ATAR Notes: Forum
Archived Discussion => 2010 => End-of-year exams => Exam Discussion => Victoria => Mathematical Methods CAS => Topic started by: milkcarton on November 08, 2010, 11:48:42 am
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How'd everyone do??
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didn't it start like.. ten minutes ago?
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Pretty tough. Especially M.C #19 and #20. Also the Markov Chains on question 3 (or 4), plus that crazy tan graph. Nobody saw that coming.
Can somebdy plz give me a Study score: pr0x:
GA1: F+
Ha2: C
Ia3: D+
thx a million. Howd everyone else find it?
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Was there yet another Tasmania Jones question this year?!
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i just got owned.
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Did anyone get the number of statues that had to be made as 18?
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Yay, I love how easily a whole years worth of hard work can so easily amount to nothing..
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Was there yet another Tasmania Jones question this year?!
Nope... but this year we had to deal with his cousin, who makes statues! (probability q, curious111 is referring to.)
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Did anyone get the number of statues that had to be made as 18?
yes i got that.
Also chavi was mcq 20 the one with e^modulus x ?
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Did anyone get the number of statues that had to be made as 18?
Ya I did. Trial and error yeah?
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Pretty tough. Especially M.C #19 and #20. Also the Markov Chains on question 3 (or 4), plus that crazy tan graph. Nobody saw that coming.
Can somebdy plz give me a Study score: pr0x:
GA1: B+
Ha2: A
Ia3: A+
thx a million. Howd everyone else find it?
You should get 50! Don't worry ;D .. worst case scenario you'll come out with a 48 raw for methods ;)
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Did anyone get the number of statues that had to be made as 18?
Did you get 18? I only got 2... (probably wrong...) I got like 2.994 or something, and rounded it down to 2
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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
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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.
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can someone upload the exam, i'd loveeee to get going on some worked sonutions..
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Did anyone get the number of statues that had to be made as 18?
Yep, 18, but then again I just got completely and utterly raped.
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Fuck you vcaa. That is all.
Probably get a fkn 35
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Might be just me but.... It was harder than specialist lol.
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Fuck you vcaa. That is all.
Probably get a fkn 35
+1
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What did people get for 'p' in the last question?!?!
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What did people get for 'p' in the last question?!?!
i only got decimal answer so i wrote 13.8/4 + 1/4
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We should bomb VCAA HQ. They will pay for this.
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I got totally ripped apart.
All that practice I had done in preparation for this exam flew out the door.
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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(.8)^n + (n 1)(.2)(.8)^(n-1)=0.10
(.8)^n + n(.2)(.8)^(n-1)=.10
Solve for n using CAS
n=17.9479
~~18.
Should be right...>.<
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Fuck you vcaa. That is all.
Probably get a fkn 35
+1
+1000000
damn! stupid vcaa
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Did people think that exam was harder than 2007 cas?
2007 saw 62.5/80 as A+ threshold, can presume hope the threshold is similar this year?
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Definitely harder than 2007.
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Did people think that exam was harder than 2007 cas?
2007 saw 62.5/80 as A+ threshold, can presume hope the threshold is similar this year?
Might just be a bit of post-exam hysteria on my part, but that DEFINTELY seemed harder than any previous VCAA exam. So perhaps the threshold will be around that.
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My CAS ruined it for me
Took ages to input the equations
then it left me with "no solutions" and "insufficient memory"
DAMN YOU CASIO
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What did people get for 'p' in the last question?!?!
i only got decimal answer so i wrote 13.8/4 + 1/4
Yeah I did it wrong. Just verifying now...>.>
I still don't know how to do it. >.>. My friend got 4...Saying something like getting 4 simultaneous equations. >.> I had no time at all to do it. >.<
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Aargh, failed to finish the last part of question 4...I thought that exam was cruelly tough.
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What did people get for 'p' in the last question?!?!
i only got decimal answer so i wrote 13.8/4 + 1/4
Yeah I did it wrong. Just verifying now...>.>
I still don't know how to do it. >.>. My friend got 4...Saying something like getting 4 simultaneous equations. >.> I had no time at all to do it. >.<
thats close to four :p
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My CAS ruined it for me
Took ages to input the equations
then it left me with "no solutions" and "insufficient memory"
DAMN YOU CASIO
THIS IS WHY TINSPIRE TRUMPS ALL.
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It should be 3..
