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marysya [2.9K]
3 years ago
6

The probability that an archer hits a target on a given shot is .7 if five shots are fired find the probability that the archer

hits the target on three shots out of the five.
Mathematics
1 answer:
VikaD [51]3 years ago
5 0
This is a problem in "binomial probability."  Either the archer hits his target or he does not.  This experiment is performed 5 times (so that n=5), and the probability that the archer will hit the target is 0.7 (so that p=0.7).

We need to find the binomial probability that x=3 when the possible outcomes are {0, 1, 2, 3, 4, 5}.

You could use a table of binomial probabilities to evaluate the following:

P(5, 0.7, 3).

Alternatively, you could use a TI-83 or TI-84 calculator and its built-in "binompdf(  " function.

I evaluated binompdf(5,0.7,3) and obtained the result 0.309.


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A 50 ft flagpole is mounted to the top of a building. If the angle of elevation from a spot on the street to the top of the pole
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Answer:

the height of the building = 91.67161722 feet.

Step-by-step explanation:

Suppose the height of the building (BC) = X feet.

A 50 ft flagpole (AB) is mounted to the top of a building.

So, height of the top of flag above the ground (AC) = (X+50) feet.

If the angle of elevation from a spot (P) on the street to the top of the pole is 58 degrees and the angle of elevation from the same spot (P) to the bottom of the pole is 46 degrees.

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Considering Right triangle ΔACP, cot(∠APC) = PC / AC.

PC = AC*cot(∠APC) = (X+50)*cot(58°)

Considering Right triangle ΔBCP, cot(∠BPC) = PC / BC.

PC = BC*cot(∠BPC) = X*cot(46°)

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X*cot(46°) = (X+50)*cot(58°)

0.965688774 * X = 0.624869351 * (X+50)

(0.965688774 - 0.624869351) * X = 0.624869351 * 50

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X = 31.2434676 / 0.340819422 = 91.67161722 feet.

Hence, the height of the building = 91.67161722 feet.

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