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Akimi4 [234]
3 years ago
15

A lifeguard is 16.2 ft above a pool in her lifeguard station when she spots a swimmer struggling 48.3 ft away from the base of h

er station. What is the angle of depression from the lifeguard station to the swimmer? Round the answer to the nearest hundredth.
Mathematics
1 answer:
Montano1993 [528]3 years ago
4 0
Pretend y is the angle of depression.
Tan y= opposite/ adjacent

Tan y = 16.2/ 48.3
Y= tan-1( 16.2/ 48.3)

Angle = 18.5416
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a) There is a 18.75% probability that the first question that she gets right is the second question.

b) There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

c) There is a 10.35% probability that she gets the majority of the questions right.

Step-by-step explanation:

Each question can have two outcomes. Either it is right, or it is wrong. So, for b) and c), we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

In this problem we have that:

Each question has 4 choices. So for each question, Robin has a \frac{1}{4} = 0.25 probability of getting ir right. So \pi = 0.25. There are five questions, so n = 5.

(a) What is the probability that the first question she gets right is the second question?

There is a 75% probability of getting the first question wrong and there is a 25% probability of getting the second question right. These probabilities are independent.

So

P = 0.75(0.25) = 0.1875

There is a 18.75% probability that the first question that she gets right is the second question.

(b) What is the probability that she gets exactly 1 or exactly 2 questions right?

This is: P = P(X = 1) + P(X = 2)

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 1) = C_{5,1}.(0.25)^{1}.(0.75)^{4} = 0.3955

P(X = 2) = C_{5,2}.(0.25)^{2}.(0.75)^{3} = 0.2637

P = P(X = 1) + P(X = 2) = 0.3955 + 0.2637 = 0.6592

There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

(c) What is the probability that she gets the majority of the questions right?

That is the probability that she gets 3, 4 or 5 questions right.

P = P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 3) = C_{5,3}.(0.25)^{3}.(0.75)^{2} = 0.0879

P(X = 4) = C_{5,4}.(0.25)^{4}.(0.75)^{1} = 0.0146

P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001

P = P(X = 3) + P(X = 4) + P(X = 5) = 0.0879 + 0.0146 + 0.001 = 0.1035

There is a 10.35% probability that she gets the majority of the questions right.

6 0
3 years ago
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