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tensa zangetsu [6.8K]
4 years ago
8

A regular hexagon is circumscribed about a circle with a radius of 6. Find the area of the shaded region shown. Give the exact a

nswer. (Do not approximate or any square root.)

Mathematics
1 answer:
Alecsey [184]4 years ago
4 0

Answer:

19.58 unit²

Step-by-step explanation:

To find the area of the shaded region, we use the formula:

A_{shaded}=A_{circle} - A_{hexagon}\\\\

As we know that,

A_{circle} = \pi r^2 = \pi 6^2

           =36πunit²

   

Next is to split up the hexagon into 6 regular triangles.

A_{hexagon} = 6. A_{triangle}

               =6 x \frac{1}{2} bh

the base is equal to the radius i.e 6 as the triangles are regular.The height can be represented by taking one of the triangles and drawing a line down the middle.

So, the newly formed triangle is a 30°-60°90° right triangle.

(see figure 1 in attachment)

Here, a= 6/2=> 3

h= a√3 => 3√3

Substituting the required values in the formula of area of hexagon, we get

A_{hexagon}= 6.\frac{1}{2}.6.3\sqrt{3} => 54√3unit²

A_{shaded}=A_{circle} - A_{hexagon}\\\\

           =36π-54√3

            = 19.58 unit²  

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3 years ago
Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 1.9 and a mean diameter of 200
ExtremeBDS [4]

Answer:

64.76% probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches

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 problem, we have that:

\mu = 200, \sigma = 1.9, n = 78, s = \frac{1.9}{\sqrt{78}} = 0.2151

What is the probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches?

This is the pvalue of Z when X = 200 + 0.2 = 200.2 subtracted by the pvalue of Z when X = 200 - 0.2 = 199.8. So

X = 200.2

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

By the Central Limit Theorem

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

Z = \frac{200.2 - 200}{0.2151}

Z = 0.93

Z = 0.93 has a pvalue of 0.8238

X = 199.8

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

Z = \frac{199.8 - 200}{0.2151}

Z = -0.93

Z = -0.93 has a pvalue of 0.1762

0.8238 - 0.1762 = 0.6476

64.76% probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches

4 0
3 years ago
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