Maths Help! Page 2

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  • Deleted user 20 February 2011 22:55:27
    If you're thick, your kid will be Thick. Learn. Otherwise you're holding your kid back. Your OP is simple.
  • chopsen 20 Feb 2011 22:56:09 16,209 posts
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    in which case

    5r^n-1 + 5(r^n - 1)/(r-1)

    is correct.

    e.g. for n = 3 and r =0.6 you get

    5*0.6^2 + 5 * (0.6^3 - 1)/(0.6-1) = 11.6

    which is what you said you got X3 as being equal to :)
  • Zomoniac 20 Feb 2011 23:05:19 7,908 posts
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    But when I did it for n=5 I got

    5*0.6^4 + 5 * (0.6^5 - 1)/(0.6-1) = 9.908, should be 12.176. Am I inputting it wrong? What does it give you?

    Actually, just tried it on n=3 and I'm getting 5.3. Will try inputting it differently. I'm having a thick day. Have to send all this in tomorrow. It's going to be a long night :/
  • Zomoniac 20 Feb 2011 23:09:11 7,908 posts
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    Got it. I'm a spaz.

    I was reading it as (5r^n-1 + 5(r^n - 1))/(r-1), when it's 5r^n-1 + (5(r^n - 1))/(r-1)

    It works now!
  • chopsen 20 Feb 2011 23:11:08 16,209 posts
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    I get 5*0.6^4 + 5 * (0.6^5 - 1)/(0.6-1) = 12.176

    I have no idea how you got 9.908
  • Zomoniac 20 Feb 2011 23:11:21 7,908 posts
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    One last thing (not that I'm attempting to use you to do my homework or anything!), how did you work that one out? I'm at a loss as to how to show my working to go from the sequence I got to to your equation. It's evidently right, I just can't work out how you worked it out.

    PS you are a lovely genius.
  • chopsen 20 Feb 2011 23:11:32 16,209 posts
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    Ah-ha!
  • Zomoniac 20 Feb 2011 23:12:10 7,908 posts
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    Chopsen wrote:
    I get 5*0.6^4 + 5 * (0.6^5 - 1)/(0.6-1) = 12.176

    I have no idea how you got 9.908

    I was doing the whole thing over (0.6-1), instead of just the last bit.
  • chopsen 20 Feb 2011 23:20:43 16,209 posts
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    It's difficult to explain on the forum due to the fact you can't do proper subscripts and stuff, so it gets confusing and meaningless if you type it out.

    To start with, you've got the sum of a geometic series
    xn = 1 + r^1 + r^2 .. + r^n-2 + r^n-1 (eqn one)

    which can be shown to be equal to

    xn = (r^n -1)/(r-1)

    by multiplying the original eqn one by r on both sides, subtracting eqn one from the result, and then dividing the result of that by (r-1).

    The example you gave was just that, with an an extra r^n-1 term and multiplied throughout by 5.

    Easy ;)
  • Zomoniac 20 Feb 2011 23:23:53 7,908 posts
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    We have differing opinions of easy, I feel well out of my depth here (though I'm sure I knew all of this stuff when I was 15, last week I had to look up how to solve simultaneous equations ffs, that was depressing, it used to all be second nature). I'm sure it'll click eventually. Thanks very much for all your help.
  • chopsen 20 Feb 2011 23:28:41 16,209 posts
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    Heh, good luck. I had to look it up and if it wasn't for the fact I was apparently doing a few months back the same as you are now, I wouldn't have a clue. Does make sense eventually, honest.
  • FairgroundTown 24 Nov 2011 08:23:18 2,522 posts
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    Bump!

