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3 years ago

At one baseball stadium, there are 4 kinds of seats .what fraction of the total seats are premium

1 answer:

7 0

Assuming that each kind of seat is unnique and different from each other
( assume 4 kinds, premium, gold, silver, bronze, or premium, not premium, not premium, not premium)

so fraciton is parts out of whole so

1 out of 4 toal parts is 1/4

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A number greater than -5 written as a problem

bazaltina [42]

The answer is x > -5

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3 years ago

Given two points (2, 7) and (1, -2) on a line, calculate the slope of the line.

Kipish [7]

Answer:

Step-by-step explanation:

Slope of line passing through (2, 7) and (1, -2) = (-2-7)/(1-2) = 9

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2 years ago

Help omg if so thank you.

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Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

How to calculate standard deviation given mean and sample size?

ad-work [718]

In addition to mean and sample size you will need the individual scores.

The formula for standard deviation is:

S^2 = E(X-M)^2/N-1

Here's an example:
Data set: 4,4,3,1
Mean: 3
Sample size: 4

First, put the individual scores one after the other and subtract the mean from it.
4 - 3 = 1
4 - 3 = 1
3 - 3 = 0
1 - 3 = -2

Second, square the answers you got from step 1.
1^2 = 1
1^2 = 1
0^2 = 0
-2^2 = 4
Third, plug the values from step 2 into the formula.

S^2 = (1+1+0+4)/(4-1) = 6/3 = 2

Standard deviation = 2

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3 years ago

An article in The Engineer (Redesign for Suspect Wiring," June 1990) reported the results of an investigation into wiring errors

GarryVolchara [31]

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 $\pi = 0.5$