If the probability of observing at least one car on a highway during any 20-minute time interval is 609/625, then the probability of observing at least one car during any 5-minute time interval is 609/2500
Given The probability of observing at least one car on a highway during any 20 minute time interval is 609/625.
We have to find the probability of observing at least one car during any 5 minute time interval.
Probability is the likeliness of happening an event among all the events possible. It is calculated as number/ total number. Its value lies between 0 and 1.
Probability during 20 minutes interval=609/625
Probability during 1 minute interval=609/625*20
=609/12500
Probability during 5 minute interval=(609/12500)*5
=609/2500
Hence the probability of observing at least one car during any 5 minute time interval is 609/2500.
Learn more about probability at brainly.com/question/24756209
#SPJ4
Answer:
t = 25.1seconds
Step-by-step explanation:
X(t)= 16cos75°t
Y(t)= 6 + (Vosin75°t - 16t^2)
Converting 78 yards to feet :
1 yard = 3 feet
78feet =?
78×3=234 feet
Y(t)= number of feet above the ground at t seconds
VoCos75° t = 234
t = 234/(VoCos75°)
At this time,4 = y(234/VoCos75°)
4 = 6 + VoSin75°(234/VoCos75) - 16(234/VoCos75°)^2
Vo= 140.66ft/s
Time,t= 234/(140.66× Cos75°)
t = 25.087seconds
t= 25.1 seconds (1 decimal place)
Multiply by 3/5 (three fifths)
Answer: I took it last year as a junior, what is the questions tho?
Answer:
(3,4) and (6,8).
Step-by-step explanation:
Rise is equal to y and run is equal to x. If the fraction is 4/3, y is equal to 4 and 3 is equal to x. 3 across is 3 and 4 upwards is 4. Just keep adding 3,4 to get the following points the line passes through. Since we need two points, 3,4 and 6,8 would be the appropriate answers.
