Suppose that 6 out of the 19 doctors in a small hospital are General Practitioners, 9 out of the 19 are under the age of 45, and 3 are both General Practitioners and under the age of 45. What is the probability that you are randomly assigned a General Practitioner or a doctor under the age of 45

12/19

Step-by-step explanation:

We have these events

A = Doctors are general practitioners

B = doctors are under the age of 45

P(A) = 6/19

P(B) = 9/19

P(A n B) = 3/19

Then the probability of being assigned a general practitioner under age 45

P(AuB)

= P(A) + P(B) - P(A n B)

= 6/19 + 9/19 - 3/19

= 12/19

So 12/19 is the probability of getting a general practitioner under age 45.

a

step-by-step explanation:

its wrong

x^2 + 81

we can factor this using difference of square binomials.

rewrite the equation:

x^2 - (-81)

factor:

(a^2 - b^2) = (√a^2 + √b^2)(√a^2 - √b^2)

(x^2 - (-81) = (x + √-81)(x - √-81)

we can remove the - from under the radical by multiplying it by i.

(x^2 - (-81) = (x + i√81)(x - i√81)

(x^2 - (-81) = (x + 9i)(x - 9i)

we can check our answer by using foil:

x^2 - 9ix + 9ix - 81i^2

x^2 - 81^2

i^2 = -1

x^2 + 81