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Nezavi [6.7K]
2 years ago
9

Taylor is converting an old house into a music shop. In order to comply with city building code, she must install a wheelchair r

amp that has an angle of elevation of no more than 4.8°. Will the dimensions shown the figure work? If not, what change should she make?

Mathematics
1 answer:
irakobra [83]2 years ago
7 0

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>

  • Whether the given dimensions will work or not .

Here Taylor wants to install a wheelchair ramp with an angle of elevation not more than 4.8° . And according to the given figure , the height of the ramp is 8ft and its base length is 100ft .

In a right angled triangle we know that ,

\longrightarrow \tan\theta =\dfrac{perpendicular}{base}

In the given figure , perpendicular is 8ft and base is 100ft . So , the given dimensions will work if ,

\longrightarrow \tan\theta =  \dfrac{perpendicular_{ramp}}{base_{ramp}}

Substitute ,

\longrightarrow \tan4.8^o = \dfrac{8ft}{100ft}

Substituting the value of tan4.8° ,

\longrightarrow 0.83 =  0.8

And ,

\longrightarrow 0.83 \approx 0.8

<u>Therefore</u><u> the</u><u> </u><u>given</u><u> </u><u>dimensions</u><u> </u><u>will </u><u>work </u><u>.</u>

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Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu
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Answer:

a) s^2 =30^2 =900

b) \frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}

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c) 23.818 \leq \sigma \leq 41.112

Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in  the variability in the number of rooms occupied per day during a particular season of the  year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per  day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

s^2 =30^2 =900

Part b

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The degrees of freedom are given by:

df=n-1=20-1=19

Since the Confidence is 0.90 or 90%, the significance \alpha=0.1 and \alpha/2 =0.05, the critical values for this case are:

\chi^2_{\alpha/2}=30.144

\chi^2_{1- \alpha/2}=10.117

And replacing into the formula for the interval we got:

\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}

567.28 \leq \sigma^2 \leq 1690.224

Part c

Now we just take square root on both sides of the interval and we got:

23.818 \leq \sigma \leq 41.112

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The depth of the bottom of the hole after the second day is 36 feet using addition operation.

<h3>What is addition?</h3>

In math, addition is the process of adding two or more integers together. Addends are the numbers that are added, while the sum refers to the outcome of the operation.

Given the depth on the first day is 26 ½ feet.

Depth on the second day = 9½ feet more than on the first day i.e. 9½ feet + depth on the first day

This implies, depth on the second day = 9½ + 26 ½

= 36 feet

Therefore, the depth of the bottom of the hole after the second day is 36 feet.

To learn more about addition, visit:

brainly.com/question/25621604

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Answer:

(maaf kalau salah

THANKS

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