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2 years ago

A flower pot has a circumference of 47 inches. What is the approximate length of the radius of the flower pot? Use 3.14 for .

Mathematics

2 answers:

Rainbow [258]2 years ago

6 0

Answer:

7.48

Step-by-step explanation:

C=2πr

Solving for

r=C

2π=47

2·π≈7.48028

Tema [17]2 years ago

4 0

Answer:

Step-by-step explanation:

Circumference = 47 inches

$2\pi r = 47\\\\2*3.14*r = 47\\\\6.28*r = 47\\\\r = \dfrac{47}{6.28}\\\\$

r = 7.48 in

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3 years ago

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The cost c, for manufacturing x flags is given by c=x? -24x +39. Find the number of flags manufactured if the cost is $120.

IRINA_888 [86]

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Answer:

Step-by-step explanation:

You want to find x when c = 120.

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3 years ago

Find the surface area of the right triangle prism(above) using the net (below)

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Answer:

144

Step-by-step explanation:

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2 years ago

In City A, the temperature rises 7 degrees F from 8AM to 9AM. Then the temperature drops 6 degrees F from 9AM to 10AM. In City B

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3 years ago

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro

Gnom [1K]

Answer:

A sample of 1068 is needed.

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 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 $1 - \frac{\alpha}{2}$ .

The margin of error is:

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

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of $\pi$ , so we use the worst case scenario, which is $\pi = 0.5$

Then

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

$0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.03\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}$

$n = 1067.11$

Rounding up

A sample of 1068 is needed.

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3 years ago