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Haha what brings you to that?

That's just stupid...

Let u=x-2, du/dx=1 When x=4 u=2 When x=0 u=-2 So I=\int^2_{-2} u^3 \ du = 0 (f(u)=u^3, f(-u) = (-u)^3 = -u^3 = -f(u), so f(u) represents an odd function ; The integration of an odd function about symmetrical limits is zero.)

LOL what are you doing on BOS?

Yeah if CI = [x bar +/- z_{1-alpha/2} rho/sqrt(n)] then the margin of error m = z_{1-alpha/2} rho\sqrt{n} ie. CI = [x bar +/- m] (or whatever respective estimator you want)

Post the question here.

Well, if one has done a bit of linear algebra, then one would conclude that it is a polynomial in P^2 (R)... ie a vector (s,t,0)^T with respect to the standard basis {1,x,x^2} for s,t real... So it still could be a quadratic. It just depends how you define one...

Use the Taylor series expansion.... (joke -- You'd require like 30 terms for 4dp accuracy)

I'd assume that you'd need to know regression well...

In maths they did (obviously before online marking).

Also, it isn't all of the parabola. It's the part of parabola for which x <= -2 or x>=2, and y >= 2...

From what I've heard, although accounting is very practical and useful in life, it is very boring.

There was no other lecture stream, or lecture recordings.

Isn't that a parabola? ...

He stopped coming to lectures after Week 3...

If deg of numerator = deg of denominator then a horizontal asymptote exists. It is then given by y=(coeff of highest term in numerator)/(coeff of highest term in the denominator). So in your example it would be y=(-1)/1 ie. y=-1. And you don't do division of graphs usually. In Graphs you mainly...

Not true. One of my friends does nothing all semester. Then gets 85+ as his final mark (usually around 87-90).

If I remember my first/second year university maths correctly, differentiability is very hard to prove (continuity, however is not). But, it should be any "smooth" curve with the endpoints defined...

I believe maths teachers can take the papers home to mark, as long as they don't mark it at school during teaching... Also, portable airconditioners come to mind...

Lol conditional probability... and yes I think the given answer is wrong too... I get 1/3 using 2 different ways Method 1 (conditional probability method): P(at least one red) = 1- (3/7)*(1/3) = 6/7 P(both red) = 4/7 * 1/3 = 4/21 So P(both red | at least one red) = (4/21)/(6/7) = 1/3...