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Sonbull [250]
3 years ago
8

What is the reciprocal of 3 and 1/5

Mathematics
2 answers:
professor190 [17]3 years ago
7 0
To find the reciprocal of 3 and 1/5, first you have to put it into a improper fraction.

3=5/5*3=15/3
15/3+1/5=16/3

3 1/5=16/3

To find the reciprocal, you simply flip the top and the bottom.
3/16

Hope this helps!
Your fortune is "Your present plans are going to succeed."
Vaselesa [24]3 years ago
4 0
1/3 and 5 because 3/1 is three and 1/3 is the opposite while 1/5 is 5/1 which is 5.
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An article in The Engineer (Redesign for Suspect Wiring," June 1990) reported the results of an investigation into wiring errors
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Answer:

a) The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b) A sample of 408 is required.

c) A sample of 20465 is required.

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

Of 1600 randomly selected aircraft, eight were found to have wiring errors that could display incorrect information to the flight crew.

This means that n = 1600, \pi = \frac{8}{1600} = 0.005

99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 - 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0005

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 + 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0095

The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b. Suppose we use the information in this example to provide a preliminary estimate of p. How large a sample would be required to produce an estimate of p that we are 99% confident differs from the true value by at most 0.009?

The margin of error is of:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

A sample of n is required, and n is found for M = 0.009. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.009 = 2.575\sqrt{\frac{0.005*0.995}{n}}

0.009\sqrt{n} = 2.575\sqrt{0.005*0.995}

\sqrt{n} = \frac{2.575\sqrt{0.005*0.995}}{0.009}

(\sqrt{n})^2 = (\frac{2.575\sqrt{0.005*0.995}}{0.009})^2

n = 407.3

Rounding up:

A sample of 408 is required.

c. Suppose we did not have a preliminary estimate of p. How large a sample would be required if we wanted to be at least 99% confident that the sample proportion differs from the true proportion by at most 0.009 regardless of the true value of p?

Since we have no estimate, we use \pi = 0.5

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.009 = 2.575\sqrt{\frac{0.5*0.5}{n}}

0.009\sqrt{n} = 2.575*0.5

\sqrt{n} = \frac{2.575*0.5}{0.009}

(\sqrt{n})^2 = (\frac{2.575*0.5}{0.009})^2

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Rounding up:

A sample of 20465 is required.

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