In a city school of 1,200 students, 40% of the students are on the honor roll, 60% have a part-time job, and 22% are on the honor roll and have a part-time job. what is the probability (rounded to the nearest whole percent) that a randomly selected student is on the honor roll, given that the student has a part-time job?

Let
A = event that the student is on the honor roll
B = event that the student has a part-time job
C = event that the student is on the honor roll and has a part-time job

We are given
P(A) = 0.40
P(B) = 0.60
P(C) = 0.22
note: P(C) = P(A and B)

We want to find out P(A|B) which is "the probability of getting event A given that we know event B is true". This is a conditional probability

P(A|B) = [P(A and B)]/P(B)
P(A|B) = P(C)/P(B)
P(A|B) = 0.22/0.6
P(A|B) = 0.3667 which is approximate

Convert this to a percentage to get roughly 36.67% and this rounds to 37%

Final 37%

Where’s the

29. because you will divide each number of months by 12 to find the number of years. which would mean, 5, 12.5, 16, 45, 50, 72.5. which is crazy. anyhow, i think you find the median in the middle. and yea hope your test goes good