several people at a local mall were surveyed about the kind of vehicle they drive and whether they own or rent their home. Use the results listed in the table to find the probability that a randomly selected person from the survey either rents their home or drives an SUV.

Answer as a fraction = 18/25

Answer in decimal form = 0.72

Answer in percent form = 72%

Explanation:

We have A = 6+6+1 = 13 people who rent their home (add the values along the bottom row). There are B = 5+6 = 11 people who drive an SUV (add up the values in the SUV column). There are A+B = 13+11 = 24 people who either rent their home or drive an SUV. Keep in mind that 6 people have been double counted if we just go with A+B without adjustment. These 6 people both rent their home and drive an SUV. To correct for the double-count, we subtract off 6 from A+B to get A+B-6 = 24-6 = 18.

So there are 18 people who rent their home, drive an SUV, or both.

This is out of 4+5+3+6+6+1 = 25 people surveyed total.

The probability of selecting someone who rents their home or drives an SUV is 18/25. In decimal form that is equivalent to 0.72 which converts to 72%

answer:

c. the third graph

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