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monitta
3 years ago
6

How would I do this one?

Mathematics
2 answers:
viva [34]3 years ago
6 0
First we would need to make a diagram, and from the given it would seem like it is a right triangle with the height as 760 ft. We have an angle given 89°18' which would need to be converted to degrees form, which is 89.3<span>°. 

Now we would need to look for the ground distance, meaning the adjacent of the triangle. Since we have been given the angle and the height (which is the opposite side from the angle) we use the trigonometric function cotangent.

</span><span>760 Cot <span>[<span><span>(<span>89.3</span>)</span>∘</span>]</span>= 9.29 ft, that is the answer.</span>
Gwar [14]3 years ago
5 0

Answer:

The headquarters is 9 ft away from a point on the ground directly below the balloon .

Step-by-step explanation:

The height of balloon is 760 ft and the angle of depression is 89°18' = 89 \frac{18}{60} =89.3°.

A right angled triangle is formed ( as shown in figure ) .Let x represent the distance of headquarters from a point on the ground directly below the balloon .

Tan 89.3=\frac{760}{x}

81.85 x= 760 .[ Tan 89.3= 81.85]

Dividing both sides by 81.85:

x= 9.3

Or x=9 ft ( rounded to nearest foot)

The headquarters is 9 ft away from a point on the ground directly below the balloon .

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The Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are
almond37 [142]

Answer:

For x = 0, P(x = 0) = 0.35

For x = 1, P(x = 1) = 0.54

For x = 2, P(x = 2) = 0.11

For x = 3, P(x = 3) = 0

Step-by-step explanation:

We are given that the Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are not. Three altimeters are randomly selected, one at a time, without replacement.

Let X = <u><em>the number that are not correctly calibrated.</em></u>

Number of altimeters that are correctly calibrated = 6

Number of altimeters that are not correctly calibrated = 2

Total number of altimeters = 6 + 2 = 8

(a) For x = 0: means there are 0 altimeters that are not correctly calibrated.

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Total number of ways of selecting 3 altimeters from a total of 8 = ^{8}C_3

The number of ways of selecting 3 altimeters from a total of 6 altimeters that are correctly calibrated = ^{6}C_3

So, the required probability = \frac{^{6}C_3}{^{8}C_3}  

                                              = \frac{20}{56}  = <u>0.35</u>

(b) For x = 1: means there is 1 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 1 is not correctly calibrated and 2 are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = ^{8}C_3

The number of ways of selecting 2 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = ^{6}C_2

The number of ways of selecting 1 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = ^{2}C_1

So, the required probability = \frac{^{6}C_2 \times ^{2}C_1 }{^{8}C_3}  

                                                = \frac{30}{56}  = <u>0.54</u>

(c) For x = 2: means there is 2 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 2 are not correctly calibrated and 1 is correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = ^{8}C_3

The number of ways of selecting 1 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = ^{6}C_1

The number of ways of selecting 2 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = ^{2}C_2

So, the required probability = \frac{^{6}C_1 \times ^{2}C_2 }{^{8}C_3}  

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(d) For x = 3: means there is 3 altimeter that is not correctly calibrated.

This case is not possible, so this probability is 0.

6 0
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