<span>You are given the waiting times between a subway departure schedule and the arrival of a passenger that are uniformly distributed between 0 and 6 minutes. You are asked to find the probability that a randomly selected passenger has a waiting time greater than 3.25 minutes.
Le us denote P as the probability that the randomly selected passenger has a waiting time greater than 3.25 minutes.
P = (length of the shaded region) x (height of the shaded region)
P = (6 - 3.25) * (0.167)
P = 2.75 * 0.167
P = 0.40915
P = 0.41</span><span />
The first fraction is 7/10.
The second bag has the same probability as the top so just copy the fractions into the boxes.
To calculate the probability of the same colour being chosen you follow these steps.
P(RR) = 3/10 x 4/9 = 12/90
P(GG) = 7/10 x 5/9 = 35/90
Then add the two fractions.
12/90 + 35/90 = 47/90
The last box is 47/90.
8/15= 0.53333333333333333333333333333333333333333333
14/25= 0.56
so 14/25 is bigger
<span>4x - 5y = 20 I think, good luck!</span>
Answer with Step-by-step explanation:
Let speed of one car=x miles/h
Speed of second car=x+8 mi/h
Time,t=2 hours
Distance between two cars=208 miles
We have to find the average speed for each car for 2-hour trip.
Distance=
Distance traveled by one car=2x
Distance traveled by second car=2(x+8)
According to question
Speed of one car=48mph
Speed of second car=48+8=56mph
