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Oksi-84 [34.3K]
3 years ago
9

Form the union for the following sets. E = {1, 2, 3} F = {101, 102, 103, 104} E ∪ F ={1, 2, 3, 101, 102, 103, 104}

Mathematics
1 answer:
Yanka [14]3 years ago
8 0
<span>E = {1, 2, 3}
F = {101, 102, 103, 104}

answer
E ∪ F ={1, 2, 3, 101, 102, 103, 104}</span>
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Answer:

Probability that the average length of a sheet is between 30.25 and 30.35 inches long is 0.0214 .

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.

Also, a sample of four metal sheets is randomly selected from a batch.

Let X bar = Average length of a sheet

The z score probability distribution for average length is given by;

                Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 30.05 inches

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So, Probability that average length of a sheet is between 30.25 and 30.35 inches long is given by = P(30.25 inches < X bar < 30.35 inches)

P(30.25 inches < X bar < 30.35 inches)  = P(X bar < 30.35) - P(X bar <= 30.25)

P(X bar < 30.35) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{30.35-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z < 3) = 0.99865

 P(X bar <= 30.25) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{30.25-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z <= 2) = 0.97725

Therefore, P(30.25 inches < X bar < 30.35 inches)  = 0.99865 - 0.97725

                                                                                       = 0.0214

                                       

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Step-by-step explanation:

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An 'Arthimetic sequence' is what is presented in this question.

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Answer:

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Step-by-step explanation:

1=-38

2=-25

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