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3 years ago

5

An engineer on the ground is looking at the top of a building. The angle of elevation to the top of the building is 22°. The eng

ineer knows the building is 450 ft tall. What is the distance from the engineer to the base of the building to the nearest whole foot?
1,114 ft

1,201 ft

1,818 ft

990 ft

2 answers:

Kobotan [32]3 years ago

3 0

Let the distance of the engineer from the base of the building be x, then
tan 22 = 450/x
x = 450/tan 22 = 450/0.4040 = 1,113.79

Therefore, the distance of the engineer from the base of the building to the nearest foot is 1,114 feet.

Ksju [112]3 years ago

3 0

Answer:

Distance, x = 1114 ft

Step-by-step explanation:

It is given that,

The angle of elevation to the top of the building, $\theta=22^0$

Height of the building, h = 450 ft

We have to find the distance from the engineer to the base of the building i.e. x

In triangle ABC, using trigonometric equations as :

$tan(22)=\dfrac{AB}{BC}$

$tan(22)=\dfrac{450}{x}$

$x=\dfrac{450}{tan(22)}$

$x=1113.7\ ft$

or

x = 1114 ft

So, the distance from the engineer to the base of the building is option (A) i.e. 1114 ft

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a) $(14.68 -18.77) - 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =-12.968$

$(14.68 -18.77) + 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =4.788$

b) $t=\frac{14.68-18.77}{\sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}}}}=-1.10$

Step-by-step explanation:

Data given and notation

$\bar X_{A}=14.68$ represent the mean for Pirates

$\bar X_{B}=18.77$ represent the mean for Splash Mountain

$s_{A}=11.87$ represent the sample standard deviation for the sample Pirates

$s_{B}=16.79$ represent the sample standard deviation for the sample Slpash Mountain

$n_{A}=32$ sample size selected for Pirates

$n_{B}=30$ sample size selected for Splash Mountain

$\alpha=0.02$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a

The confidence interval would be given by:

$(\bar X_A -\bar X_B) \pm t_{\alpha/2} \sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}$

The degrees of freedom are given by:

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Since we want 98% of confidence the significance level is $\alpha =1-0.98 =0.02$ and $\alpha/2 =0.01$ , we can find in the t distribution with df =60 a critical value that accumulates 0.01 of the area on each tail and we got:

$t_{\alpha/2}= 2.39$

And replacing we got for the confidence interval:

$(14.68 -18.77) - 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =-12.968$

$(14.68 -18.77) + 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =4.788$

Part b

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the means are equal, the system of hypothesis would be:

Null hypothesis: $\mu_{A} = \mu_{B}$

Alternative hypothesis: $\mu_{A} \neq \mu_{B}$

the statistic is given by:

$t=\frac{\bar X_{A}-\bar X_{B}}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{14.68-18.77}{\sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}}}}=-1.10$

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