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CaHeK987 [17]
2 years ago
5

The circumference of the Earth model at the Globe Arena is 56.52 centimeters. What is

Mathematics
1 answer:
ankoles [38]2 years ago
7 0

Answer:

3052 cm^3 (Answer C)

Step-by-step explanation:

The formula for the circumference of a circle of radius r is C = 2(pi)(r).  We know that C = 56.52 cm and want to find the model's radius:

                                       C                                                56.52 cm

Solving for r, we get ----------   so that in this case   r = ------------------  or

                                      2(pi)                                                 6.28

r = 9 cm.

The formula for the volume of a sphere of radius r is V = (4/3)(pi)(r^3), which here comes out to:

V = (4/3)(3.14)(9 cm)^3 = 3052 cm^3 (Answer C)

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The number of pieces of popcorn in a large movie theatre popcorn bucket is normally distributed, with a mean of 1610 and a stand
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A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 40
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a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

a. How large was the sample used in this survey?

We have that s = 25, \sigma = 400. We want to find n, so:

s = \frac{\sigma}{\sqrt{n}}

25 = \frac{400}{\sqrt{n}}

25\sqrt{n} = 400

\sqrt{n} = \frac{400}{25}

\sqrt{n} = 16

(\sqrt{n})^2 = 16^2[tex][tex]n = 256

A sample of 256 was used in this survey.

b. What is the probability that the point estimate was within ±15 of the population mean?

15 is the bounds with want, 25 is the standard error. So

Z = 15/25 = 0.6 has a pvalue of 0.7257

Z = -15/25 = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

45.14% probability that the point estimate was within ±15 of the population mean

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3 years ago
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