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IrinaK [193]
2 years ago
12

7/8 - 4/5 what is the answer to this thing because i don't know it at all duhhhh

Mathematics
1 answer:
wlad13 [49]2 years ago
4 0
7/8-4/5. When subtracting two fractions, we must have a common denominator. A common denominator is a factor in which goes into both 8 and 5.

1.) Common factor: 45. 45 goes into and 8 and 5. If we list out all the factors of 8 and 5, 45 will be the lowest common factor.

2.) We need to multiply by a number to get 45 for each denominator. We can a one-step equation to figure out what we need to multiply 8 by and 5 by to get 45.
X=the factor of 8 that will get us 45.
8x=45
8x/8=45/8
x=5.625.
Y=factor of 5 that will get us 45.
5y=45
5y/5=45/5
y=9.

3.) We now know that we need to multiply 8 by 5.625 and 5 by 9 in order to get 45 as the denominator for both fractions.
7/8x5.625-4/5x9
7/45-4/45.

4.) Whatever is done to the denominator must be done to the numerator.
7x5.625/45-4x9/45
39.375/45-36/45

5.) We now have common denominators, so we simple subtract the numerators ONLY. We DO NOT subtract the denominators.
39.375/45-36/45=3.375/45.

Therefore, 3.375/45 is your answer.
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An article in The Engineer (Redesign for Suspect Wiring," June 1990) reported the results of an investigation into wiring errors
GarryVolchara [31]

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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

n = 20464.9

Rounding up:

A sample of 20465 is required.

8 0
3 years ago
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