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Darina [25.2K]
3 years ago
14

Given: Line AC is parallel to DF, Line BE is perpendicular to DF, and angle AEB is congruent to angle CEB, prove angle BAE is co

ngruent to angle BCE. Will give Brainliest if explained thoroughly.

Mathematics
1 answer:
Anastasy [175]3 years ago
5 0

9514 1404 393

Explanation:

There are several ways you can go at this. Here are a couple. All proofs will start with the given relations being repeated as part of the proof. Here are the next steps.

<h3>Angle Sum</h3>

∠AED ≅ ∠BAE . . . . alternate interior angles are congruent

∠AED +∠AEB = 90° . . . . angle sum theorem

∠BAE +∠AEB = 90° . . . . substitution property of equality

∠CEF ≅ ∠BCE . . . . alternate interior angles are congruent

∠CEF +∠CEB = 90° . . . . angle sum theorem

∠BCE +∠CEB = 90° . . . . substitution property of equality

∠BAE +∠AEB = ∠BCE +∠CEB . . . . substitution property of equality

∠BAE +∠AEB = ∠BCE +∠AEB . . . . substitution property of equality

∠BAE = ∠BCE . . . . addition property of equality

<h3>Congruent Triangles</h3>

∠ABE = ∠CBE = 90° . . . . BE ⊥ AC

BE ≅ BE . . . . reflexive property of congruence

ΔBEA ≅ ΔBEC . . . . ASA congruence theorem

∠BAE ≅ ∠BCE . . . . CPCTC

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Answer:

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To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

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