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jonny [76]
2 years ago
11

from a window 24 feet above the ground, the angle of elevation to the top of another building is 38 degrees. The distance betwee

n the buildings 63 feet. Find the height of the building to the nearest tenth of a foot.

Mathematics
1 answer:
Delvig [45]2 years ago
6 0

Answer:

The  height of the building is 73.2\ ft

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

In the right triangle ABC

tan(38\°)=\frac{BC}{AB}

we have

AB=63\ ft

substitute and solve for BC

tan(38\°)=\frac{BC}{63}

BC=(63)tan(38\°)=49.2\ ft

Find the height of the building

The  height of the building (h) is equal to

h=BC+24=49.2+24=73.2\ ft

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Identify the amount, base, and percent in the problem: What is 60% of 485?
IrinaK [193]

Hey there! I'm happy to help!

The percent in the problem is 60% (as it has a percent time).

The base is the number we start with, so it is 485.

The amount is the percent of the base. As a decimal, 60% is 0.6. So, we just multiply this by 485.

0.6×485=291

Therefore, the amount is 291.

AMOUNT: 291

BASE: 485

PERCENT: 60%

Have a wonderful day! :D

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3 years ago
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Answer:

A. t=66d

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since distance = time x velocity , or distance = time x rate

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2 years ago
Alana has 100 mL of water in her water bottle. She needs to share it with 3 other students so that all 4 of them have the same a
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Answer:

each person (including Alana) gets 25mL of water

Step-by-step explanation:

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3 years ago
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The thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters. (a) Determine the
maxonik [38]

Answer:

a) 0.5 = 50% of flanges exceed 1 millimeter.

b) A thickness of 0.96 millimeters is exceeded by 90% of the flanges

Step-by-step explanation:

A distribution is called uniform if each outcome has the same probability of happening.

The uniform distributon has two bounds, a and b, and the probability of finding a value higher than x is given by:

P(X > x) = \frac{b - x}{b - a}

The thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters.

This means that a = 0.95, b = 1.05

(a) Determine the proportion of flanges that exceeds 1.00 millimeters.

P(X > 1) = \frac{1.05 - 1}{1.05 - 0.95} = \frac{0.05}{0.1} = 0.5

0.5 = 50% of flanges exceed 1 millimeter.

(b) What thickness is exceeded by 90% of the flanges?

This is x for which:

P(X > x) = 0.9

So

\frac{1.05 - x}{1.05 - 0.95} = 0.9

1.05 - x = 0.9*0.1

x = 1.05 - 0.9*0.1

x = 0.96

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Answer:

the answer is 25

Step-by-step explanation:

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