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    (Original post by L'Evil Fish)
    I don't have one

    I always have to derive my inverse trig from first principles as a result...
    Oh, you don't get one? You're supposed to, for WJEC.

    http://www.wjec.co.uk/uploads/publications/10908.pdf
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    (Original post by QwertyG)
    Hi can someone help me with S2 hypothesis testing please.






    I done the part to the get the first 3 marks but im not seeing where to get the 2 and the 10.

    I would assume P(X<=A) < 0.025 but I dont know to go from there.
    You want to scan the tables for where P(X<=Critical value1) is as close to 0.05 as possible. Similarly, scan the table for where P(X>=Critical value2) is as close to 0.05 as possible. As the tables have P(X<=x), you want to look for when 1-P(X<=Critical value2 - 1) ~ 0.05. Your critcal regions will be X<=critical value 1 or X>= critical value 2
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    (Original post by L'Evil Fish)
    Like for two marks if they ask:

    Differentiate sin^-1(7x) I do it from scratch

    Yeah :/ it's never that high though, most I've done is... About 6 of them
    Well, to be fair, just say u = 7x and then you should know the derivative of arcsin anyway and don't forget to multiply by 7, or just skip the substitution like I do cos I'm a bad ass

    And really? Binomial tables ftw, stats is tedious enough as it is
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    (Original post by tigerz)
    Okays thank youu I'll try it then post it
    Okey dokey, I may think up another one if you want
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    Hi guys i'm sitting Ocr stats 1 exam and am stuck on these perms and combs questions - can anyone please help me , I'd really appreciate it , thanks perms and combs.ppt
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    (Original post by AmirHabeeb)
    You want to scan the tables for where P(X<=Critical value1) is as close to 0.05 as possible. Similarly, scan the table for where P(X>=Critical value2) is as close to 0.05 as possible. As the tables have P(X<=x), you want to look for when 1-P(X<=Critical value2 - 1) ~ 0.05. Your critcal regions will be X<=critical value 1 or X>= critical value 2

    Oh I see, thanks so much!

    How come it its x<=2 not x<=1 since 0.0500 is below
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    (Original post by joostan)
    Haha
    Here's a nice logs question.
    Show that the solution for x in the equation.
    6^x = 5 \times 9^x
    Can be written as x=\dfrac{\log_3(5)}{\log_3(2)-1}
    Can't get further than this:

    6^x = 5 \times 9^x



xlog6=log(5 x 9^x)



x=\dfrac{\log5\times 9^x}{\log6}

    How do you change the base from 10 to 3? :s

    (Original post by joostan)
    Okey dokey, I may think up another one if you want
    Haha thank you :P i'm not too good at it though!
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    (Original post by QwertyG)
    Oh I see, thanks so much!

    How come it its x<=2 not x<=1 since 0.0500 is below
    X~Po(6)
    P(X<=2) = 0.0620
    P(X<=1) = 0.0174
    The question says specifically as close to 0.05 as possible. 0.0620 is closer to 0.05 than 0.0174 so lower critical value is 2. But sometimes they won't say "as close to..". If that's the case then 1 would be the critical value because P(X<=2) >0.05
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    (Original post by mia_hilton)
    Hi guys i'm sitting Ocr stats 1 exam and am stuck on these perms and combs questions - can anyone please help me , I'd really appreciate it , thanks perms and combs.ppt
    Hi



    For the first one, the word THURSDAY has 8 letters, and two vowels (U and A).

    Four letters are picked at random.

    The number of possible combinations of 4 letters you can get from THURSDAY would be 8C4 = 70.

    Now, they want the probability that there is at least one vowel among the letters.

    There are 6 consonants/non-vowels in THURSDAY - T, H, R, S, D, Y.

    So the number of possible combinations of 4 letters you can get from the consonants alone would be 6C4 = 15.

    So the probability that you'd have NO vowels would be \dfrac{15}{70} = \dfrac{3}{14}

    And thus the probability that you'd have at least one vowel would be 1 - \dfrac{3}{14} = \dfrac{11}{14}



    For the second one, for the number to be greater than 4000, the first digit would have to be 5, 6 or 8.

