30% (or 3/10) is the answer
it wont matter if your boy tommy puts it back in the box, the odds would still be the same
2007-02-22 13:58:25
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answer #1
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answered by charlotte's web 3
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He has a (3/10) 30% chance of getting a blue crayon on the first attempt. Then he puts the crayon back and has a 30% chance the second time. So you need to take 30% of 30.
30 * .3 = 9
He has a 9% chance of getting a blue crayon BOTH times.
2007-02-22 13:39:36
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answer #2
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answered by hatevirtual 3
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well, 3+5+2=10
2 crayons out of a box of 10
30% chance of blue, 50% yellow, 20% green.
2007-02-22 12:46:38
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answer #3
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answered by Anonymous
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You have a 3/10 chance to get a blue the first time.
You have a 2/9 chance to get it the second time.
3/10 X 2/9= 6/90 = 3/45 chance.
2007-02-22 12:59:04
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answer #4
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answered by Ace 5
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3 out of 10 chance it was blue. 0.30. 9 crayons left. 9 divided by .3 is
30. so 1 out of 30. now multiply 10 times 30 and 1 times 3. 3 out of 300 or 100. Also 1%
2007-02-22 13:18:27
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answer #5
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answered by fpsfreak101 2
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2/10 or 1/5
2007-02-22 12:42:10
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answer #6
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answered by Brea 2
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3+5+2=10
2/10 = 1/5
The answer is 1/5.
2007-02-22 13:01:26
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answer #7
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answered by Destiny C 3
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Well there is a 50/50 chance of both getting a blue crayon...either they do or they don't : ]
Or 30% percent if you want the technical answer.
2007-02-22 13:00:39
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answer #8
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answered by War Veteran 3
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1. Because if u times a negative number by itself, the negative becomes a positive. ( -1 X -1 = 1 ) 2. 10 miles: 2X5=10 3. 5.2 X 10^11: 520000000000 take out the #'s that r not zero and make sure the one in front is in the one's places. count the other # places and times the # that's in the one's place by 10^other#places 4. 2.34 X 10^6: same as above, if u left out a zero Hope ur able to pass math and hope this helps! XD
2016-05-24 00:33:34
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answer #9
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answered by Anonymous
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i think it is 9/100 because chosin the first blue will give probability of 3/10 so to get 2 consecutive blues u times 3/10 and 3/10 to give 9/100
o w8 that can't b ryt can it??????
2007-02-22 12:53:37
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answer #10
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answered by Anonymous
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