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#1 Re: Help Me ! » Need help on percentage. » 2007-03-06 06:29:12
For six people, the top 20% is the top [0.2×6] = [1.2] = 1 people (or rather person). ([x] denotes the greatest integer less than or equal to x.)
So only 1 person in the any group of 6 people can be in the top 20%; in this case, its A.
Perhaps something like this?
((99+93+91+85+60+50)/6 ) *1.2
(1/6) <-for the six grades
(99+93+91+85+60+50) <- add the grades
* 1.2 <- to get the top 20%
#3 Re: Help Me ! » Need help on percentage. » 2007-03-05 08:21:11
That would be quite difficult w/o the use of variables (assigning each grade a variable and then sorting numerically). I can think of a solution by writing a computer program, but I'm not sure how I'd do it mathematically in a cut and dried method. Mind you, I'm just starting Trig. 
#4 Help Me ! » finding X in fraction » 2007-02-11 10:38:50
- mykel
- Replies: 3
I have the following problem.
1/x-3 + 2/x-5 = 3/x^2 - 8x + 15
I know the answer is x = 14/3, but what's the procedure for that conclusion?
Thanks!
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