help, another hard math question?
827635
Sue prob.: 9/10\nAnn prob.: 3/5\n\nThese are the fractional forms of .9 and .6, respectively.\n\nFor both passing, multiply the probabilities.\nFor neither passing, multiply the probabilities that Sue and Ann both do not pass.  To get this, subtract the probability they do pass from one.\n\nBoth pass: 9/10 times 3/5 = 27/50 \nNeither: (1-9/10) times (1-3/5) = 1/10 times 2/5 = 2/50 = 1/25\n\nThe probability that Sue passes, as you have said, does not depend on whether Ann passes or not.  That means that the probability Sue passes (whether Ann passes or not) is .9
891781
Hi I have tried to help you but keep coming up with nothing. So it look`s like the previous answer is the way to go, good luck I hope you find what your looking for.
448749
Lisa = L, Jen = J\nL+J=25\nwe know from the question L=4J  \nso L+J=25 becomes\n         \n          4J+J=25\n          5J=25\n           J=25/5\n            J=5  so we get Jen catches 5 frogs \n                           Lisa catches 20 frogs
186766
Difficult question to answer.  Don't really understand the question.  Please try again.
912865
bring the a's to one side... u then get 9a = 27 ... divide and answer is a = 3
221530
Add them together and divide by 5.\n\nWow. I could seriously have you people do my homework for me. Every night. I'll type out the problems and see how you all do. :P JP
849885
I understand your frustration in receiving no Answer yet to this question, but since I do not have any direct knowledge of how to perform the calculation(s) you need ... I hope the link below serves you well.  It appears to explain the mathematics behind all the concepts you mention.
909352
This question has been asked so many times!?!?!\n\nGood challenge question!\n\nIt's a tough one, too.\n\nI'm taking it to work with me and asking my colleges to help me out on it.\n\nSomeone "did" crack this CHALLENGE Question.\n\nSo here's the answer. . .\n\nhttp://answers.yahoo.com/question/index;_ylt=AvnsMpyNhtwrHV1JVMvXrE4jzKIX?qid=20061130045858AAB57A7\n\nANSWER:  ABCDEFGHJ = 25,693,148\n\nA=2\nB=5 \nC=6\nD=9\nE=7\nF=3\nG=1\nH=4\nJ=8
880981
Sorry, but your question is written very poorly.  You might want to try looking at how some of the other math questions are written out, then rewriting your questions so that it makes more sense.
774792
Let the length of side "b" be x ft.\nTherefore side a=2x and side c=2x+3\nAccording to the problem,\nx+2x+2x+3=93\n=>5x+3=93\n=>5x=93-3=90\n=>x=90/5=18\nTherefore.side a=2X18=36 and side c=36+3=39\nAnswer a=36ft. b=18 ft and c=39 ft
780025
x ≤ 5\n\n-- - - - - - -s-
763076
1st Problem:\n\nLet's see, if we sell 198 thousand candy bars, we are giving them away, which is terrific for volume, but bad for profit.    \n\nRevenue = x P(x), = 9x-(x^2/22)  differentiate to get 9-(2x/22) so x/11=9 x=99\n\nPrice would be  4.50
276680
Ok, If A is a point on the circle and C is the middle of the circle then AC is the RADIUS of the circle.  Which means any point on the circle to C WILL ALWAYS be 12.  So if the excess of the circle to B is 8, then Add the RADIUS to the distance to B, and we arrive at the result of 20. Which is AC + (Distance of circle to B)\n\n20
263164
I won 10 points!\n\nThis is pretty easy. Try to stump me with harder ones.
849743
This is the third time that this question has been asked!?!?!\n\nIt's a tough one, too.\n\nI'm taking it to work with me and asking my colleges to help me out on it.\n\nBut someone "did" crack this CHALLENGE Question.\n\nSo here's the answer. . .\n\nhttp://answers.yahoo.com/question/index;_ylt=AvnsMpyNhtwrHV1JVMvXrE4jzKIX?qid=20061130045858AAB57A7\n\nANSWER:  ABCDEFGHJ = 25,693,148\n\nA=2\nB=5 \nC=6\nD=9\nE=7\nF=3\nG=1\nH=4\nJ=8
144487
the answer is exponents.
1132233
this question will require calcs using logarithm.\nhence will take time