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horrorfan [7]
3 years ago
12

Daniel wants to prove that the interior angles of any triangle sum to 180°. He draws a line through one vertex parallel to the o

pposite side, and then he labels all the angles formed.
Drag a reason to match each statement in Daniel's proof in the table below.

Mathematics
1 answer:
Serggg [28]3 years ago
4 0

Alternate interior angles theorem.

<u>Step-by-step explanation:</u>

By using the definition of Alternate Interior Angles theorem, we can say that If a transversal cuts any two parallel lines, then those pairs of alternate interior angles are seems to be  congruent.

So in the given triangle,

∠4 ≅ ∠2

∠5 ≅ ∠3

Sum of all the angles in a triangle = 180°

∠1 + ∠2 + ∠3 = 180°

Since from the above congruence, we can write that

∠1 + ∠4 + ∠5 = 180°

Hence proved.

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Answer: we reject the initial claim

(50. 3802, 51.6198)

Step-by-step explanation:

A)

The initial statement is that the mean temperature throughout history is 50°

This is the null hypothesis and it is denoted as H'

H': u = 50.

After taking a sample of 40 years, we realized that the mean temperature is 51°.

This is the alternative hypothesis, in contradiction to the alternative.

The alternative hypothesis is denoted as H1

H1: u > 50 ( upper tailed).

Sample size (n) = 40

Sample mean (x) = 51

Population standard deviation (σ) = 2

We use a z test to get the value of the test statistics.

We are using a z test because sample size is greater than 30 ( n = 40) and population standard deviation is given.

Z score = x - u/ (σ/√n)

Z score = 51 - 50 / (2/√40)

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Our level of significance is 5%, and the critical value at this level of significance is 1.645.

By comparing the z score relative to the critical value, since our z score is greater than 1.645, it implies that we are in the rejection region hence we reject the initial claim.

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We are to construct a 95% confidence interval for population mean temperature.

For upper limit

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Where Zα/2 = 1.96 which is the critical value for a two tailed test at 5% level of significance.

For upper limit, we have that

u = 51 + 1.96 × (2/√40)

u = 51 + 1.96 (0.3162)

u = 51 + 0.6198

u = 51.6198.

For lower limit, we have that

u = 51 - 1.96 × (2/√40)

u = 51 - 1.96 (0.3162)

u = 51 - 0.6198

u = 50. 3802.

Hence the 95% confidence level for mean temperature is given as (50. 3802°, 51.6198°)

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