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kondaur [170]
2 years ago
14

At a school fair, students were challenged to hit one of the small congruent circles on the large rectangular board with a ball.

Find the probability of hitting any small circle. Type answer as a percent.

Mathematics
1 answer:
shtirl [24]2 years ago
6 0

Answer:

~4.7%

Step-by-step explanation:

Area of the rectangle: 35*52=1820

Area of 1 small circle: A=πr2=π·32≈28.27433

28.27433*3=84.82299

84.82299/1820=0.04660603846

Therefore, about 4.7%

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We would like to create a confidence interval.
Vlada [557]

Answer:

c.A 90% confidence level and a sample size of 300 subjects.

Step-by-step explanation:

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

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

In which

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In this problem

The proportions are the same for all the options, so we are going to write our margins of error as functions of \sqrt{\pi(1-\pi)}

So

a.A 99% confidence level and a sample size of 50 subjects.

n = 50

99% confidence interval

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 margin of error is

M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.3642\sqrt{\pi(1-\pi)}

b.A 90% confidence level and a sample size of 50 subjects.

n = 50

90% confidence interval

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

The margin of error is

M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.2623\sqrt{\pi(1-\pi)}

c.A 90% confidence level and a sample size of 300 subjects.

n = 300

90% confidence interval

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

The margin of error is

M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.0950\sqrt{\pi(1-\pi)}

This produces smallest margin of error.

d.A 99% confidence level and a sample size of 300 subjects.

n = 300

99% confidence interval

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 margin of error is

M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.1487\sqrt{\pi(1-\pi)}

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