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mihalych1998 [28]
3 years ago
6

Troy drives 65 miles west and then drives 72 miles north. How far is he from his starting point?

Mathematics
1 answer:
iragen [17]3 years ago
4 0

Answer:

132 miles away from the start

Step-by-step explanation:

First Troy went to the left 65 miles so he starts off 65 miles from the start then he goes 72 miles right  so then he is 132 miles away

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Angle G = 76 degrees

Step-by-step explanation:

66 + 2x + x = 180

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3x = 114

x = 38

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What is the 1st quartile? (median of lower<br>half)<br>240<br>235<br>230​
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lower half

Step-by-step explanation:

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What is the greatest common factor for 30,48,60 and whats the GCF ??????
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A.) find the test statistic
Arada [10]

Regarding the hypothesis tested in this problem, it is found that:

a) The test statistic is: t = 0.49.

b) The p-value is: 0.6297..

c) The null hypothesis is not rejected, as the p-value is greater than the significance level.

d) It appears that the students are legitimately good at estimating one minute, as the sample does not give enough evidence to reject the null hypothesis.

<h3>What is the test statistic?</h3>

The test statistic of the t-distribution is given by the equation presented as follows:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

In which the variables of the equation are given as follows:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the hypotheses.
  • s is the standard deviation of the sample.
  • n is the sample size.

The value tested at the hypothesis is:

\mu = 60

From the sample, using a calculator, the other parameters are given as follows:

\overline{x} = 62.47, s = 19.2, n = 15

Hence the test statistic is obtained as follows:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{62.47 - 60}{\frac{19.2}{\sqrt{15}}}

t = 0.49.

<h3>What are the p-value and the conclusion?</h3>

The p-value is obtained using a t-distribution calculator, with a two-tailed test, as we are testing if the mean is different a value, in this case 60, with:

  • 15 - 1 = 14 df, as the number of degrees of freedom is one less than the sample size.
  • t = 0.49, as obtained above.

Hence the p-value is of 0.6297.

Since the p-value is greater than the significance level of 0.01, the null hypothesis is not rejected, attesting to the ability of the students to estimate one minute.

More can be learned about the test of an hypothesis at brainly.com/question/13873630

#SPJ1

5 0
1 year ago
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