According to a center for disease control, the probability that a randomly selected person has hearing problems is 0.157. the probability that a randomly selected person has vision problems is 0.096. can we compute the probability of randomly selecting a person who has hearing problems or vision problems by adding these probabilities? why or why not?

According to a center for disease control. We can’t compute the probability.

SOLUTION:

Given that, According to a center for disease control,  

The probability that a randomly selected person has hearing problems is 0.157.  

The probability that a randomly selected person has vision problems is 0.096.  

We have to find whether we can compute the probability of randomly selecting a person who has hearing problems or vision problems by adding these probabilities or not?  

The answer is no, because hearing and vision problems are not mutually exclusive.

So, some people have both hearing and vision problems.

These people would be included twice in the probability.

Hence, we can’t compute the probability.

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

about 0.75 hours

step-by-step explanation: