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nevsk [136]
3 years ago
5

A. The distance between the towers is 7350 ft, and the angles of elevation are given in the diagram. Find the distance from Towe

r 2 to the plane, to the nearest tenth of a foot.
11099 ft
9448 ft
4894 ft
10032 ft

B. Find the height of the plane from the ground, to the nearest foot. Hint: Use your answer from part a.
8563 ft
9170 ft
10135 ft
11300 ft

Mathematics
2 answers:
melomori [17]3 years ago
7 0
Hello,


A.9448


B.8563

Have a amazing day and Good luck with school!

(I need one more brainliest to reach the expert rank! Thank you!)
Mademuasel [1]3 years ago
4 0
The angle T1PT2 is 65-37=28 degree
use law of sine: sin37/PT2=sin28/7370, PT2=7370*sin37/sin28
Use a calcuator, I got the answer 9448

B. sin65=PA/PT2, PA=PT2*sin65=9448*sin65=8563
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Answer:

\hat p \pm Z*\sqrt{\frac{\hat p\times(1-\hat p)}{n} }

=0.5546\pm1.96*\sqrt{\frac{0.5546\times(1-0.5546)}{458} }

=0.5546\pm1.96*\sqrt{\frac{0.5546\times(0.4454)}{458} }

=0.5546\pm1.96*\sqrt{\frac{0.2470}{458} }

=0.5546\pm1.96*\sqrt{0.00053934}

=(0.5091,0.6001)

Lower limit for confidence interval=0.5091

             

Upper limit for confidence interval=0.6001

Step-by-step explanation:

We have given,              

             

x=254          

n=458          

Estimate for sample proportion= \bar p = 0.5546

Level of significance is =1-0.95=0.05      

Z critical value(using Z table)=1.96      

Confidence interval formula is

\hat p \pm Z*\sqrt{\frac{\hat p\times(1-\hat p)}{n} }

=0.5546\pm1.96*\sqrt{\frac{0.5546\times(1-0.5546)}{458} }

=0.5546\pm1.96*\sqrt{\frac{0.5546\times(0.4454)}{458} }

=0.5546\pm1.96*\sqrt{\frac{0.2470}{458} }

=0.5546\pm1.96*\sqrt{0.00053934}

=(0.5091,0.6001)

Lower limit for confidence interval=0.5091

             

Upper limit for confidence interval=0.6001

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Answer:

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