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Illusion [34]
3 years ago
13

When the sun is at 31 above the horizon, a tree casts a 42 ft. shadow. how tall is the tree?

Mathematics
1 answer:
Fantom [35]3 years ago
5 0
Answer: 25.2361459992ft

This question can be solved by a trigonometric equation. Shadow formed when something with height like a tree, so it can be pictured as a vertical line. Shadow is a length of and can be pictured as a horizontal line.  
In trigonometric, to find the vertical line length with horizontal line you need tan function. The sun angle will be used in the function, so the calculation would be: tan(31) x 42ft = 25.2361459992ft

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

a)\hat \sigma^2 =s^2 =30^2 = 900

b) 567.277 \leq \sigma^2 \leq 1690.224

Rounded to the nearest number would be:

567 \leq \sigma^2 \leq 1690

c) 23.818 \leq \sigma \leq 41.112

And rounded :

23.8 \leq \sigma \leq 41.1

Step-by-step explanation:

Data given and notation

s=30 represent the sample standard deviation

\bar x=290 represent the sample mean

n=20 the sample size

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

Part a

The best point of estimate for the population variance is the sample variance, so on this case:

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

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=20-1=19

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,19)" "=CHISQ.INV(0.95,19)". so for this case the critical values 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 \leq \frac{(19)(30)^2}{10.117}

567.277 \leq \sigma^2 \leq 1690.224

Rounded to the nearest number would be:

567 \leq \sigma^2 \leq 1690

Part c

In order to find the confidence interval for the deviation we just need to take the square root for the interval of the variance, and we got:

23.818 \leq \sigma \leq 41.112

And rounded :

23.8 \leq \sigma \leq 41.1

7 0
3 years ago
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