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bekas [8.4K]
2 years ago
5

Please help!!!!!!!!!!

Mathematics
1 answer:
Ostrovityanka [42]2 years ago
8 0

Using the together rate, it is found that it will take 18 hours for Erick to build the garden box himself.

<h3>What is the together rate?</h3>

The together rate is given by the sum of each separate rate.

In this problem, the rates are given as follows:

  • Eric's: 1/x.
  • Drew's: 2/x.
  • Together: 1/6.

Hence:

\frac{1}{x} + \frac{2}{x} = \frac{1}{6}

\frac{3}{x} = \frac{1}{6}

x = 18.

Hence, it will take 18 hours for Erick to build the garden box himself.

More can be learned about the together rate at brainly.com/question/25159431

#SPJ1

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A manufacturer considers his production process to be out of control when defects exceed 3%. In a random sample of 100 items, th
pochemuha

Answer:

The p-value of the test is of 0.2776 > 0.01, which means that the we accept the null hypothesis, that is, the manager's claim that this is only a sample fluctuation and production is not really out of control.

Step-by-step explanation:

A manufacturer considers his production process to be out of control when defects exceed 3%.

At the null hypothesis, we test if the production process is in control, that is, the defective proportion is of 3% or less. So

H_0: p \leq 0.03

At the alternate hypothesis, we test if the production process is out of control, that is, the defective proportion exceeds 3%. So

H_1: p > 0.03

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.03 is tested at the null hypothesis

This means that \mu = 0.03, \sigma = \sqrt{0.03*0.97}

In a random sample of 100 items, the defect rate is 4%.

This means that n = 100, X = 0.04

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.04 - 0.03}{\frac{\sqrt{0.03*0.97}}{\sqrt{100}}}

z = 0.59

P-value of the test

The p-value of the test is the probability of finding a sample proportion above 0.04, which is 1 subtracted by the p-value of z = 0.59.

Looking at the z-table, z = 0.59 has a p-value of 0.7224

1 - 0.7224 = 0.2776

The p-value of the test is of 0.2776 > 0.01, which means that the we accept the null hypothesis, that is, the manager's claim that this is only a sample fluctuation and production is not really out of control.

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\frac{\sqrt{2} }{3} r = 22 ( multiply both sides by 3 to clear the fraction )

\sqrt{2} r = 66 ( divide both sides by \sqrt{2} )

r = \frac{66}{\sqrt{2} } × \frac{\sqrt{2} }{\sqrt{2} } ( rationalising the denominator )

r = \frac{66\sqrt{2} }{2} = 33\sqrt{2}

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Answer:

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