Wow, 150% of her pens are red. You should note that Claire has more red pens than total pens, which makes no sense at all. If you meant she has 4 red pens out of 6 pens total, then here you go:
(4 red pens)/(6 pens total) = .6666666
.6666666 x 100% = 66.666% (two thirds of her pens are red)
If you meant she has 6 red pens and 4 non-red pens, then this is what you want:
(6 red pens)/(10 pens total) = .6
.6 x 100% = 60% (three fifths of her pens are red)
2006-06-06 16:15:51
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answer #1
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answered by Galatix27 2
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P(r)=6/24=1/4=25%
2006-06-15 04:40:07
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answer #2
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answered by JAMES 4
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Ok. the first part of the question was funny, considering that the total pen was 4, and red was 6.Now, we know the total pen is 24.
Total pen = 24, red pens = 6,
% is simply this: 100 x (amount of specific pens)/ (total pens)
that gives 100 x 6 = 600/24 = 25%
Also, there are other approaches to it. But basic line is this: understand the relationships btw fractions, percentages & decimals.
Take care.
2006-06-19 21:36:49
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answer #3
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answered by zzzlordcharmyzzz 1
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P(r) = 6/24
= 1/4
= 25%
2006-06-18 20:38:38
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answer #4
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answered by rajesh_583 1
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25%
24 Pens, 6 Red
6/24 or 1/4
1 divide 4 = .25 or 25%
2006-06-17 13:02:33
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answer #5
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answered by thedude2005 3
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your work should be clear and simple
total number of pens = 24
number of red pens = 6
fraction of red pens = 6 / 24
percentage number of red pens = 6/24 x 100
=1/4 x 100
= 25 %
2006-06-20 07:06:31
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answer #6
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answered by shylady_nuz 3
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24/6=4
4=1/4
1/4=25%
2006-06-20 02:57:58
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answer #7
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answered by Chris C 3
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6 out of 24 pens are red
To calculate percentage of red = No. of red pens divided by total number of pens multiplied by 100
Or that is, 6 / 24 X 100 = 100 / 4 = 25 %
Hence, 25% of Claire's pens are red !
2006-06-19 21:44:08
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answer #8
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answered by young_friend 5
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out of 24 pens, 6 are red. If you write it as a simplified fraction you get 1/4. Multiply 1/4 by 100, because percent is out of a hundred, to get 25%
2006-06-19 03:58:48
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answer #9
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answered by quickster94 3
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6/24 = 1/4 or 25%
2006-06-20 08:19:32
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answer #10
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answered by anklegno 2
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