An Uber driver carries riders to and from a local airport and a neighboring suburb. Due to traffic and the time of day the drive
r works, the trip to the airport takes 60 minutes longer than the trip to the suburb. The car’s average speed in traffic is 30 miles per hour. Without traffic, the car travels 15 miles per hour. The equation can be used to determine how many miles apart the two locations are shown. What is the value of m, the distances between the two locations?
A) 0.5 miles
B) 22.5 miles
C) 30 miles
D) 90 miles
A) 0.5 miles
B) 22.5 miles
C) 30 miles
D) 90 miles
1 answer:
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Answer:
P(X>4)= 0.624
Step-by-step explanation:
Given that
n = 10
p= 0.5 ,q= 1 - p = 0.5
Two fifth of 10 = 2/5 x 10 =4
It means that we have to find probability P(X>4).
P(X>4)= 1 -P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)
We know that
P(X>4)= 1 -P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)
P(X>4)= 1 -0.0009 - 0.0097 - 0.043 - 0.117-0.205
P(X>4)= 0.624
Answer:
The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.
Step-by-step explanation:
Using Bayes' Theorem
P(A|B) =
where
P(B|A) is probability of event B given event A
P(B|a) is probability of event B not given event A
and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.
For this problem,
Let P(B1) = Probability of machine B1 = 0.3
P(B2) = Probability of machine B2 = 0.2
P(B3) = Probability of machine B3 = 0.5
Let P(D) = Probability of a defective product
P(N) = Probability of a Non-defective product
P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003
P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006
P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010
Likewise,
P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997
P(N|B2) be probability of a non-defective product produced by machine 2 = 1 - P(D|B2) = 1 - 0.006 = 0.994
P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990
For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)
P(B1|N) = =
= 0.1138
Hence the probability that a non-defective product is produced by machine B1 is 11.38%.