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What did people get for 'p' in the last question?!?!
i only got decimal answer so i wrote 13.8/4 + 1/4
Yeah I did it wrong. Just verifying now...>.>
I still don't know how to do it. >.>. My friend got 4...Saying something like getting 4 simultaneous equations. >.> I had no time at all to do it. >.<
thats close to four :p
Yeah means you did it correctly xD. How did you do it? D:
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i got 10,000,000 not 18 :P
Haha best answer yet
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GOOD BYE my preferred uni course...... GOOD BYE….
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How did people do 21 in MC? Mutually exclusive....
Pr(A)=p and Pr(B)=q..
Find Pr(A' (intersect) B')
>.> I did C...as a last minute thing. LOL I shaded the box when the lady said PENS DOWN >.>
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What did people get for 'p' in the last question?!?!
i only got decimal answer so i wrote 13.8/4 + 1/4
Yeah I did it wrong. Just verifying now...>.>
I still don't know how to do it. >.>. My friend got 4...Saying something like getting 4 simultaneous equations. >.> I had no time at all to do it. >.<
I got p=4, and only had 2 simultaneous equations?
How did people do 21 in MC? Mutually exclusive....
Pr(A)=p and Pr(B)=q..
Find Pr(A' (intersect) B')
>.> I did C...as a last minute thing. LOL I shaded the box when the lady said PENS DOWN >.>
Draw a Karnaugh map? Thats how i did it.
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How did people do 21 in MC? Mutually exclusive....
Pr(A)=p and Pr(B)=q..
Find Pr(A' (intersect) B')
>.> I did C...as a last minute thing. LOL I shaded the box when the lady said PENS DOWN >.>
I got 1 - (Pr(A)+Pr(B))
Dunno what option that was nor if it was right...
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what a waste of a year.. what do you guys think the b+/b cutoff will be? that's the best I can aim for now. killed the 09 exam and failed '10. fuck vcaa
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How did people do 21 in MC? Mutually exclusive....
Pr(A)=p and Pr(B)=q..
Find Pr(A' (intersect) B')
>.> I did C...as a last minute thing. LOL I shaded the box when the lady said PENS DOWN >.>
I got 1 - (Pr(A)+Pr(B))
Dunno what option that was nor if it was right...
XD. Did you have a method? Cos I just drew a karnaugh map...and that got me nowhere. lawl. epic fail. ><
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I think it is 1- (p+q)
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that was the stupidest/hardest exam vcca has eva dished out...them writers of this paper should get shot :(
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plus that crazy tan graph.
what tan graph????
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Yay, I love how easily a whole years worth of hard work can so easily amount to nothing..
agreed.
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I think it is 1- (p+q)
+1, the events do not intersect so the intersection of A' and B' is just 1 minus both the probabilities of A and B.
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I think it is 1- (p+q)
+1, the events do not intersect so the intersection of A' and B' is just 1 minus both the probabilities of A and B.
Yep got this. WOO!
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plus that crazy tan graph.
what tan graph????
LOL A tan graph??? where what question??
Yep I definitely failed now....
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Pretty tough. Especially M.C #19 and #20. Also the Markov Chains on question 3 (or 4), plus that crazy tan graph. Nobody saw that coming.
Can somebdy plz give me a Study score: pr0x:
GA1: B+
Ha2: A
Ia3: A+
thx a million. Howd everyone else find it?
err what q was the tan graph?
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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?
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There was none? Question 1 was a MC and the answer was 3pi because the period is pi/(1/3) (Referring to tan graph).
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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
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I HATE VCAAA
WHAT A BITCHH
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Ripper exam. Loved every minute of it. Goodbye commerce/law at Monash hello commerce at VU.
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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
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seriously fucked on.
First two short answers were good, except for the minimum number of statues produced one, forgot how to do that on my cas. Then the derivitave with cos^6 etc, that was just fucked, i had no idea. And pretty much the last 2 thirds of the last questoin where it was asking for values of a and stuff where there are S.P's, i thought it has something to do with a discriminate, but cubics dont have discriminates ><
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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
+1.
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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.
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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.
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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.
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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. :]
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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
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i drew a pretty flower
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What u guys get for that expected number of statues question?
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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
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What u guys get for that expected number of statues question?
1.2, something like that.
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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...)
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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(.8)^n + (n 1)(.2)(.8)^(n-1)=0.10
(.8)^n + n(.2)(.8)^(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.
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121/100 for E(X) one i think
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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. ^_^
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I for some reason decided to round it up... reakon i will still get some marks?
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awesome i got 1.21 as well.
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BAh i forgot to add one of the probabilties got 0.999 :|
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I for some reason decided to round it up... reakon i will still get some marks?
YOU CANT ROUND E(X)
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Darn, I got 1.9 statues I think. This isn't a good sign.
