Answer:
Velocity = displacement / time
Distance = 3t^2 +2t +6 to find distance you need to plug in time = 2 sec
Distance = 3(2)^2 +2(2) +6
Distance = 3(4) +4 + 6
Distance = 12 + 4 +6
Distance = 22 meters
In this problem the displacement is the distance.
so
Velocity = 22 meters / 2 second
Velocity = 11 meters/second
Does this help?
Step-by-step explanation:

a. 9:00 AM is the 60 minute mark:

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

The probability of doing so for at least 2 of 5 days is

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.
d. Integrate the PDF to obtain the CDF:

Then the desired probability is

|2(-5) - 1| + 10
|-10-1| + 10
|-11| + 10
11 + 10
21 is your answer
Hope this helped!
Tbh i dont really remember how to this this but the only way i could think of solving it is saying how a spuare/rectangle have all equal sides so if they have all equal sides you can divide that number by 4 so 4/30=7.5 so i would say x = 7.5 tell me if its wrong
~ Good Luck :D ~