    If the number starts with 5, 6 or 8, we'd have 4 numbers left.

    The number of permutations of these 4 numbers in the 3 spots left would be 4P3 = 24.

    So we'd have 24 numbers starting with 8, 24 starting with 6, 24 starting with 5.

    So the total would be 24+24+24 = 72
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    (Original post by tigerz)
    Can't get further than this:

    6^x = 5 \times 9^x



xlog6=log(5 x 9^x)



x=\dfrac{\log5\times 9^x}{\log6}

    How do you change the base from 10 to 3? :s



    Haha thank you :P i'm not too good at it though!
    Why not just start by taking \log_3?
    Spoiler:
    Show
    For the record the change of base formula is given by:
    \log_a(x) = \dfrac{\log_b(x)}{\log_b(a)}
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    (Original post by justinawe)
    Oh, you don't get one? You're supposed to, for WJEC.

    http://www.wjec.co.uk/uploads/publications/10908.pdf
    Well that's useful :mmm:
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    (Original post by L'Evil Fish)
    Well that's useful :mmm:
    Your school is supposed to supply you with one for every maths exam :sly:

    Do they not do this?
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    (Original post by joostan)
    Why not just start by taking \log_3?
    Spoiler:
    Show
    For the record the change of base formula is given by:
    \log_a(x) = \dfrac{\log_b(x)}{\log_b(a)}
    How do you do that we've never changed it lool

    so 6^x=xlog6=\log_{3}2 How did you figure it out :s is it because:  log6= 2(log3)= \log_{3}2?
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    (Original post by justinawe)
    Your school is supposed to supply you with one for every maths exam :sly:

    Do they not do this?
    They do have a yellow booklet there... I've just never thought to use it :ninja:
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    (Original post by tigerz)
    How do you do that we've never changed it lool

    so 6^x=xlog6=\log_{3}2 How did you figure it out :s is it because:  log6= 2(log3)= \log_{3}2?
    No . . .
    It's fairly straight forward to derive:
    Take:
    y=\log_ax \Rightarrow a^y = x 

\Rightarrow y\log_b(a) = \log_b(x)

\Rightarrow y = \dfrac{\log_b(x)}{\log_b(a)}
    As we said that \ y=\log_ax
    \Rightarrow  \log_ax = \dfrac{\log_b(x)}{\log_b(a)}

    Also \log(6) \not= 2\log(3)
    Instead: \log(6) = \log(3) + \log(2)
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    (Original post by L'Evil Fish)
    They do have a yellow booklet there... I've just never thought to use it :ninja:
    :facepalm2: you must be joking!
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    (Original post by L'Evil Fish)
    They do have a yellow booklet there... I've just never thought to use it :ninja:
    Here we were, labouring under the assumption that you were a clever, sensible chap :facepalm:
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    (Original post by justinawe)
    :facepalm2: you must be joking!
    (Original post by joostan)
    Here we were, labouring under the assumption that you were a clever, sensible chap :facepalm:
    I've done okay then :cool:

    If I forget things, I derive them...

    Proof of series and that
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    (Original post by L'Evil Fish)
    I've done okay then :cool:

    If I forget things, I derive them...

    Proof of series and that
    Good practice, though tbh, there's not actually anything that I look up in the formula book other than the tables, and on occasion to double check the Maclaurin Series
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    (Original post by joostan)
    No . . .
    It's fairly straight forward to derive:
    Take:
    y=\log_ax \Rightarrow a^y = x 

\Rightarrow y\log_b(a) = \log_b(x)

\Rightarrow y = \dfrac{\log_b(x)}{\log_b(a)}
    As we said that \ y=\log_ax
    \Rightarrow  \log_ax = \dfrac{\log_b(x)}{\log_b(a)}

    Also \log(6) \not= 2\log(3)
    Instead: \log(6) = \log(3) + \log(2)
    Ahh totally forgot about the last bit! Okay I get the derivation of the formula and how to use it now, but I still can't figure out how

    log 5 \times 9^x=\log_{3}(5)

    also where did the -1 come from? Sorry i've never done anything like this looool
 
 
 
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