According to a center for disease control, the probability that a randomly selected person has hearing problems is 0.155. The probabiity that a randomly selected person has vision problems is 0.084. Can we compute the probabiitiy of randomly selecting a person who has hearing problems or vision problems by adding these probabilities

23.9% probabiitiy of randomly selecting a person who has hearing problems or vision problems.

Step-by-step explanation:

We have these following probabilities:

0.155 = 15.5% probability that a randomly selected person has hearing problems.

0.084 = 8.4% probabiity that a randomly selected person has vision problems.

Can we compute the probabiitiy of randomly selecting a person who has hearing problems or vision problems by adding these probabilities

These probabilities are independent(a person having hearing problems is independent from having vision problems), which means that we can add these probabilities.

15.5 + 8.4 = 23.9%

23.9% probabiitiy of randomly selecting a person who has hearing problems or vision problems.

493 would be the correct answer

step-by-step explanation:

answer: h=5.6 i don't really know how to put into words, so i though i would let you do that. i hope i was able to some

step-by-step explanation:

18 = 6(h)+1/2

18 = 6h+1/2

18-1 = 6h/2

17 = 3h

17/3 = h

h = 5.6