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attashe74 [19]
2 years ago
11

Find the area of the shaded region. Round your answer to the nearest hundredth.

Mathematics
1 answer:
PilotLPTM [1.2K]2 years ago
4 0

Answer:

4.56

Step-by-step explanation:

i see a sector hosting 2 semi circles.

this sector area will be .25 of area of circle with radius 4ft.

area of sector = .25*π*16= 4π ft^2

area of semi circle of radius 2ft = .5π*4= 2πft^2

area of 2 semicircles = 4π

the total area of sector = area of 2 semicircles - intersection area + area not covered by semi circles

4π = 4π - intersection area+area not covered by semi circles

intersection area = area not covered by semi circles

now draw a square starting from the intersection point of 2 semi circles at their upper portion. the side will be 2ft

the area of intersection will be covered in the square.

area of intersection = 2*quarter area of circle of radius 2ft - area of square

= 2*(.25π4) - 4

2π-4 = 2.28

area of intersection = area not covered by semicircle

so area of shaded region = 4.56

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<h3>Answer: Choice A) circle</h3>

Explanation:

Imagine that white rectangle as a blade that cuts the cylinder as the diagram shows. If you pull the top cylinder off and examine the bottom of that upper piece, then you'll see a circle forms. It's congruent to the circular face of the original cylinder. This is because the cutting plane is parallel to both base faces of the cylinder. Any sort of tilt will make an ellipse form. Keep in mind that any circle is an ellipse, but not vice versa.

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Time spent using​ e-mail per session is normally​ distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that
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Answer:

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c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 11, \sigma = 3

a. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 10.8 and 11.2 ​minutes?

Here we have that n = 25, s = \frac{3}{\sqrt{25}} = 0.6

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

Z = \frac{11.2 - 11}{0.6}

Z = 0.33

Z = 0.33 has a pvalue of 0.6293.

X = 10.8

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

Z = \frac{10.8 - 11}{0.6}

Z = -0.33

Z = -0.33 has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 10.5 and 11 ​minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

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

Z = \frac{11 - 11}{0.6}

Z = 0

Z = 0 has a pvalue of 0.5.

X = 10.5

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

Z = \frac{10.5 - 11}{0.6}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 ​sessions, what is the probability that the sample mean is between 10.8 and 11.2 ​minutes?

Here we have that n = 100, s = \frac{3}{\sqrt{100}} = 0.3

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

Z = \frac{11.2 - 11}{0.3}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486.

X = 10.8

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

Z = \frac{10.8 - 11}{0.3}

Z = -0.67

Z = -0.67 has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

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Answer:

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