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October 04, 2025, 09:58:33 pm

Author Topic: VCE Methods Question Thread!  (Read 5713915 times)  Share 

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Sanguinne

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Re: VCE Methods Question Thread!
« Reply #2265 on: July 17, 2013, 08:48:19 pm »
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I think it is a typo because it doesn't state the standArd deviation
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zhe0001

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Re: VCE Methods Question Thread!
« Reply #2266 on: July 17, 2013, 08:58:10 pm »
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da hello,
this unit 2 question about log graphs
Interested in finding a logs asymptote, when its not the vertical asymptote along the y-axis.
eg. How would you find the asymptope of this function,


Thanks!

lzxnl

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Re: VCE Methods Question Thread!
« Reply #2267 on: July 17, 2013, 09:25:34 pm »
+1
Asymptote when argument of log function = 0
8x + 10 = 0
x = -5/4

If you want, transpose the equation to yield x = 1/8*(e^(y/3) - 10)
Asymptotes are formally found by letting x, y approach +- infinity.
Nothing of note happens if you let x approach +- infinity. If y approaches -inf, the exponential term drops to zero. We're just left with x = -10/8 = -5/4
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Professor Polonsky

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Re: VCE Methods Question Thread!
« Reply #2268 on: July 17, 2013, 10:06:45 pm »
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I think it is a typo because it doesn't state the standArd deviation
Well it is 0.2 according to the original question.

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Re: VCE Methods Question Thread!
« Reply #2269 on: July 20, 2013, 07:06:52 pm »
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does anyone know how to divide something like 5x-5 by x-5 on the ti-inspire
thanks

lzxnl

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Re: VCE Methods Question Thread!
« Reply #2270 on: July 20, 2013, 07:10:42 pm »
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Control divide to get to the fraction display
Or, put brackets around (5x-5) and (x-5) when dividing.
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Re: VCE Methods Question Thread!
« Reply #2271 on: July 20, 2013, 07:32:50 pm »
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Control divide to get to the fraction display
Or, put brackets around (5x-5) and (x-5) when dividing.

that was a bad example I mean to change the form of something like
5x-3  divided by x-7

lzxnl

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Re: VCE Methods Question Thread!
« Reply #2272 on: July 20, 2013, 07:34:47 pm »
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expand(fraction) will decompose (5x-3)/*(x-1) into 5 + 2/(x-1)
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Re: VCE Methods Question Thread!
« Reply #2273 on: July 20, 2013, 07:39:04 pm »
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expand(fraction) will decompose (5x-3)/*(x-1) into 5 + 2/(x-1)

thanks

zvezda

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Re: VCE Methods Question Thread!
« Reply #2274 on: July 20, 2013, 10:10:22 pm »
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Hey, how would i go about doing this q:
A spinner is numbered from 0 to 5, and each of the six numbers has an equal chance of coming up. A player who bets 1$ on any number win $5.00 if that number comes up, otherwise the 1$ is lost. What is the player's expected profit on the game?

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Re: VCE Methods Question Thread!
« Reply #2275 on: July 20, 2013, 10:16:27 pm »
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Let be the amount of money won (or lost), then (below is a discrete probability distribution table)

                                         
                          


That is the expected profit is a loss of

EDIT: That is interpreting it as you lose the $1 you bet, but get $5 back if you win, i.e. profit from that result of $4, they may have interpreted it differently, in which case you'd replace the 4 with a 5, and the expected profit would be zero.

e.g
                                         
                          


That is the expected profit is , i.e. a fair game.
« Last Edit: July 20, 2013, 10:19:28 pm by b^3 »
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zvezda

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Re: VCE Methods Question Thread!
« Reply #2276 on: July 20, 2013, 10:25:16 pm »
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Ahh ok. What i did was instead of having -1 in the distribution table i had 0.
Thanks !
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jimmy22

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Re: VCE Methods Question Thread!
« Reply #2277 on: July 21, 2013, 12:01:13 pm »
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How would you find an antiderivative of x/(x+1)   ?

cheers

Jeggz

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Re: VCE Methods Question Thread!
« Reply #2278 on: July 21, 2013, 12:09:18 pm »
+4
How would you find an antiderivative of x/(x+1)   ?

cheers

If we divide using long division, we will get it in the form




And because it says "an" you don't need the "+ c" at the end.
« Last Edit: July 21, 2013, 12:11:54 pm by Jeggz »
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jimmy22

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Re: VCE Methods Question Thread!
« Reply #2279 on: July 21, 2013, 12:13:04 pm »
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If we divide using long division, we will get it in the form




And because it says "an" you don't need the "+ c" at the end.

ohh ok. Thank you very much!