14 divided by 17 is 14/17.
Hope this helps~
Answer:
Step-by-step explanation:
Time varies inversely with speed if the distance is constant. This means that if speed increases, the time it will take to cover the same distance will reduce
t = k/s
Where t = time
k = constant of inverse proportionality
s = speed
It takes James 4 hours to get to his destination when he travels at 65 miles per hour
4 = k/65
k = 65×4 = 260
t = 260/s
If he drove 15 miles/ hour faster, the new speed is 65 + 15 = 80miles/hour.
The time it will take him is
t = 260/s
t = 260/80 = 3.25 hours
The amount of time he would save is
4 hours - 3.25 hours = 0.75 hours
Converting 0.75 hours to minutes, we multiply by 60
Amount if time saved in minutes = 0.75×60 = 45 minutes
Im not sure this question is complete but the answer should be <u><em>152.4</em></u>
Answer:
a. 0.4
b. 0.6
c. 0.6493
Step-by-step explanation:
p(checking work email) = p(A) = 0.40
p(staying connected with cell phone) = p(B) = 0.30
p(having laptop) = p(c) = 0.35
p(checking work mail and staying connected with cell phone) = p(AnB) = 0.16
p(neither A,B or C) = p(AuBuC)
= 1-42.8%
= 0.572
p(A|C) = 88% = 0.88
p(C|B) = 70% = 0.7
a. What is the probability that a randomly selected traveler who checks work email also uses a cell phone to stay connected?
p(B|A) = p(AnB)/p(A)
= 0.16/0.4
= 0.4
b. What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected?
p(B|C) = P(C|B)p(B)/p(C)
= 0.7x0.3/0.35
= 0.6
c. If the randomly selected traveler checked work email and brought a laptop, what is the probability that he/she uses a cell phone to stay connected?
p(A|BnC)
= P(BnAnC)/p(AnC)
= p(AnC) = p(A|C).p(C)
= 0.88x0.35
= 0.308
p(AnBnC) = p(AuBuC)-p(a)-p(b)+ p(AnB)+p(AnC)+p(BnC)
p(BnC) = 0.7x0.3
= 0.21
p(AnBnC) = 0.572-0.4-0.3-0.35+0.16+0.308+0.21
= 0.2
p(A|BnC) = 0.2/0.308
= 0.6493
Answer:
B
Step-by-step explanation:
long but easy have to be menso to not know this
