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elena55 [62]
2 years ago
7

Question :

Mathematics
1 answer:
Elza [17]2 years ago
8 0

Answer:

C) 405.3 ft

Step-by-step explanation:

A triangle is a polygon with three angles and three sides. Types of triangle are scalene, isosceles, equilateral, right angled triangle.

Given a triangle with angles A, B, C and the corresponding sides directly opposite to the angles as a, b, c. The sine rule states that:

\frac{a}{sin(A)}= \frac{b}{sin(B)} =\frac{c}{sin(C)}

In the triangle formed by point C, ship B and ship A, we have ∠A = 82°, ∠B = 78°, c = AB = 140 ft. Hence:

∠A + ∠B + ∠C = 180° (sum of angles in a triangle)

82 + 78 + ∠C = 180

∠C + 160 = 180

∠C = 20°

Using sine rule:

\frac{c}{sin(C)}=\frac{a}{sin(A)}\\\\\frac{140}{sin(20)}=\frac{a}{sin(82)}\\\\a=    \frac{140*sin (82)}{sin(20)}\\\\a=405.3\ ft

a =  distance from Ship B to the signal fire at point C

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Answer:

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Step-by-step explanation:

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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
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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.

This means that all three selected altimeters 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 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}  

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(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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