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3 years ago

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An engineer makes a model of a bridge using a scale of 1 inch = 3 yards. The length of the actual bridge is 50 yards. What is th

e length of the model?

2 answers:

Scrat [10]3 years ago

8 0

Answer:

$16\frac{2}{3}\approx 16.67$ inches.

Step-by-step explanation:

We have been given that an engineer makes a model of a bridge using a scale of 1 inch = 3 yards. The length of the actual bridge is 50 yards.

To find the length of the model, we will use proportions as:

$\frac{\text{Length of model}}{\text{Actual length}}=\frac{1\text{ inch}}{\text{3 yards}}$

$\frac{\text{Length of model}}{50\text{ yards}}=\frac{1\text{ inch}}{\text{3 yards}}$

$\frac{\text{Length of model}}{50\text{ yards}}*50\text{ yards}=\frac{1\text{ inch}}{\text{3 yards}}*50\text{ yards}$

$\text{Length of model}=\frac{50}{3}\text{ inch}$

$\text{Length of model}=16\frac{2}{3}\text{ inch}$

Therefore, the length of model is $16\frac{2}{3}\approx 16.67$ inches.

AlladinOne [14]3 years ago

7 0

<span>16.6666666667

To solve the problem, do 50 divided by 3. This will give you how many inches the model is.

</span>

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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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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.

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

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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