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need trial solutions (1 Viewer)
- Thread starter sleepycat
- Start date
freaking_out
Saddam's new life
well at the moment i guess its best if u post up the hard questions here...and get 'em answered.
N
ND
Guest
Ok i just did Q8, here are the solutions:
a) (ok i'm not so good at probability, so i'm not certain that this is correct, can someone please verify this?)
i)9P3
ii) Ok now consider one particular code, the prob that the numbers are increasing (i.e. 123) is 1/6, so the prob that it isn't increasing is 5/6.
.'. there are 5/6*9P3 codes where the number order isn't increasing.
b)
i) simply differentiate, get it as one fraction, then factorise the numerator.
ii) f'(x) = ((1-cosx)/(2+cosx))^2, .'. gradient is always positive.
stationary pt where cosx=1
x = 0.
subbing into f(x) gives 0.
'.' at x=0, y=0, and f'(x) is always positive for x>0:
f(x) > 0
x - (3sinx)/(2+cosx) > 0
x > (3sinx)/(2+cosx)
c)
i) sin(2r+1)@ - sin(2r-1)@ = 2sin@cos2r@ from sum to product rule.
2sin@((sum r=1-->n)cos2r@) = sin3@ - sin@ + sin5@ - sin3@ +...+ sin(2n+1)@ - sin2n@
= sin(2n+1)@ - sin@ (as everything else cancels)
ii) sin@((sum r=1-->n)cos2r@) = sin(2n+1)@ - sin@
(sum r=1-->n)(cos2r@)^2 = (sin(2n+1)@ - sin@)^2/(sin@)^2
now let n=100 and @=pi/200:
(sum r=1-->100)(cosr*pi/100)^2 = (sin(201*pi/200) - sin(pi/200))^2/(sin(pi/200))^2
Nothing like a bit of maths before hitting the clubs.
a) (ok i'm not so good at probability, so i'm not certain that this is correct, can someone please verify this?)
i)9P3
ii) Ok now consider one particular code, the prob that the numbers are increasing (i.e. 123) is 1/6, so the prob that it isn't increasing is 5/6.
.'. there are 5/6*9P3 codes where the number order isn't increasing.
b)
i) simply differentiate, get it as one fraction, then factorise the numerator.
ii) f'(x) = ((1-cosx)/(2+cosx))^2, .'. gradient is always positive.
stationary pt where cosx=1
x = 0.
subbing into f(x) gives 0.
'.' at x=0, y=0, and f'(x) is always positive for x>0:
f(x) > 0
x - (3sinx)/(2+cosx) > 0
x > (3sinx)/(2+cosx)
c)
i) sin(2r+1)@ - sin(2r-1)@ = 2sin@cos2r@ from sum to product rule.
2sin@((sum r=1-->n)cos2r@) = sin3@ - sin@ + sin5@ - sin3@ +...+ sin(2n+1)@ - sin2n@
= sin(2n+1)@ - sin@ (as everything else cancels)
ii) sin@((sum r=1-->n)cos2r@) = sin(2n+1)@ - sin@
(sum r=1-->n)(cos2r@)^2 = (sin(2n+1)@ - sin@)^2/(sin@)^2
now let n=100 and @=pi/200:
(sum r=1-->100)(cosr*pi/100)^2 = (sin(201*pi/200) - sin(pi/200))^2/(sin(pi/200))^2
Nothing like a bit of maths before hitting the clubs.
N
ND
Guest
I figure i may as well put up the question too.
freaking_out
Saddam's new life
can u scan up ya solutions as welll.... coz i hate reading the mathematical notations and having 2 convert 2 paper.
N
ND
Guest
I don't think you'll have much hope reading my handwriting. i don't have a scanner anyway.
i don't have a scanner anyway.
freaking_out
Saddam's new life
Originally posted by ND
I don't think you'll have much hope reading my handwriting.
...same here(abt the handwriting bit):mad1:
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