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devlian [24]
2 years ago
12

Liliana is making a vase with a circular base. She wants the area of the base to be between 135 cm2 and 155 cm2. Which circle co

uld represent the base of the vase? Use 3. 14 for Pi.
Mathematics
1 answer:
Sergeeva-Olga [200]2 years ago
5 0

The area of the vase lies between the area 135\ cm^2 and 155\ cm^2 will be 153.86\ cm^2.

<h3>What will be the area?</h3>

As we know that the area of the circle is given by the formula

A=\pi r^2

So now from the question, we will find the radius for both the given areas.

Radius for the first area

r=\dfrac{A}{\pi} =\dfrac{135}{\pi} =6.56\ cm

Radius for the second area

r=\dfrac{A}{\pi} =\dfrac{155}{\pi} =7.03\ cm

Now we can see the range of the radius lies

6.56\leq r\leq 7.03

The radius must be r=7 cm lie between two quantities.

Then the area at r= 7cm will be

A=\pi r^2

A=\pi\times 7^2=153.86\ cm^2

Thus the area for the vase will be A=153.86\ cm^2 ar radius r= 7 cm

To know more about the Area of the circle follow

brainly.com/question/24375372

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The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6.
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Answer:

a) There is a 10.75% probability of observing less than 60 hits in an hour.

b) The 99th percentile of the distribution of the number of hits is 95.21 hits.

c) There is a 24% probability of observing between 80 and 90 hits an hour

Step-by-step explanation:

Problems of normally distributed samples can be 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}

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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem, we have that

The web logs of a certain website show that the average number of hits in an hour is 75 with a standard deviation equal to 8.6, so \mu = 75, \sigma = 8.6.

a) What’s the probability of observing less than 60 hits in an hour? Use the normal approximation

This is the pvalue of Z when X = 60. So

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

Z = \frac{60 - 75}{8.6}

Z = -1.74

Z = -1.74 has a pvalue of 0.1075. This means that there is a 10.75% probability of observing less than 60 hits in an hour.

b) What’s the 99th percentile of the distribution of the number of hits?

What is the value of X when Z has a pvalue of 0.99.

Z = 2.35 has a pvalue of 0.99

So

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

2.35 = \frac{X - 75}{8.6}

X - 75 = 20.21

X = 95.21

The 99th percentile of the distribution of the number of hits is 95.21 hits.

c) What’s the probability of observing between 80 and 90 hits an hour?

This is the pvalue of the zscore of X = 90 subtracted by the pvalue of the zscore of X = 80.

For X = 90

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

Z = \frac{90 - 75}{8.6}

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Z = 1.74 has a pvalue of 0.95907

For X = 80

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

Z = \frac{80 - 75}{8.6}

Z = 0.58

Z = 0.58 has a pvalue of 0.71904

So

There is a 0.95907 - 0.71904 = 0.24003 = 24% probability of observing between 80 and 90 hits an hour

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