"All complex solutions... have non-zero... imaginary parts"
So
will yield an imaginary number, which would make q complex, where q should instead be a real number.
While \(\dfrac{\sqrt{b^2-4ac}}{2a}\) might be non-real, \(\left|\dfrac{\sqrt{b^2-4ac}}{2a}\right|\) is most certainly non-negative and real. Your answer is equivalent to mine, and the modulus brackets take care of it.
Multiple choice, Question 4: i^1! = i, and both i^2! and i^3! = -1, then i^n! = 1, for all n ≥ 4, since n! is a multiple of 4 for all n ≥ 4. This gives i 2 + 97*1.
I knew this doing the question, but I couldn't be bothered carefully counting when it's an MCQ.
Extended response 3a ii, I'm confused about your inequalities. I got, when 0 ≤ b < a, we have r > s (so that a b > 0 and r/s > 1).
\(k\) is also positive when \(a-b<0\) and \(0<r/s<1\), making the log negative as well. There are two possibilities, and I think both should be stated.
Extended response 6a. I think your calculation gives the chance of getting at least one sample with mean mass less than 370 or greater than 375. I did 1 (1 Pr(370 ≤ X ≤ 375))^2.
Yep, I screwed this up. My bad. In my defence, I had my Human Structure and Function exam 1 just before I did the solutions (so I was pretty tired) and I had my Differential Equations exam this morning, so I wasn't exactly taking my time with the solutions