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astraxan [27]
3 years ago
13

Suppose you live 4.4 miles from a hill. From your home you see a plane directly above the hill. Your angle of elevation to the p

lane is 30°. What is the plane’s altitude?
Mathematics
1 answer:
IrinaVladis [17]3 years ago
4 0
█ Explanation <span>█

Let's call the missing altitude 'x'. 

</span><span>✦ To find the plane's altitude, find the tan of 30 degrees. The side opposite is going to be x, while the side adjacent is going to be 4.4 miles. 
</span>✦ Tan(30)= \frac{x}{4.4}
✦ 5.7735= \frac{x}{4.4}
✦ x=4.4*.57735
✦ x=4.4*.55735
✦ x=2.54 mi
<span>✦ We usually write altitude in feet. 
</span>
Answer: 2.54 * 5280 ~ <span>13,400 ft

</span><span>Hope that helps! ★ If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia</span>
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Answer:

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Step-by-step explanation:

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earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

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P (C) = 0.001

The events <em>A</em>, <em>B</em> and <em>C</em> are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

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