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Dima020 [189]
3 years ago
13

2. CTfastrak bus waiting times are uniformly distributed from zero to 20 minutes. Find the probability that a randomly selected

passenger will wait the following times for a CTfastrak bus. b. Between 5 and 10 minutes. c. Exactly 7.5922 minutes. d. Exactly 5 minutes. e. Between 15 and 25 minutes.
Mathematics
1 answer:
Juliette [100K]3 years ago
6 0

Answer:

b. 0.25

c. 0.05

d. 0.05

e. 0.25

Step-by-step explanation:

if the waiting time x follows a uniformly distribution from zero to 20, the probability that a passenger waits exactly x minutes P(x) can be calculated as:

P(x)=\frac{1}{b-a}=\frac{1}{20-0} =0.05

Where a and b are the limits of the distribution and x is a value between a and b. Additionally the probability that a passenger waits x minutes or less P(X<x) is equal to:

P(X

Then, the probability that a randomly selected passenger will wait:

b. Between 5 and 10 minutes.

P(5

c. Exactly 7.5922 minutes

P(7.5922)=0.05

d. Exactly 5 minutes

P(5)=0.05

e. Between 15 and 25 minutes, taking into account that 25 is bigger than 20, the probability that a passenger will wait between 15 and 25 minutes is equal to the probability that a passenger will wait between 15 and 20 minutes. So:

P(15

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With Pythagorean theorem, we'll be finding one of the legs. If we split the original triangle in 1/2 from the vertex to the base, we now have two right triangles. Both of these right triangles have a 30 cm hypotenuse and one 13 cm leg (remember we divided the entire original triangle in half).

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b^2 = 731

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Now that we know the height of the triangle, we can find the area.

A = 1/2 x 26 x \sqrt{731}

A = 13 x \sqrt{731}

A = 351.48

The final area, rounded to the nearest whole number is 351 centimeters squared.

Note: Please know that depending on if the answer rounded \sqrt{731} before using it to find the area, the final answer may then be rounded to 352.

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Answer:

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