

200-104=

96

Then you convert to hrs and mins

96-60=36

1hr 36mins

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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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3 years ago
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Answer:

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Step-by-step explanation:

For a recursive sequence of reasonable length, it is convenient to use a suitable calculator for figuring the terms of it. Since each term not only depends on previous terms, but also depends on the term number, it works well to use a spreadsheet for doing the calculations. The formula is easily entered and replicated for as many terms as may be required.

__

The result of executing the given algorithm is shown in the attachment. (We have assumed that g_1 means g[-1], and that g_2 means g[-2]. These are the starting values required to compute g[0] when k=0.

That calculation looks like ...

  g[0] = (0 -1)×g[-1] +g[-2} = (-1)(9) +5 = -4

The attachment shows the last term (for k=8) is 61,940.

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Answer:

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Step-by-step explanation:

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Answer:

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