Got any Maths questions? Post and I'll try and help.
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Ok, the hint here is that each robot will aim for the guy with the highest accuracy cos hes the biggest threatOriginally posted by CoolMyth:Okay, I don't know the answer. Care to enlighten me by showing the answer?
Three scenarios for A to survive
1. B miss C, then C kills B instantly(100%), then A kills C
0.5x1x0.1 = 0.05
2 B kills C, then A miss B, then B miss A, then A kills B(cos aim second time)
0.5x0.9x0.5x0.1 = 0.0225
3 B kills C, then A kills B (quite heng)
0.5x0.1 = 0.05
Total probabilty = 0.05 +0.0225+0.05 = 0.1225 aka 12.25%
Phew cant believe I typed all that
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Actually i not too sure how to use now alsoOriginally posted by CoolMyth:Chocolate bar charts are just a way to get into the introduction of simultanous equations.
Of course, my maths is not that good in knowing whether there can be any differences in solving using different methods. Will you like to show me? Thanks.
My tenant at home used to stump me with questions from his primary 6 maths book... thats where i learned that u can solve anything with chocolate bar... One of the questions is something like this which must be solved with choco bar.... but i managed only after painful hours -
Not bad.....learned something new. But strictly speaking, if you want to be more throughout, shall add one more:Originally posted by DriftingGuy:Ok, the hint here is that each robot will aim for the guy with the highest accuracy cos hes the biggest threat
Three scenarios for A to survive
1. B miss C, then C kills B instantly(100%), then A kills C
0.5x1x0.1 = 0.05
2 B kills C, then A miss B, then B miss A, then A kills B(cos aim second time)
0.5x0.9x0.5x0.1 = 0.0225
3 B kills C, then A kills B (quite heng)
0.5x0.1 = 0.05
Total probabilty = 0.05 +0.0225+0.05 = 0.1225 aka 12.25%
Phew cant believe I typed all that
B miss A, A hits C, B miss A again, A hits B.
Then, it's 0.5 x 0.1 x 0.5 x 0.1.....
I was kinda confused about where B will aim first....that got me stuck.....
Oh yeah, what about the case C is the second to shoot.....?
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Originally posted by CoolMyth:
er.. nope.. i purposely put there that B or A wont miss the second time cos they are robots.. also cos it will more simpler
Not bad.....learned something new. But strictly speaking, if you want to be more throughout, shall add one more:
[b]B miss A, A hits C, B miss A again, A hits B.
Then, it's 0.5 x 0.1 x 0.5 x 0.1.....
I was kinda confused about where B will aim first....that got me stuck.....
Oh yeah, what about the case C is the second to shoot.....? [/b] -
I see.....Originally posted by DriftingGuy:Actually i not too sure how to use now also
My tenant at home used to stump me with questions from his primary 6 maths book... thats where i learned that u can solve anything with chocolate bar... One of the questions is something like this which must be solved with choco bar.... but i managed only after painful hours
Quite true.....primary school maths can sometimes be harder than secondary school maths....not because it's really hard, but you are restricted by what they know and also the difficulty of how to show them then they know.....
Maybe there's a third condition that is miss out in the fowl question. Everytime, I always use simultanous first then, do back to use chart method.
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Oh yeah....doh! I miss that condition.....Originally posted by DriftingGuy:er.. nope.. i purposely put there that B or A wont miss the second time cos they are robots.. also cos it will more simpler
So either way, if C go second, then A sure die.... -
Maybe la, if u can make a third equation that makes sure x,y and z is whole number then i can pray to you as godOriginally posted by CoolMyth:I see.....
Quite true.....primary school maths can sometimes be harder than secondary school maths....not because it's really hard, but you are restricted by what they know and also the difficulty of how to show them then they know.....
Maybe there's a third condition that is miss out in the fowl question. Everytime, I always use simultanous first then, do back to use chart method. -
Hey! u guys 4got about me?
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Why not? I just assume an answer I want then I can get third equation.....Originally posted by DriftingGuy:Maybe la, if u can make a third equation that makes sure x,y and z is whole number then i can pray to you as god -
let x=hen, y=rooster, z=chickOriginally posted by CoolMyth:Why not? I just assume an answer I want then I can get third equation.....
x+y+z=100 (1)
6x+3y+0.1z=100 (2)
i will make a third eqn, taking z away in this case...
taking (1)...
z=100-x-y (3)
substitude (3) into (2)
6x+3y+0.1(100-x-y)=100
braaahhh...braahhh...i will get
59x+29y=900
looking at x, the possible range is from 0 to 15...tabulating a table..only x =1 is valid....
so when x = 1, y=29, therefore z = 70
so 1 hen, 29 rooster and 70 chicks from geylang... -
help me prove that greene's theorem and stoke's theorem are bullshit!
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use delta epsilon to prove that for any IQ of anyone, the.raven's IQ is always higher.
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Well, they are not crap, but so far, I haven't reached the level where they are really essential, mainly just something to take note of.Originally posted by hum kah leng:help me prove that greene's theorem and stoke's theorem are bullshit!
Therefore, sorri har, cannot help you.... -
wad are those??? never hear before... haha...Originally posted by CoolMyth:Well, they are not crap, but so far, I haven't reached the level where they are really essential, mainly just something to take note of.
Therefore, sorri har, cannot help you.... -
How about this?Originally posted by the.raven:use delta epsilon to prove that for any IQ of anyone, the.raven's IQ is always higher.
For any person that has a lower IQ than the raven, there will also be at least one person whose IQ is higher than the raven's IQ.
This can be easily seen through the density theorem. -
Okay, I'm feeling bo-liao right now, so I'll list out their definition:Originally posted by starblue:wad are those??? never hear before... haha...
Green's Theorem: Let C be a positive oriented, piecewise-smooth, simple closed curve in the plane and let D be the region bounded by C. If P and Q have continuous partial derivatives on an open region that contains D, then:
|(C) P dx + Q dy = ||(D) (@Q/@x - @P/@y) dA
Stokes' Theorem: Let S be an oriented piecewise smooth surface that is bounded by a simple, closed, piecewise smooth boundary curve C with positive orientation. Let F be a vector field whose components have continuous partial derivatives on an open region in R^3 that contains S, then:
|(C) F * dr = ||(S) curl F * dS
Come to think, peharps I did do some questions based on this before.....hahahaha.....but now forget already.....I must go do some revision....