Igive a medal to anyone to !
3 party-goers are in the corner of the ballroom having an intense argument. you walk over to settle the debate. they are discussing a function g(x). you take out your notepad and jot down their statements.
• professor mccoy: she says that 2 is a zero of g(x) because long division with (x + 2) results in a remainder of 0.
• ms. guerra: she says that 2 is a zero of g(x) because g(2) = 0.
• mr. romano: he says that 2 is a zero of g(x) because synthetic division with 2 results in a remainder of 0.

Well, the way I see it is that both Mr. Romano, and Ms. Guerra are correct but Professor McCoy is incorrect because he said (x+2) when it should be (x-2).

The factor theorem states that:
If f(a)=0, then (x-a) is a factor

The remainder theorem states that:
If (x-a) is a factor of f(x), then f(x) / (x-a)  = 0

So if 2 is indeed a zero of f(x), then a factor must be (x-2) according to the fist which supports Ms. Guerra and also if (x-2) is indeed a factor as Ms. Guerra says then we know that f(x) / (x-a) = 0 which supports Mr. Romano

Professor McCoy is wrong because he used (x + 2) when it should be (x-2). I know this because according to the factor theorem if f(a)=0, then (x-a) is a factor. And the remainder theorem says if (x-a) is a factor of f(x), then f(x)/x-a =0.

a) 0.82

b) 0.18

Step-by-step explanation:

We are given that

P(F)=0.69

P(R)=0.42

P(F and R)=0.29.

a)

P(course has a final exam or a research paper)=P(F or R)=?

P(F or R)=P(F)+P(R)- P(F and R)

P(F or R)=0.69+0.42-0.29

P(F or R)=1.11-0.29

P(F or R)=0.82.

Thus, the the probability that a course has a final exam or a research paper is 0.82.

b)

P( NEITHER of two requirements)=P(F' and R')=?

According to De Morgan's law

P(A' and B')=[P(A or B)]'

P(A' and B')=1-P(A or B)

P(A' and B')=1-0.82

P(A' and B')=0.18

Thus, the probability that a course has NEITHER of these two requirements is 0.18.