Probability that Carl arrives first is ¹/₃ = 33.33% since their arrival times is uniformly distributed. The same probability applies to Bob and Alice.

b. What is the probability that Carl will have to wait more than 10 minutes for one of the others to show up?

Assuming that Carl arrived on time, 1:10 PM, we must determine the probability that Alice or Bob arrive between 1:20 and 1:30 (half the remaining time)

P = [3 · (¹/₂ - ¹/₃)] · [3 · (¹/₂ - ¹/₃)] = (3 · ¹/₆) · (3 · ¹/₆) = ¹/₂ · ¹/₂ = ¹/₄ = 25% chance that either Alice or Bob arrive more than 10 minutes later

c. What is the probability that Carl will have to wait more than 10 minutes for both of the others to show up?

P = 1 - 25% = 75%

d. What is the probability that the person who arrives second will have to wait more than 5 minutes for the third person to show up?

I divided the 20 minutes by 5 to get ¹/₄:

P (|S - T| ≤ ¹/₄) = {[(x + ¹/₄)²] / 2} + (1 / 2x) + {[(⁵/₄ - x)²] / 2} = 0.09375 + 0.25 + 0.09375 = 0.4375 = 43.75%

4 0

3 years ago

Factorise 2p^2 - p - 10<br> (2p squared)

velikii [3]

2p² - p - 10 = 0

2 = a

- 1 = b

- 10 = c

To factor, find two numbers that multiply to equal a·c and also that have a sum of b

2p² + 4p - 5p - 10 = 0

2p (p + 2) - 5 (p + 2) = 0

<h3>Answer: ( 2p - 5 ) ( p + 2 ) = 0</h3>

3 0

3 years ago