2011 hsc mx1 marathon (3 Viewers)

nightweaver066

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there is only one variable x and the constant so just let x = 0

definite integrals wont work if you hav to integrate twice
Thanks for that, had slipped my mind after i couldn't find a question where i had actually used it..
 

taeyang

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Woops, Next Question.... Uhh let me just make one up..



May require some -------------------lateral thinking---------------------
 

artosis

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Woops, Next Question.... Uhh let me just make one up..



May require some -------------------lateral thinking---------------------
thats a 4u q1 question.
but nonetheless its still adequate for 3u students
 

artosis

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since no one posted a question heres a fun one:
<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{-1}^{1}\frac{1}{x^2} dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{-1}^{1}\frac{1}{x^2} dx" title="\int_{-1}^{1}\frac{1}{x^2} dx" /></a>
 

hup

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since no one posted a question heres a fun one:
<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{-1}^{1}\frac{1}{x^2} dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{-1}^{1}\frac{1}{x^2} dx" title="\int_{-1}^{1}\frac{1}{x^2} dx" /></a>
lol
invalid but still works out
 

acemusic415

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Find the probability that of 5 cards chosen from a 52-card deck that 2 will be of the same suit.
 

apollo1

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Screen shot 2011-10-14 at 4.06.56 AM.png

i got the number of objects as n = 7. can someone confirm my answer.
 

Comeeatmebro

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Don't you reckon you guys should be doing a bit of English??? :S :S :S
 

someth1ng

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Find the probability that of 5 cards chosen from a 52-card deck that 2 will be of the same suit.
P(pair)=1

Okay,
Ways to pick the loner=7
Ways to create 3 pairs= 6C2 * 4C2
Total ways=7 * 6C2 * 4C2 = 630 ways

Ways that not together = 630 - Ways together
Ways together = 5 * 4C2 = 30 ways
.`. Ways that not together = 600 - 30 = 600 ways.
 
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