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Sophie [7]
3 years ago
8

A conjecture and the two-column proof used to prove the conjecture are shown. Given: segment A B is congruent to segment B D, se

gment B D is congruent to segment C E, and segment C E is congruent to segment A C. Prove: triangle A B C is an isosceles triangle. Segment A D with endpoints D and A moving from left to right. Segment A D is diagonally down to the left from point A. B is the midpoint of segment A D. Segment A E shares endpoint at point A with segment A D. Segment A E is diagonally down to the right from point A. C is the midpoint of segment A E. Segment B C is horizontal between segment A D and segment A E. Drag an expression or phrase to each box to complete the proof. Statement Reason 1. AB¯¯¯¯¯≅BD¯¯¯¯¯ Given 2. BD¯¯¯¯¯≅CE¯¯¯¯¯ Given 3. Transitive Property of Congruence 4. Given 5. AB¯¯¯¯¯≅AC¯¯¯¯¯ 6. △ABC is an isosceles triangle.

Mathematics
2 answers:
rusak2 [61]3 years ago
5 0

Answer:

3.Transitive Property Of Congruence

daser333 [38]3 years ago
4 0

Answer:

Step-by-step explanation:

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A researcher was interested in comparing the resting pulse rates of people who exercise regularly and people who do not exercise
DiKsa [7]

Answer:

We Reject H₀ if t calculated > t tabulated

But in this case,

0.83 is not greater than 2.056

Therefore, we failed to reject H₀

There is no difference between the mean pulse rate of people who do not exercise and the mean pulse rate of people who do exercise.

Step-by-step explanation:

Refer to the attached data.

The Null and Alternate hypothesis is given by

Null hypotheses = H₀: μ₁ = μ₂

Alternate hypotheses = H₁: μ₁ ≠ μ₂

The test statistic is given by

$ t = \frac{\bar{x}_1  - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } }  $

Where \bar{x}_1 is the sample mean of people who do not exercise regularly.

Where \bar{x}_2 is the sample mean of people who do exercise regularly.

Where s_1 is the sample standard deviation of people who do not exercise regularly.

Where s_2 is the sample standard deviation of people who do exercise regularly.

Where n_1 is the sample size of people who do not exercise regularly.

Where n_2 is the sample size of people who do exercise regularly.

$ t = \frac{72.7  - 69.7}{\sqrt{\frac{10.9^2}{16} + \frac{8.2^2}{12} } }  $

t = 0.83

The given level of significance is

1 - 0.95 = 0.05

The degree of freedom is

df = 16 + 12 - 2 = 26

From the t-table, df = 26 and significance level 0.05,

t = 2.056 (two-tailed)

Conclusion:

We Reject H₀ if t calculated > t tabulated

But in this case,

0.83 is not greater than 2.056

Therefore, We failed to reject H₀

There is no difference between the mean pulse rate of people who do not exercise and the mean pulse rate of people who do exercise.

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2 years ago
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Alex_Xolod [135]

Answer:

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Step-by-step explanation:

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