ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE General & Further Mathematics => Topic started by: Mikey123 on October 21, 2008, 10:11:42 pm
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Well I do this subject via distance ed. Need help with the question 12 from the 2007 exam.
If anyone can do it on the calculator please spill :D.
Love ya ;)
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post the question here :P
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the three median line one? had problems with that also..the only way i think you can do it is by figuring out the median points of the outside groups.
either that or you input all the data on the graph on the calculator and then use the med-med function.
either way takes quite a bit of time..
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- Split the points into three groups (so that there is an even number of points in each - should be 12)
- Find the median points for the lower and upper group by putting in all the X values in L1 and all the Y values in L2... simply use a 1-variable statistic to find the median for each list.. you should get: (6.5, 3000) and (30.75, 7000)
- use the gradient formula to work out the slope
7000 - 3000 / 30.75 - 6.5
= 167
= C
*I went over that pretty quick so I can explain it in more steps if you still don't understand
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It's very easy when explain cleary (jsimmo you did an excellent job). I found it difficult and got it wrong on my first trial. However, my error was only a minor miss-calculation and i eventually got the right answer by looking at the assessors reports - it's probably one of the hardest concepts in core IMO. I think they will definately test it this year.
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it's annoying trying to work out the position of those points
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Do you think? I got mine wrong initially because i drew my two lines over the points that are extremely high in that particular graph lol. I don't know why i did it, probably because i'm an idiot haha. But, finding the median point on both the x and y axis in the 1sr and 3rd quadran will be extremely easy if done correctly - i know this for a fact, and when i got the answer right for this questions the media points were easily visable. Also, the examiners don't want to be fiddling around for a long time, so they make the points easy to recognise.
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Heh, I found that one of the more straight-forward questions from last year. Funny how some people find certain areas easiser/more difficult than others. :/
The question that really fucked me up was the scale factor one... in Geo and Trig(Or graphs and relations, can't remember). It involved removing a proportion of height of a pyramid and then finding the ratio of remaining volume to original volume.
I found that much harder.
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Yeah, I find the geo+trig ratio questions to hard too.
This is a question I found to be quite challenging :P Have a go at it.
(http://img411.imageshack.us/img411/5662/ratioqhf6.jpg)
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^ E ?
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Ummm is it C?
EDIT: apparently I'm wrong...*off to work this out on paper*
EDIT2: ok don't mind me, I did 2^3:3^3 instead of 2^3:5^3, answer is E
doesn't look like it on the graph though :P
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Volume = Linear^3
Linear ratio (part removed:entire cone) = 2:5
Volume ratio = 2^3:5^3 = 8:125
Percentage removed of total volume = 8/125 * 100 = 6.4%
Therefore percentage remaining = 100% - 6.4% = 93.6%
Answer is E.
I did that really quickly so I may be wrong ???
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(http://img65.imageshack.us/img65/6104/solutionii7.jpg)
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lol thank god im not doing geo+trig
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(http://img65.imageshack.us/img65/6104/solutionii7.jpg)
this was on my in school trial exam yesterday lol