    From my son's maths book:

    x / 8 * 2 = x

    I can't make this work? Am I being thick? Or is the problem wrong?
  • Grunk 24 Nov 2011 08:42:35 4,718 posts
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    Dammit, I keep trying to post, but its not showing up. Anyway:

    0
  • chopsen 24 Nov 2011 08:57:14 16,209 posts
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    is that
    (x/8) * 2
    or
    x/(8*2)

    Either way, it basically reduces to

    xk=x

    where k is a number, the only solution possible is 0, which is a bit daft.
  • FairgroundTown 24 Nov 2011 09:16:27 2,522 posts
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    Thanks chaps. I guess that is the answer. It was the last question on the exercise, so it was probably supposed to be a teaser.
  • White_Shadow 24 Oct 2012 21:09:08 2,537 posts
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    bump!

    Mathsy people of EG, I need your help! How would I work out (as an example) how many people went to see both displays in a gallery in 1 visit if I know a) the total number of people who went to each display and b) the percentage of people overall who saw both?
  • White_Shadow 24 Oct 2012 21:34:25 2,537 posts
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    Anyone?
  • chopsen 24 Oct 2012 21:45:49 16,209 posts
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    Shut it, I'm thinking!

    (I'm not sure you've got enough info, as you don't know how many of the intersection of the two groups is made up from the two group. Unless you assume the people who visit both are equally likely to be from either group of people who visited one of the displays)
  • X201 24 Oct 2012 21:54:05 15,696 posts
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    I think there's enough info. Its just the way the he's written it that is confusing the matter
  • White_Shadow 24 Oct 2012 21:58:43 2,537 posts
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    Maybe I didn't word it particularly well? The total for each individual display includes the number of people who went to both
  • X201 24 Oct 2012 22:01:11 15,696 posts
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    Gallery 1 = A
    Gallery 2 = B
    Both = C

    We know A+B and so we know C is a percentage of that.

    Edit. Shadow give us the actual numbers. Think this can be done

    Edited by X201 at 22:03:11 24-10-2012
  • White_Shadow 24 Oct 2012 22:03:00 2,537 posts
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    But the issue is complicated by the fact that not everyone went to both, which I probably should have made clear
  • monkehhh 24 Oct 2012 22:07:47 3,477 posts
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    Numbers would help, the wording isn't clear.
  • chopsen 24 Oct 2012 22:10:23 16,209 posts
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    @X201

    you don't know A + B, as some of the members of A are in C, and some of the members of B are in C. It's a set theory problem. You know the number of people in set A, and the number of people in set B, and you the proportion of the total number of the total who are in the intersection (C). What you don't know is how many of set A are also members of C and how many of B are also members of C. As you only have the percentage of the overall total.
  • X201 24 Oct 2012 22:16:28 15,696 posts
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    It was the way it was written that confused me.

    Give us the numbers. :-)
  • White_Shadow 24 Oct 2012 22:18:02 2,537 posts
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    Ok. 1354 people went to A, 1644 went to B, and 79 per cent went to both. I've tried coming up with a formula or equation but it's been too long since I did any proper maths.
  • mrpon 24 Oct 2012 22:31:43 29,427 posts
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    2368

    Give yourself 5 or gig, you're worth it.

  • chopsen 24 Oct 2012 22:35:37 16,209 posts
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    a = total number who saw exhib A
    b = total number who saw exhib B
    p = proportion (as a fraction) of the total visitors who saw both
    t = total number of visitors
    c = total number of visitors who saw exhib A and B = pt

    From conisdering them as sets:

    a + b -pt = t
    a + b = t + pt
    a + b = t (1 + p)
    t = (a+b)/(1 + p)
    pt = p (a + b)/(1 + p)
    c = p (a + b)/(1 + p)
  • chopsen 24 Oct 2012 22:40:35 16,209 posts
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    Which gives 1323 approx. I'm assuming 79 isn't exact and the result of rounding.
  • ED209 24 Oct 2012 22:46:18 505 posts
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    You're all idiots. The answer is 1.

    (Only the man was going to the gallery in St Ives. He left his wives and children at home.)

    HTH.

    Samantha Janus?

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