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    (Original post by Boy_wonder_95)
    Nah , it's just that TSR has 'colourful' underlined in red which buggers me :rolleyes:
    That would be your American internet browser's doing
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    (Original post by purplemind)
    Ok, so P(B|A)=1/5
    P(B|A')=3/7
    I add them and get P(B)=22/35
    My textbook says it should be 9/35
    So I don't know if it is right and the answer in the textbook is wrong or the other way round. :confused:
    I'm probably wrong. Look as justinawe's post. I haven't done stats in a long time.
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    (Original post by Scientific Eye)
    I'm probably wrong. Look as justinawe's post. I haven't done stats in a long time.
    Ok, thank you for trying to help me anyway.
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    It's not too late for me, got 75/75 June C3 :cool:

    Although this took me for ever to do... I only got the answer because it seemed to be what they'd have wanted as opposed to what I got.

    - asec^2(at)/(acos(at))

    = - a sec^3(at)
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    (Original post by Boy_wonder_95)
    Still they make my S1 notes look rusty in comparison , who me or justin?
    LOOL, my friend didn't know anything about s1 so I made them for her and they turned out to be better than I thought so I took a pic haha
    Both of you! :P
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    (Original post by justinawe)
    That would be your American internet browser's doing
    Mozilla Firefox? :eyeball:
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    (Original post by Boy_wonder_95)
    Mozilla Firefox? :eyeball:
    yeah? Mozilla Foundation is based in the USA
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    In C2 does it matter if you use a method to find the area under a curve if its not on the mark scheme however you still get the same answer?

    (For example them integration questions where you do the (line-curve) then integrate or find area of the curve - area of a shape)
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    (Original post by L'Evil Fish)
    It's not too late for me, got 75/75 June C3 :cool:

    Although this took me for ever to do... I only got the answer because it seemed to be what they'd have wanted as opposed to what I got.

    - asec^2(at)/(acos(at))

    = - a sec^3(at)
    Wouldn't the 'a's cancel out?
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    (Original post by osman705)
    In C2 does it matter if you use a method to find the area under a curve if its not on the mark scheme however you still get the same answer?

    (For example them integration questions where you do the (line-curve) then integrate or find area of the curve - area of a shape)
    No. it doesn't matter, if the working and answer are correct, I'm pretty sure you attain full credit.
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    (Original post by justinawe)
    Wouldn't the 'a's cancel out?
    That's not the point

    It was just the fractions in fractions... How do I know whether it's being multiplied through the bottom or top?
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    (Original post by osman705)
    In C2 does it matter if you use a method to find the area under a curve if its not on the mark scheme however you still get the same answer?

    (For example them integration questions where you do the (line-curve) then integrate or find area of the curve - area of a shape)
    It'll be accepted if they asked you to "find the area under the curve".

    If they specifically asked you to integrate to find the area, then no.
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    Thanks for the speedy reply (repped)
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    (Original post by L'Evil Fish)
    That's not the point

    It was just the fractions in fractions... How do I know whether it's being multiplied through the bottom or top?
    What? Seems like you're not comfortable with basic fraction operations :naughty:

    \dfrac{-a\sec^2 (at)}{a\cos (at)} = \dfrac{-a}{\cos^2 (at)} \times \dfrac{1}{a\cos (at)}
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    (Original post by justinawe)
    What? Seems like you're not comfortable with basic fraction operations :naughty:

    \dfrac{-a\sec^2 (at)}{a\cos (at)} = \dfrac{-a}{\cos^2 (at)} \times \dfrac{1}{a\cos (at)}
    I multiplied top and bottom by 1/(acos(at))

    And the as aren't meant to cancel out :confused:

    My final answer was:

    -Sec^3 (at)

    Oh :facepalm: :lol:

    I know what happened... My initial answer was right (thankfully I didn't change itbefore I marked it)

    I had second thought and just multiplied and it went wrong. If I deduct that anyway, it's still an A*. 72 marks.

    But I had it first!
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    (Original post by L'Evil Fish)
    I multiplied top and bottom by 1/(acos(at))

    And the as aren't meant to cancel out :confused:

    My final answer was:

    -Sec^3 (at)

    Oh :facepalm: :lol:
    I could have cancelled the as out immediately, but whatever :dontknow:
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    (Original post by justinawe)
    I could have cancelled the as out immediately, but whatever :dontknow:
    Don't worry, my mistake
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    Guys, need some quick help with C2 here, it's kinda important. :O

    Is Log(28x-9) the same as log28x - log9 ??? I would guess not but need some one to confirm. Can we multiply out logs like that?
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    (Original post by GCSE-help)
    Guys, need some quick help with C2 here, it's kinda important. :O

    Is Log(28x-9) the same as log28x - log9 ??? I would guess not but need some one to confirm. Can we multiply out logs like that?
    No.

    \log (28x) - \log (9) = \log \left( \dfrac{28x}{9} \right), using basic log rules.

    So definitely not the same thing.
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    This has got to be the worst a level maths question I have seen lmfao. I scored 1/14 on it.

    http://filestore.aqa.org.uk/subjects...2-QP-JAN12.PDF

    Question 8. I can't make any progress past part (a). Can anyone explain what to do at all?
 
 
 
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