Question

A teacher gave her class two exams; 60% of the class passed the second exam, but only 48% of the class passed both exams. What percent of those who passed the second exam also passed the first exam?

Answer 1

@AllTehMaffs

Answer 2

So this is exactly the inverse of the street light one. Remember how you set that one up?

Answer 3

80%

Answer 4

A diner has collected data about customer coffee-drinking habits. They have calculated that P(cream) = 0.5, P(sugar) = 0.6, and P(cream or sugar) = 0.7. Determine the P(cream and sugar).

Answer 5

correct ^_^

Answer 6

@AllTehMaffs

Answer 7

@dan815

Answer 8

iddnt read that right

Answer 9

\[ \textrm{P(a and b)} \ = \ P(a \cap b)
\\ \textrm{P(a or b)} \ = \ P(a \cup b)
\\ \
\\ P(a \cup b) = P(a) + P(b) - P(a \cap b)
\]

Answer 10

yo explain why 80% of who passed 2nd exam passed the first one

Answer 11

n * 6/10 = 48/100
n = 8/10

Answer 12

oh, read that wrong

Answer 13

yeah, 80% is totally not right. THat's just the percent of students who passed the first exam. oops.