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Sauron [17]
3 years ago
13

What is the difference between the areas of a circular target region with radius 5 m and a target region with radius 7 m? Round

your answer to the nearest square meter. 36 m2 6 m2 75 m2 12 m2
Mathematics
2 answers:
Talja [164]3 years ago
8 0

Answer:

Hence, the required area is:

                               75 m²

Step-by-step explanation:

The difference of the area of circular target  region with radius 5 m and a target region with radius 7 m is calculated as:

Area of circle of radius 7 m(A)- Area of a circle of radius 5 m(A')

Now, we know that the area of a circle of radius 7 m is given by:

A=\pi (7)^2\\\\A=49\pi\ m^2

Similarly the area of a circle of radius 5 m is given by:

A'=\pi (5)^2\\\\A'=25\pi\ m^2

Hence, the difference in the areas is given by:

=49\pi-25\pi\\\\=24\pi\\\\=75.36\ m^2

which is approximately equal to:

                      75 m²

         Hence, the difference in the area is:

                              75 m²

Oliga [24]3 years ago
6 0
That would be 7²π-5²π=49π-25π=24π≈75
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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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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

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to check work

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