If you're thick, your kid will be Thick. Learn. Otherwise you're holding your kid back. Your OP is simple. 
Maths Help! • Page 2

Page
of 3 First / Last 

chopsen 16,937 posts
Seen 46 minutes ago
Registered 10 years agoin which case
5r^n1 + 5(r^n  1)/(r1)
is correct.
e.g. for n = 3 and r =0.6 you get
5*0.6^2 + 5 * (0.6^3  1)/(0.61) = 11.6
which is what you said you got X3 as being equal to 
Zomoniac 8,343 posts
Seen 5 hours ago
Registered 11 years agoBut when I did it for n=5 I got
5*0.6^4 + 5 * (0.6^5  1)/(0.61) = 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 8,343 posts
Seen 5 hours ago
Registered 11 years agoGot it. I'm a spaz.
I was reading it as (5r^n1 + 5(r^n  1))/(r1), when it's 5r^n1 + (5(r^n  1))/(r1)
It works now! 
chopsen 16,937 posts
Seen 46 minutes ago
Registered 10 years agoI get 5*0.6^4 + 5 * (0.6^5  1)/(0.61) = 12.176
I have no idea how you got 9.908 
Zomoniac 8,343 posts
Seen 5 hours ago
Registered 11 years agoOne 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 16,937 posts
Seen 46 minutes ago
Registered 10 years agoAhha! 
Zomoniac 8,343 posts
Seen 5 hours ago
Registered 11 years agoChopsen wrote:
I get 5*0.6^4 + 5 * (0.6^5  1)/(0.61) = 12.176
I have no idea how you got 9.908
I was doing the whole thing over (0.61), instead of just the last bit. 
chopsen 16,937 posts
Seen 46 minutes ago
Registered 10 years agoIt'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^n2 + r^n1 (eqn one)
which can be shown to be equal to
xn = (r^n 1)/(r1)
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 (r1).
The example you gave was just that, with an an extra r^n1 term and multiplied throughout by 5.
Easy 
Zomoniac 8,343 posts
Seen 5 hours ago
Registered 11 years agoWe 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 16,937 posts
Seen 46 minutes ago
Registered 10 years agoHeh, 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 2,522 posts
Seen 2 years ago
Registered 10 years agoBump!
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 4,717 posts
Seen 9 months ago
Registered 10 years agoDammit, I keep trying to post, but its not showing up. Anyway:
0 
chopsen 16,937 posts
Seen 46 minutes ago
Registered 10 years agois 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 2,522 posts
Seen 2 years ago
Registered 10 years agoThanks 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 2,538 posts
Seen 3 months ago
Registered 10 years agobump!
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 2,538 posts
Seen 3 months ago
Registered 10 years agoAnyone? 
chopsen 16,937 posts
Seen 46 minutes ago
Registered 10 years agoShut 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 16,388 posts
Seen 1 minute ago
Registered 9 years agoI think there's enough info. Its just the way the he's written it that is confusing the matter 
White_Shadow 2,538 posts
Seen 3 months ago
Registered 10 years agoMaybe I didn't word it particularly well? The total for each individual display includes the number of people who went to both 
X201 16,388 posts
Seen 1 minute ago
Registered 9 years agoGallery 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 24102012 
White_Shadow 2,538 posts
Seen 3 months ago
Registered 10 years agoBut the issue is complicated by the fact that not everyone went to both, which I probably should have made clear 
monkehhh 3,795 posts
Seen 6 minutes ago
Registered 7 years agoNumbers would help, the wording isn't clear. 
chopsen 16,937 posts
Seen 46 minutes ago
Registered 10 years ago@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 16,388 posts
Seen 1 minute ago
Registered 9 years agoIt was the way it was written that confused me.
Give us the numbers. 
White_Shadow 2,538 posts
Seen 3 months ago
Registered 10 years agoOk. 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 30,279 posts
Seen 12 hours ago
Registered 9 years ago2368 
chopsen 16,937 posts
Seen 46 minutes ago
Registered 10 years agoa = 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 16,937 posts
Seen 46 minutes ago
Registered 10 years agoWhich gives 1323 approx. I'm assuming 79 isn't exact and the result of rounding. 
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. 
Page
of 3 First / Last