High school students often take more than one standardized test as a means to ensure their acceptance into college. At one high school all 900 seniors report which test they take to their guidance counselor. 250 seniors reported taking the SAT, 200 seniors reported taking the ACT, and 90 seniors reported taking both the ACT and SAT. If a senior is selected at random, what is the probability they took the ACT or the SAT?
A) 0.50
B) 0.40
C) 0.30
D) 0.20

B) 0.40

Step-by-step explanation:

Let's say ACT is A and SAT is B.

There are 200 seniors taking ACT out of 900 students, so P(A) = 200/900. There are 250 seniors taking SAT, so P(B) = 250/900. There are 90 seniors taking both, so P(A∩B) = 90/900

We are asked the probability for the ACT or SAT, in other words P(AUB). We already have the necessary number, so the calculation will be:

P(AUB) = P(A) + P(B) - P(A∩B)

P(AUB) = 200/900 + 250/900 - 90/900

P(AUB) = 360/900=0.4

idk lol xd

step-by-step explanation:

ez do it yourself lazy

so, converting firstly the mixed fractions to improper fractions and then "adding".