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xxTIMURxx [149]
3 years ago
10

2. The Geo Air pilot is looking at SCCA from the plane. From the aircraft the angle of depression is 17 degrees. If the plane is

at an altitude of 10,000 feet, approximately how far is the plane to SCCA? Round your answer to the nearest tenth. The image is not drawn to scale. (2 points)

Mathematics
1 answer:
lana66690 [7]3 years ago
7 0

Answer:

The distance  from the plane to SCCA is 34,203.0 feet approximately,  and the horizontal distance is 32,708.5 feet approximately.

Step-by-step explanation:

You can draw a right triangle like the one shown in the figure attached, where:

x: horizontal distance.

y: distance from the plane to SCCA.

You can calculate x as following:

tan\alpha=\frac{opposit}{adjacent}

Where:

\alpha=17\°\\opposite=10,000\\adjacent=x

Substitute and solve for x:

tan(17\°)=\frac{10,000}{x}\\\\x=\frac{10,000}{tan(17\°)}\\\\x=32,708.5ft

You can calculate y as following:

sin\alpha=\frac{opposit}{hypotenuse}

Where:

\alpha=17\°\\opposite=10,000\\hypotenuse=y

Substitute and solve for y:

sin(17\°)=\frac{10,000}{y}\\\\y=\frac{10,000}{sin(17\°)}\\\\y=34,203.0ft

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gtnhenbr [62]

Answer:

a) r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857  

The correlation coefficient for this case is very near to 1 so then we can ensure that we have linear correlation between the two variables

b) m=\frac{64.5}{2000}=0.03225  

Now we can find the means for x and y like this:  

\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50  

\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425  

b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27  

So the line would be given by:  

y=0.3225 x -0.27  

Step-by-step explanation:

Part a

The correlation coeffcient is given by this formula:

r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}  

For our case we have this:

n=4 \sum x = 200, \sum y = 5.37, \sum xy = 333, \sum x^2 =12000, \sum y^2 =9.3501  

r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857  

The correlation coefficient for this case is very near to 1 so then we can ensure that we have linear correlation between the two variables

Part b

m=\frac{S_{xy}}{S_{xx}}  

Where:  

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}  

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}  

With these we can find the sums:  

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=12000-\frac{200^2}{4}=2000  

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=333-\frac{200*5.37}{4}=64.5  

And the slope would be:  

m=\frac{64.5}{2000}=0.03225  

Now we can find the means for x and y like this:  

\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50  

\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425  

And we can find the intercept using this:  

b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27  

So the line would be given by:  

y=0.3225 x -0.27  

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