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121/100 for E(X) one i think
+1 except as 1.21
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Aw damn I rounded mine =/
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got 1.21 aswell~~drew out the prob distribution table= =
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Darn, I got 1.9 statues I think. This isn't a good sign.
Apparently, you had to draw up a distribution table according to how many S popped up in the tree diagram...then find E(X) from that...i got something like 2.68___...but the answer using this methods got 1.21..strange. >.<
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i got 1.61
but you have to round down right??? cause you cant have .61 of a statue
so final answer is 1
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I for some reason decided to round it up... reakon i will still get some marks?
YOU CANT ROUND E(X)
It's sort of hard to produce 1.2 statues in that size though..
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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.
Chavi hasn't even seen the methods exam LOL
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Would i get method marks if i did it the right waY? drawing the table wrting Pr(s=0) = blajh blah and so on?
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Horrible exam. But only need 25 (melbourne), so hopefully good enough, I think i got 1.9 for the number of statues, probs wrong tho. didn't answer the las part of question 4 ehhhhhhhh HORRIBLE
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I for some reason decided to round it up... reakon i will still get some marks?
YOU CANT ROUND E(X)
It's sort of hard to produce 1.2 statues in that size though..
I'm assuming this was a question about E(X) because of the previous post..
If so, you must leave it unrounded. You leave the mean as is, no matter if it is physically possible.
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I got 2.something for number of statues and rounded down to 2 lol. Think I'm gonna get 0/4 for that :(
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same got 2.39 :(
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I got 1.98 and rounded to 2, urgh exam nerves got me there.
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I think I got 2 as well, how did u do that question?
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i got 1.1, hopefully i wrote in such a hurry that my 1.1 looks like a 1.2 :)
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I think I know where I went wrong...the 3 statues was INCLUDING the first one that wasn't Superior, wasn't it? I did it for the 3 statues after the one that wasn't superior. Bummer.
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The statue question was 1.21. The expect value could be found by finding the probablities for the transition matrix given that p = 0.75 and then drawing a tree diagram. Once this was implemented, it was only a matter of multiplying down the branches and adding the three different probabilities together, which give you the expected value.
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I can't even remember how i did it, LOL
I think I used a transition matrix to find the probability of superior after 3 trials then use that probability in biniomnalCDF/PDF
F**K IF I REMEMBER BUT I GOT THAT BULLSHIT ANSWER which is most probably wrong :(
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I thought expected meant mean?
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The easiest way to handle the statue question would have to made 4 outcomes: RRR, RRS, RSR, RSS, then work out the probabilities of each, then make a discrete random variable X which is the number of Superior statues...
So Pr(RRR) = Pr(X = 0), Pr(RRS) + Pr(RSR) = Pr(X = 1) and Pr(RSS) = Pr(X = 2)
Then you apply the mean formula to get 1.21... so in the end your mean should have been:
Pr(RRS) + Pr(RSR) + 2 * Pr(RSS)
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^^
+1
Pretty hectic.
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I rounded to one. is that wrong?
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The easiest way to handle the statue question would have to made 4 outcomes: RRR, RRS, RSR, RSS, then work out the probabilities of each, then make a discrete random variable X which is the number of Superior statues...
So Pr(RRR) = Pr(X = 0), Pr(RRS) + Pr(RSR) = Pr(X = 1) and Pr(RSS) = Pr(X = 2)
Then you apply the mean formula to get 1.21... so in the end your mean should have been:
Pr(RRS) + Pr(RSR) + 2 * Pr(RSS)
+1
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Guys its just one q. settle. the whole thread about this one q. And lets be honest. that q wasnt hard in relation to the whole paper. >.<
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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
Yeah i got 18 but i used pr(X>2)= 1-Pr(X=0)-Pr(X=1)
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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.
Chavi hasn't even seen the methods exam LOL
Oh LOL
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for the first 3 parts of the probability question i left the answer in exact form. ie 7/8. I dont remember them asking for decimal, but it was probability do you think i would lose marks?
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What did people get for the volume?
I got (4000(root) 3)/27 or something like that...
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What did people get for the volume?
I got (4000(root) 3)/27 or something like that...
i got that :D :D :D :D :D :D :D :D
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That was a BITCH of a question!
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What did people get for the volume?
I got (4000(root) 3)/27 or something like that...
ditto. fucku vcaa
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The last question screwed me over. SOOOO bad. That is all.
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The last question screwed me over. SOOOO bad. That is all.
ditto. vcaa are fags of the highest order.
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I lost nine marks from the last question ALONE. I need to find a nice hole to die in now.
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I lost nine marks from the last question ALONE. I need to find a nice hole to die in now.
same. vcaa go and get fucked.