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Alona [7]
3 years ago
10

Given that the diameter of Circle A is 6 cm , and the radius of Circle B is 18 cm , what can be concluded about the two circles?

Mathematics
1 answer:
ELEN [110]3 years ago
8 0

Answer:

The radius of circle B is 6 times greater than the radius of circle A

The area of circle B is 36 times greater than the area of circle A

Step-by-step explanation:

we have

<em>Circle A</em>

D=6\ cm

The radius of circle A is

r=6/2=3\ cm -----> the radius is half the diameter

<em>Circle B</em>

r=18\ cm

Compare the radius of both circles

3\ cm< 18\ cm

18=6(3)

The radius of circle B is six times greater than the radius of circle A

Remember that , if two figures are similar, then the ratio of its areas is equal to the scale factor squared

All circles are similar

In this problem the scale factor is 6

so

6^{2}=36

therefore

The area of circle B is 36 times greater than the area of circle A

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An HP laser printer is advertised to print text documents at a speed of 18 ppm (pages per minute). The manufacturer tells you th
anastassius [24]

Answer:

0.288

Step-by-step explanation:

To solve this question, we need to understand 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 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.

In this question, we have that:

\mu = 17.48, \sigma = 3.25, n = 10, s = \frac{3.25}{\sqrt{10}} = 1.027740

Find the probability that the mean printing speed of the sample is greater than 18.06 ppm.

This is 1 subtracted by the pvalue of Z when X = 18.06. So

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

By the Central Limit Theorem

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

Z = \frac{18.06 - 17.48}{1.027740}

Z = 0.56

Z = 0.56 has a pvalue of 0.712

1 - 0.712 = 0.288

The answer is 0.288

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3 years ago
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vagabundo [1.1K]

Answer:

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GalinKa [24]

Answer:

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Step-by-step explanation:

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8 0
4 years ago
What equation(s) represent joint variations? check all that apply c=2pi z=3x/y y=3x+2 w=abc/4 v=lwh
Alinara [238K]

Answer:

Step-by-step explanation:

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