MATHS METHODS EXAM 2 THOUGHTS/DISCUSSION

[letsride]

Crazy... tan... graph? Whaaaat?
Also, with the multiple choice question which described a derivative function (I can't remember which number but it was towards the end)- how can f'(2)=0 and f'(4)=0 and f'(x) between 2 and 4 be positive? How does the graph (of f'(x)) end up back down at 4 if it is increasing in that region? Or is there something I missed in that, which seems to be the only possible explanation?

the answer to that was at x=4 it was a stationary point of inflection

Thanks but I still don't get it. This is how I interpreted it: the gradient graph is shaped like _/_ but the only possible way to do that is to have a point of discontinuity, which would be undifferentiable so you wouldn't be able to get a second derivative to work out the nature of the stationary point.

jsut draw the graph, they gave you enough data to get a rough estimate of what it looked like.
yea i got x=4 stationary pt.

[sooeyy]

I think it is 1- (p+q)

I choose this too.

I subbed in points. so Pr(A)=.4 and Pr(B)=.2 , therefore (Pr A' and B') = .4 so to get Pr( A' and B' ) as 0.4 = 1 - (.4 +.2).

Hopefully that's right :S

Itute helper said it was C...I guessed C. THANK GOD. :]

[qshyrn]

I think the so called crazy tan graph was on of the options in the MC where it gave you the derivative and you gotta find the original function.

WHAT??? which one was this  dont remember anything about tan except q1 MC

[grace10111]

i drew a pretty flower

[Elnino_Gerrard]

What u guys get for that expected number of statues question?

[Whatlol]

What u guys get for that expected number of statues question?

i wrote down like 1.1 or 1.3 something like that.... lol probs wrong

[curious111]

What u guys get for that expected number of statues question?

1.2, something like that.

[slothpomba]

What u guys get for that expected number of statues question?

1 (1.3 but i rounded it down to 1, you can only have whole statues afterall...)

[bubsy]

Did anyone get the number of statues that had to be made as 18?

Ya I did. Trial and error yeah?

i did it by using a table in graph and tab and got 18

Same thing, same thing. All good though.

X~Bi(n,.20)
Pr(x>=2)=0.90
Pr(x<2)=0.10
Pr(X=0)+Pr(X=1)=0.10
(n 0)(.2)^0(.^n + (n 1)(.2)(.^(n-1)=0.10
(.^n + n(.2)(.^(n-1)=.10

Solve for n using CAS
n=17.9479
~~18.

Should be right...>.<

RAWR I put that into my calculator and it wouldn't solve! Should've thought of the table of values but I was too busy being raped by pyramids.

[98.40_for_sure]

121/100 for E(X) one i think

[YDragon]

Crazy... tan... graph? Whaaaat?
Also, with the multiple choice question which described a derivative function (I can't remember which number but it was towards the end)- how can f'(2)=0 and f'(4)=0 and f'(x) between 2 and 4 be positive? How does the graph (of f'(x)) end up back down at 4 if it is increasing in that region? Or is there something I missed in that, which seems to be the only possible explanation?

the answer to that was at x=4 it was a stationary point of inflection

Thanks but I still don't get it. This is how I interpreted it: the gradient graph is shaped like _/_ but the only possible way to do that is to have a point of discontinuity, which would be undifferentiable so you wouldn't be able to get a second derivative to work out the nature of the stationary point.

You had a turning point at x = 4 for the derivative graph which would indicate its a stationary point of inflection... I don't think it was taught in the course, but if you derived the derivative to get the second derivative, you would notice that f''(5) > 0 and f''(3) < 0, and so concavity changes which means its a stationary point of inflection I think.

Also I got 1.21 statues. ^_^

[Whatlol]

I for some reason decided to round it up... reakon i will still get some marks?

[Linkage1992]

awesome i got 1.21 as well.

[Elnino_Gerrard]

BAh i forgot to add one of the probabilties got 0.999 :|

[qshyrn]

I for some reason decided to round it up... reakon i will still get some marks?

YOU CANT ROUND E(X)