Answer:
a. What is the probability that Carl arrives first?
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%
A.
Angles 1 and 7 will add up to 180 degrees.
Hope this helps!!!
Answer:
I wanna say it is A but i am not positive
Step-by-step explanation:
If you would like to write 1386/1000 in the simplest form, you can do this using the following steps:
1386/1000 = 693/500 = 1 193/500
The correct result would be <span>693/500.</span>
The anser i got was 3.33333 but i think you would plot it as 3.25
