Ineed answer for this math question, If you find it you have a prize to win.?
367390
I think either this is easy or hard.  Being lazy, I first consider the possibility that this is easy.\n\nIf it's easy, then x and y are both perfect squares of integers, y is bigger than x and neither is larger than 16.\n\nSo, my first guess is that y = 9 and x = 4, which turns out to be a solution.\n\nLucky me.\n\n*********************************\n\nOK, Tariq, here's a mathematical solution:\n\nFirst of all, while high school students and engineers think that the more algebra you write on a page, the more mathematical you get, acutally real mathematicians hate piles of algebra and strive for simple solutions.  If there is any way to simplify a problem with logic, and avoid lots of algebra where it is easy to make mistakes, this is good.  Some of the best mathematicians never use much complicated algebra at all -- either they find a clever simplificaiton or they tell their graduate students hack through the mess.\n\nThinking about underlying properties of number, I realized that 11 is a rational number.  What if x is not a perfect square?  Then the root of x is an irrational number, which means that y must be an irrational number.  I can see that if I start squaring stuff and substituting things, I will get a fourth-order equation.  You might have seen fourth order euqations in class that can be factored, but in general, they can never be solved by brute force.  In fact, the only way to solve a general fourth-order equation is to hope that it has rational roots.\n\nAt this point, there was a loud noise in the back yard and I went out to see what it was.  In the cool air, I stared up in the sky for some time and began to wonder what would happen if the roots were rational instead of irrational, and to wonder how I could simplify the problem.  (This is an extremely important part of the mathematical process.)  I realized that this would imply that x and y were both perfect squares.  However, clearly neither x nor y is greater than 11.  Also, clearly, y is greater than x.\n\nComing inside I realized that if y and x were perfect squares less than 11 and if y were bigger than x, then there are only a few possibilities:\n\ny = 9; x = 4\ny = 9; x = 1\ny = 9; x = 0\ny = 4; x = 1\ny = 4; x = 0\ny = 1; x = 0\n\nI can try all of these possibilities in my head in a minute, so it didn't take long to find the solution.\n\nOf course, If the solution hadn't been there, the problem would have been much different.\n\nBut that's a one way to solve a mathematical problem:  use your understanding of mathematics to reduce the problem to something simpler.\n\nNo real mathematician is too proud to turn down blind luck, especially after he has been clever enough to stumble upon it.\n\nSo, what is there to see in Natchez?
262906
First of all, you know that the total money collected was 57444 * 2, which is 114888.\n\nNext, you know that you have only 202 prizes.  Of those, three of them were valued: 2000, 1500 and 1000.  Keep in mind that after those three prizes, there are 199 prizes left, I'll get to that later.\n\nThe first three prizes will take a total of 4500 from the initial pot of 114888.\n\nNow, since each following prize is 5 less than the one before it, it is a summation of the next 199 prizes starting at an initial prize value of 1000 minus 5.  (Remember, each following was 5 less than the one before it.)\n\nSo, the summation is the sum of each value of the equation: (1000 - (5 * x)) where x starts at the value of 1 and increases to 199.  So this is the sum of: 995 + 990 + 985 + ... + 5.  The result of which is 99500.\n\nSo, 99500 + 4500 is the total amount paid out in prizes, being 104000.  Subtracting that from the initial intake of funds on $2 tickets leaves a total pure profit of 10888.\n\nNot a bad profit, it was almost like asking about one out of six of those people to just 'give' you one dollar. :)
655518
a) the number of possible tickets is C(42,6) = 5245786\n1/5245786 = .000000190629\nb) the number of tickets with exactly 4 is \nC(6,4)C(36,2)\n= 15*630 = 9450\n9450/5245786 = .00180\nc) You don't win if you have less than 3 winning numbers, so either you can have 0, 1, or 2 winning numbers and not win\nP(0 numbers) = C(36,6)/5245786 = .37130\nP(1 number) = C(6,1)C(36,5)/5245786 = .43119\nP(2 numbers) = C(6,2)C(36,4)/5245786 = .16843\nP(no prize) = .37130 + .43119 + .16843\n= .97092
453954
You get certain prizes, which Yahoo has listed as "Coming Soon".\n\nhttp://answers.yahoo.com/info/scoring_system;_ylt=AgB5RHH1ym3COKXDo5ELFS0azKIX
422761
Hi,\n\nYou could try to enter as much as possible contests to win ticket's
1010896
me phycic u win\n\nphyc\n \ngood luck***
635789
Even if the prize was stupid, like a two-year supply of dishwasher soap or something, I'd still bid on it and hope to win. Even though you end up paying taxes, you could sell the prize(s). People will buy ANYTHING, especially if it's benn on tv or is somehow connected to tv -- just look at what's on e-Bay. Then I'd use the money I made to buy something I really wanted instead.
861500
No - You need to match at least one of the main numbers to win a prize.
858530
it is possible when X=33.02163079..... try it
611239
i agree with everybody here the answer is 10 so whats my prize?
420071
TAKE AWAY THE PHRASE 'WIN' FROM YOUR QUESTION.\nTHATS THE ANSWER.
894091
None. Two odds make an even
329951
Yahoo should send you something that you may win one.
467318
Win a competition of course! Surprised you need to ask!
646575
look in ebay after people win its.
9946
Win the lottery obviously lol
234077
Well, if 1st place received $6000 which is 1/2 of the prize money, then the prize money must be $12,000. The second part of the problem (prize for 2nd place) is irrelavent information.