Ask Question
16 September, 11:24

According to an automobile association of america report, 9.6% of americans traveled by car over the 2011 memorial day weekend and 88.09% stayed home. what is the probability that a randomly selected american stayed home or traveled by car over the 2011 memorial day weekend? p (stayed home or travelled by car) =

+2
Answers (1)
  1. 16 September, 11:50
    0
    p (traveled by car) = 9.6%

    p (stayed home) = 88.09%

    p (stayed home or travelled by car) = p (stayed home) + p (traveled by car) - p (stayed home and traveled by car)

    p (stayed home and traveled by car) = 0%

    p (stayed home or travelled by car) = 88.09%+9.6%-0%

    p (stayed home or travelled by car) = 97.69%
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “According to an automobile association of america report, 9.6% of americans traveled by car over the 2011 memorial day weekend and 88.09% ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers