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oksano4ka [1.4K]

4 years ago

15

The angle of depression from a helicopter to a landing pad is 65°. If the horizontal distance from the helicopter to the landing

pad is 1,000 ft how high up is the helicopter

2 answers:

Simora [160]4 years ago

6 0

Answer:

Helicopter is 2144.50 feet high.

Step-by-step explanation:

In the figure attached.

A helicopter is flying at point A and landing pad is at point c.

Angle of depression of the helicopter is 65° so angle of elevation of this helicopter from landing pad will be same as 65°.

Horizontal distance of landing pad c from point B is 1000 feet.

Now we have to calculate length of AB.

Since tan 65° = $\frac{AB}{BC}$

2.14450 = $\frac{AB}{1000}$

AB = 2.14450 × 1000

= 2144.50 feet.

Helicopter is 2144.50 feet high.

goblinko [34]4 years ago

3 0

H/1000 = tan(65°)
h = 1000*tan(65°) ≈ 2145 . . . . ft

The helicopter is about 2145 ft high.

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4 years ago

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Step-by-step explanation:

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a).

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∝ = P (Reject H₀ | H₀ is true)

= P (p°>0.63 | p=0.6)

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

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= P $[\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ]$

= P $[Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ]$

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= 1 - P [Z ≤ 1.37]

= 1 - Ф (1.37)

= 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)

≅ 0.0853

b)

The probability of Type II error β is stated as:

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= P [p° ≤ 0.63 | p = 0.75]

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

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= P [Z ≤ -6.20]

= Ф (-6.20)

≅ 0.0000 (from Cumulative Standard Normal Distribution Table).

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