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    Hello for part C, I got an answer of 14, they got 15, so please have a go at the question yourself and see what you get
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    (Original post by Assmaster2)
    Hello for part C, I got an answer of 14, they got 15, so please have a go at the question yourself and see what you get
    15 is correct.

    Your equation should be 2\times 0.95^{n-1}=1
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    nope i got 15.
    final stage should get you to:

    14.51 < n

    therefore n = 15 because it must be an integer.
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    Buuuut if you do ar^14 you get 58 minutes which is less than an hour??
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    (Original post by Assmaster2)
    Buuuut if you do ar^14 you get 58 minutes which is less than an hour??
    And that is the fifteenth term. Don't forget the first term is just "a".
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    (Original post by Assmaster2)
    Buuuut if you do ar^14 you get 58 minutes which is less than an hour??
    no because:
    n = 14 would give ar^13 = 61.60 minutes
    n = 15 would give ar^14 = 58.52 minutes

    and so n = 15 because you want a time under 60 minutes.

    [[[because remember you have to put ar to the power of n-1]]]
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    (Original post by ghostwalker)
    15 is correct.

    Your equation should be 2\times 0.95^{n-1}=1
    I just am a bit confused, why is it to the power of n-1, if we put it in inequality form:

    the term in a geometric series is ar^n, so ar^n < 60 , then i sub in the values and get n > log(0.5)/log(0.95) - inequality sign flipped due to negative log.
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    (Original post by Assmaster2)
    I just am a bit confused, why is it to the power of n-1, if we put it in inequality form:

    the term in a geometric series is ar^n, so ar^n < 60 , then i sub in the values and get n > log(0.5)/log(0.95) - inequality sign flipped due to negative log.
    no the term is ar^n-1
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    (Original post by imundercover)
    no because:
    n = 14 would give ar^13 = 61.60 minutes
    n = 15 would give ar^14 = 58.52 minutes

    and so n = 15 because you want a time under 60 minutes.

    [[[because remember you have to put ar to the power of n-1]]]
    Ahhhh i completely forget that a geometric series starts with just a... I am so sorry for wasting ur guys time, thanks for the help.
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    (Original post by Assmaster2)
    Ahhhh i completely forget that a geometric series starts with just a... I am so sorry for wasting ur guys time, thanks for the help.
    its fine, I'm happy to help - it's helping me with my c2 revision :-)
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    (Original post by ghostwalker)
    15 is correct.

    Your equation should be 2\times 0.95^{n-1}=1
    ghostwalker can you confirm that q3 in this mark scheme is wrong? I'm looking at the row totals of the expected values.
    http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
    ty x
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    (Original post by k.russell)
    ghostwalker can you confirm that q3 in this mark scheme is wrong? I'm looking at the row totals of the expected values.
    http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
    ty x
    Sorry, not clear what you're looking at.

    Which table? Do you mean the total for the column headed "E" in the second table?
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    (Original post by ghostwalker)
    Sorry, not clear what you're looking at.

    Which table? Do you mean the total for the column headed "E" in the second table?
    The table right at the top of question 3, the columns add up to whole numbers i.e. 13.51 + 8.49 = 22 but the rows don't i.e. 13.51 + 41.77 + 30.71 = 85.99, which seems like a tiny difference but it actually changes the value of chi squared.. the mark scheme says awrt 4.91, but if you use values which add up to 86 and 54 (which are in my opinion, correct) your chi squared would round to 4.93.
    Isn't the whole point of a contingency table that the totals are the same in the observed and expected tables?
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    (Original post by k.russell)
    The table right at the top of question 3, the columns add up to whole numbers i.e. 13.51 + 8.49 = 22 but the rows don't i.e. 13.51 + 41.77 + 30.71 = 85.99, which seems like a tiny difference but it actually changes the value of chi squared.. the mark scheme says awrt 4.91, but if you use values which add up to 86 and 54 (which are in my opinion, correct) your chi squared would round to 4.93.
    Isn't the whole point of a contingency table that the totals are the same in the observed and expected tables?
    Bear in mind that the expected value rounded to 2 dec.pl. If you could see it to a greater degree of accuracy you would find the discrepancy disappearing. E.g 13.51 is actually 13.51428....

    I suspect they did their actual calculations to a higher degree of accuracy, then rounded for the sake of the table.
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    (Original post by ghostwalker)
    Bear in mind that the expected value rounded to 2 dec.pl. If you could see it to a greater degree of accuracy you would find the discrepancy disappearing. E.g 13.51 is actually 13.51428....

    I suspect they did their actual calculations to a higher degree of accuracy, then rounded for the sake of the table.
    I see. Do you think it's a good idea to start doing my c. tables to 3 d.p. to improve accuracy? The tables they show in the mark scheme are always 2 d.p. but I have noticed sometimes the chi squared they end up with is (very slightly) different to mine
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    (Original post by k.russell)
    I see. Do you think it's a good idea to start doing my c. tables to 3 d.p. to improve accuracy? The tables they show in the mark scheme are always 2 d.p. but I have noticed sometimes the chi squared they end up with is (very slightly) different to mine
    Their calculations are probability to the accuracy of a calculator 8,10, or more significant figures, and rounded just for presentation.

    I see no harm in using 3 rather than 2 dec.pl. I doubt it will resolve all the mismatches though.
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    (Original post by ghostwalker)
    Their calculations are probability to the accuracy of a calculator 8,10, or more significant figures, and rounded just for presentation.

    I see no harm in using 3 rather than 2 dec.pl. I doubt it will resolve all the mismatches though.
    Do you have any advice for removing all the mismatches?
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    (Original post by k.russell)
    Do you have any advice for removing all the mismatches?
    Other than to work to the same degree of accuracy. A spreadsheet is quite useful for that sort of thing. You can then check the rounded values in the markscheme against the actual values.
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    (Original post by ghostwalker)
    Other than to work to the same degree of accuracy. A spreadsheet is quite useful for that sort of thing. You can then check the rounded values in the markscheme against the actual values.
    mm, not sure I'll be able to take an excel equipped laptop into the exam on Wednesday ( so basically I should just use as many significant figures as I can in my calculations in the exam?
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    (Original post by k.russell)
    mm, not sure I'll be able to take an excel equipped laptop into the exam on Wednesday ( so basically I should just use as many significant figures as I can in my calculations in the exam?
    Wouldn't like to speak for what's expected in exams - years since I did my A-levels.
    Mr M could probably give you a definite answer, or any other teacher on here - I'm not familiar with who is one.
 
 
 
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