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bogdanovich [222]
2 years ago
11

Find the area of the polygon​

Mathematics
1 answer:
Evgesh-ka [11]2 years ago
3 0

Answer:

784m^2

Step-by-step explanation:

is a rectangle with sides of 40m and 20m, in the upper right corner a right triangle has been removed with the legs of: 40 - 32 = 8m and 20 - 16 = 4m, we find the area of ​​the rectangle (b * h).

40 * 20 = 800m ^ 2, then we find the area of ​​the right triangle

1/2 b * h: 1/2 8 * 4 = 16 m ^ 2.

we remove the area of ​​the triangle from the rectangle and we have the area of ​​the figure: 800 - 16 = 784m^2

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Answer:

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Step-by-step explanation:

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Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

The home team therefore wins 50% of its games

This means that p = 0.5

Determine the probability that the home team would win 65% or more of its games in a simple random sample of 80 games

Sample of 80 means that n = 80 and, by the Central Limit Theorem:

\mu = p = 0.65

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5*0.5}{80}} = 0.0559

This probability is 1 subtracted by the pvalue of Z when X = 0.65. So

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

By the Central Limit Theorem

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

Z = \frac{0.65 - 0.5}{0.0559}

Z = 2.68

Z = 2.68 has a pvalue of 0.9963

1 - 0.9963 = 0.0037

0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games

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