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Levart [38]
3 years ago
6

HELP PLEASE!! As John walks 16 ft towards a chimney, the angle of elevation from his eye to the top of the chimney changes from

30° to 45°. Identify the height of the chimney from John's eye level to the top of the chimney rounded to the nearest foot.

Mathematics
1 answer:
rjkz [21]3 years ago
6 0

Answer: 22 feet.

Step-by-step explanation:

Note that there are two right triangles in the figure attached: ACD and BCD. Where "h" is the height of the chimeney  from John's eye level to the top of the chimney.

You need to use the trigonometric identity tan\alpha=\frac{opposite}{adjacent} for this exercise.

  • For the triangle BCD:

tan(45\°)=\frac{h}{x}

Solve for h:

h=xtan(45\°)\\h=x

  • For the triangle ACD:

tan(30\°)=\frac{h}{x+16}

Substitute h=x and solve for h:

tan(30\°)=\frac{h}{h+16}\\\\(h+16)(tan(30\°))=h\\\\0.577h+9.237=h\\\\9.237=h-0.577h\\\\9.237=0.423h\\\\h=\frac{9.237}{0.423}\\\\h=21.836ft

Rounded to the nearest foot:

h=22ft

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The hypothesis statements are: H0: p = 0.88 versus HA : p < 0.88

The p-value of the test is 0.2483 > 0.1, which means that the data does not provide sufficient evidence to conclude that the proportion of times that luggage is returned within 24 hours is less than 0.88.

Step-by-step explanation:

Test if the proportion of times that luggage is returned within 24 hours is less than 0. 88

At the null hypothesis, we test if the proportion is of 0.88, that is:

H_0: p = 0.88

At the alternate hypothesis, we test if this proportion is less than 0.88, that is:

H_a: p < 0.88

The hypothesis statements are: H0: p = 0.88 versus HA : p < 0.88

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.88 is tested at the null hypothesis:

This means that \mu = 0.88, \sigma = \sqrt{0.88*0.12}

A consumer group who surveyed a large number of air travelers found that 138 out of 160 people who lost luggage on that airline were reunited with the missing items by the next day.

This means that n = 160, X = \frac{138}{160} = 0.8625

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The p-value of the test is the probability of finding a sample proportion below 0.8625, which is the p-value of z = -0.68.

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The p-value of the test is 0.2483 > 0.1, which means that the data does not provide sufficient evidence to conclude that the proportion of times that luggage is returned within 24 hours is less than 0.88.

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