At a school 60% of students buy a school lunch. Only 10% of students buy lunch and dessert. What is the probability that a student who buys lunch also buys dessert?

1/6

Step-by-step explanation:

P(A) = probability a student buys lunch = 60% = 0.6

P(B) = probability a student buys lunch and dessert = 10% = 0.1

We are looking at the probability that a student buys lunch and dessert on the condition that they already bought lunch. Therefore, we are looking at a conditional probability.

P(B given A) = P(A and B) / P(A)

= 0.1 / 0.6

= 1/6 ≈ 0.167

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer