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.

step by step explanation: when u removing the parentheses and combined them that would be : x+3y-x-y=2y

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: