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 Answer: 37%
Answer:
90/100
Step-by-step explanation:
if you look at the graph on the bottom part it tells you the study time if you line it the left side which says the test score. so if they studied for 80 mins they are expected to have the score of between 85 and 90.
His total load is 4*125 + 2*75 + 2*50, which is 750
now, is that 750 95% of 800?
well 95% of anything is 95/100 * anything
thus, 95% of 800 is 95/100 * 800 or 0.95 * 800
now, is that less than more or equals to 750? if it's, then he's golden, because his load is less or equals to 95% of 800, if not, then he needs to cut down some
The answer is -20x, because the parentheses show they are being multiplied, so you multiply -10x and 2, which would be -20x.
Answer:
52.25 MPH
Step-by-step explanation:
Using the formula given D=RT we simply plug in values,
313.5=6R
313.5/6=R
52.25=R
