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Oksana_A [137]
3 years ago
13

To construct the regular hexagon, the radius length AB is used to mark the vertices of the hexagon. Can this method be used to m

ake any regular inscribed polygon, such as a square?
Mathematics
2 answers:
Kisachek [45]3 years ago
6 0

Answer:

<h2>This method cannot make any regular inscribed polygon such as a square.</h2>

Step-by-step explanation:

The main reason why it won't work to other regular polygons is beacuse <em>the regular hexagon is the only polygon that its side is equal to the radius of the circumscribed circle.</em> That's why won't work to other regular figures.

Therefore, the answer is no, it won't work to other regulat polygons such as a square.

True [87]3 years ago
4 0
No, it cannot be used to create other regular inscribed polygons.

This plan only works with a hexagon. If you inscribe a square into a circle, the lengths of the sides will not be equal to the radius.
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Answer:

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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 = 1300, \sigma = 350, n = 10, s = \frac{350}{\sqrt{10}} = 110.68

What is the probability that the mean rent is more than $1100?

This is 1 subtracted by the pvalue of Z when X = 1100. So

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

By the Central Limit Theorem

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

Z = \frac{1100 - 1300}{110.68}

Z = -1.81

Z = -1.81 has a pvalue of 0.0351

1 - 0.0351 = 0.9649

96.49% probability that the mean rent is more than $1100

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