Acertain group of women has a 0.61% rate of red/green color blindness. if a woman is randomly selected, what is the probability that she doesn't have red/green blindness? what is the probability that the woman selected does not have red/green color blindness?

99.39%

Step-by-step explanation:

Given : A certain group of women has a 0.61% rate of red/green color blindness.

We can write 0.61%= 0.0061 [by dividing it by hundred]

Let E be the event of a woman has red/gree.n blindness

Then, the probability that a woman has red/green blindness :

P(E)= 0.0061

Now, there are only two possible events , one is "E" and the other is "E not" .

Hence, the probability that the woman selected does not have red/green color blindness = 0.9939=99.39%

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.