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    Anyone got any tips for getting full marks on stats - after a dreadful C2 and a nervy C1 I need 100UMS desperately to get an A - what can I do ?


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    (Original post by reubenkinara)
    Has anyone given any thought to the pi vs tau debate? The fact that 2pi crops up numerously in maths and physics and that it'd make more sense for tau (2pi) to be used.
    I try to avoid these sort of debates. The arguments tend to be quite...

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    (Original post by justinawe)
    I try to avoid these sort of debates. The arguments tend to be quite...

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    Nice pun; so true though!
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    (Original post by krishkmistry)
    Anyone got any tips for getting full marks on stats - after a dreadful C2 and a nervy C1 I need 100UMS desperately to get an A - what can I do ?


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    Um, practice weak topics. Look at spec to ensure you know everything and do Solomons. Try and do a few 'regular' papers as mocks closer to the exam!
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    can I be a special helper? CHEERS.
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    does anyone know if we need to know the proofs of cosine and sine rules for edexcel for c2?
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    (Original post by reubenkinara)
    Just have 25(1-r^n)>24 substitute in r. Simplify a little. Tell me when you get stuck again!
    I subbed in 1/5 and now I'm stuck again



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    (Original post by charlieejobson)
    does anyone know if we need to know the proofs of cosine and sine rules for edexcel for c2?
    I don't think so, just need to know now to manipulate them. I'd get a second opinion just in case, but I'm not learning them


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    (Original post by hello.cupcake)
    I subbed in 1/5 and now I'm stuck again



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    Divide both sides by 25.
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    (Original post by reubenkinara)
    Divide both sides by 25.
    Wooo!! Thank you


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    (Original post by krishkmistry)
    Anyone got any tips for getting full marks on stats - after a dreadful C2 and a nervy C1 I need 100UMS desperately to get an A - what can I do ?


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    Try to relax... Not easy but trust me when I say (from far too much experience unfortunately) that stressing out only makes things worse.
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    BINOMIAL EXPANSION HELP:::::

    In the expansion of (1 + x )^n , the coefficient of x^5 is twice the coefficient of x^4. What is the value of n ?

    I keep getting different answers everytime I do it -__-
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    (Original post by Xx4L3x)
    BINOMIAL EXPANSION HELP:::::

    In the expansion of (1 + x )^n , the coefficient of x^5 is twice the coefficient of x^4. What is the value of n ?

    I keep getting different answers everytime I do it -__-
    What's the general form?

    1 + nx + n(n-1)/2! x^2 + ...

    So

    (2 n(n-1)(n-2)(n-3)) / 4! = (n(n-1)(n-2)(n-3)(n-4)) / 5!
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    (Original post by L'Evil Fish)
    What's the general form?

    1 + nx + n(n-1)/2! x^2 + ...

    So

    (2 n(n-1)(n-2)(n-3)) / 4! = (n(n-1)(n-2)(n-3)(n-4)) / 5!
    wow what ???
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    Just to reiterate - standard deviation shows how far/close the values are from the mean? So i low standard deviation suggests that the values are close to the mean and a high standard deviation suggests that the values are close to the mean?
    Does the graph always look like a symmetrical bell shape?
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    (Original post by Xx4L3x)
    wow what ???
    Expand it generally.

    Then equate co efficients of x^4 and x^5 making sure you multiply the co efficient of x^4 by two.

    Then solve as normal
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    (Original post by L'Evil Fish)
    Expand it generally.

    Then equate co efficients of x^4 and x^5 making sure you multiply the co efficient of x^4 by two.

    Then solve as normal

    I get confused.

    In the expansion of (1+x)^n does the pattern go like this?

    1 + nx + n (n - 1) / 2 + n (n - 2) / 6 .....

    OR

    1 + nx + n(n - 1) / 2 + n ( n - 2)(n - 3) / 6 ....

    ?
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    (Original post by Xx4L3x)
    I get confused.

    In the expansion of (1+x)^n does the pattern go like this?

    1 + nx + n (n - 1) / 2 + n (n - 2) / 6 .....

    OR

    1 + nx + n(n - 1) / 2 + n ( n - 2)(n - 3) / 6 ....

    ?
    It goes:

    1 + nx + n(n-1)/2! x^2 + n(n-1)(n-2)/3! x^3 + n(n-1)(n-2)(n-3)/4! x^4 + ...
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    (Original post by L'Evil Fish)
    It goes:

    1 + nx + n(n-1)/2! x^2 + n(n-1)(n-2)/3! x^3 + n(n-1)(n-2)(n-3)/4! x^4 + ...
    Ok can you try and work out my question please I can't do it ((
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    (Original post by Xx4L3x)
    Ok can you try and work out my question please I can't do it ((
    (5)(2) = (n-4)
    n = 14 I get

    ?
 
 
 
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