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Firlakuza [10]
3 years ago
10

I WILL MARK BRAINLIEST :) Determine which segment is shortest in the diagram.

Mathematics
2 answers:
erastovalidia [21]3 years ago
8 0

Answer:

C.

Step-by-step explanation:

Nana76 [90]3 years ago
7 0

Answer:

C. Segment AC

Step-by-step explanation:

I am going to figure all the angles first. Triangles always add up to 180 degrees so I am going to use that.

50 + 49 = 99

180 - 99 = 81

<C = 81°

99 + 37 = 136

180 - 136 = 44

<A = 44°

The shortest segment is always going to be opposite the smallest angle. The smallest angle is 37°. Therefore, Segment AC is the shortest.

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Testing the hypothesis in this problem which is a two-tailed test, we can conclude that there is not sufficient evidence to conclude that the mean time of installation has changed, since the p-value of the test is 0.0364 > 0.01,

At the null hypothesis, we test if the <u>mean is of 25 minutes</u>, that is:

H_0: \mu = 25

At the alternative hypothesis, we test if the mean has changed, that is, if it is <u>different than 25 minutes</u>.

H_1: \mu \neq 25

We have the <u>standard deviation for the sample</u>, thus, the t-distribution is used. The value of the test statistic is:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which:

  • X is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

In this problem:

  • 25 is tested at the null hypothesis, thus \mu = 25.
  • Sample mean of 26.2 minutes, thus X = 26.2.
  • Sample of 51, thus n = 51.
  • Variance of 16, thus s = \sqrt{16} = 4.

The <u>value of the test statistic</u> is:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{26.2 - 25}{\frac{4}{\sqrt{51}}}

t = 2.14

  • The p-value of the test is found using a <u>two-tailed test</u>(test if the mean is different of a value), with <u>t = 2.14 and 50 degrees of freedom</u>.
  • Using a t-distribution calculator, this p-value is of 0.0364.

Since the p-value of the test is 0.0364 > 0.01, there is not sufficient evidence to conclude that the mean time of installation has changed.

A similar problem is given at brainly.com/question/23777908

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here is the answer to your question

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