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mario62 [17]
3 years ago
5

3. Classify the polygon below by its sides,convex/concave, and regular/irregular​

Mathematics
1 answer:
wolverine [178]3 years ago
7 0

Answer:

The polygon is a concave polygon

The polygon is irregular polygon

Step-by-step explanation:

Convex polygon is polygon that has all interior angles less than 180°

In a convex polygon, every diagonal of the figure passes only through interior points of the polygon

A line drawn through a convex polygon will intersect the polygon in exactly two points

Some of convex polygons can be regular polygons

Concave polygon is a polygon that has one or more interior angles greater than 180°

In a concave polygon, at least one diagonal of the figure contains points that are exterior to the polygon

A line drawn through a concave polygon, can intersect the polygon in more than two points

Concave polygon never be a regular polygon

From the attached figure

∵ The polygon has one diagonal lies out the interior of the polygon

∴ The polygon is a concave polygon

∵ All the concave polygons are regular polygons

∴ The polygon is irregular polygon

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Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

To solve this problem, 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 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 \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33

What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

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

By the Central Limit Theorem

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

2.33 = \frac{X - 117}{4.33}

X - 117 = 2.33*4.33

X = 127.1

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

6 0
4 years ago
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