What is the answer for POW 19 Pool Pockets?
488091
just a minute we did this one last week......just experiment with different lenghth pool tables and talk about it
805535
http://www.exrx.net/Lists/ExList/WaistWt.html
855670
Boy, this is tricky.  I think the probability is (8! * 41c2 * 10c2)/49!, where 41c2 = 41!/39!2! and 10c2 = 10!/8!2!  In this number, 8! represents the correct picks, 41c2 the incorrect picks, and 10c2 the number of ways to order the picks.\n\nFor exactly seven correct numbers in 10 picks , only a minor change is needed.  It becomes (7! * 42c3 * 10c3)/49!.  Follow this pattern for 6 of 10, 5 of 10, etc
410658
@ http://www.generalmonitors.com/downloads/application_guides/CONFINED_SPACE.PDF#search='standards%20of%20explosive%20gases%20manholes'\n\nor\n\nwww.iowadnr.com/waste/policy/files/gas.pdf
326871
http://www.mailorderexpress.com/Intex.asp that's what you should go on. All pools and fliters are on that website. Try it!!!!!!
712274
overunder is 38.5 jax are favorites by 4 points i am looking for them to win but as far as you winning money in a pool good luck i had that happen to me too and i came up short. i was stuck in between 2  people like you.look for a low scoring game. my opinion jax 21 giants 10
960222
here's the results for today's draw:\nhttp://www.myit.com.my/fdresult/default.aspx
947255
Let X be mean fill of container. \n\nThe distribution is X~N(16, 1.5) i.e. it is normally distributed as stated in the question.\n\nWant to find P(X>17), so standardise.\nP(X>17) = P(Z>(17-16)/(1.5/2)) = P(Z>1.33)\n\nThen you look up your normal table. The answer is approximately 0.09.
1122511
dont make it wet that wont do anything make a fist with your hand and punch it into the pocket and push back the strings it should loosen up for girls sticks you cna undo the strings at the bottom MAKE SURE ITS NOT ILLEGAL (even though a lot of peoples are :) )
917242
I have a pool available.
30360
you have a better probability of winning if you buy all the squares!\n\nBut seriously, the pools I have been in, people sign their name BEFORE the numbers are drawn. That is so there is no better square on the sheet- its all in how the numbers are drawn- seems like it would be hard to fill a sheet if the numbers were already written on the sheet. Everyone who bought the first squares would take the 0's, 3's 4's and 7's - a lot of people wont buy squares if they are stuck with numbers like 2 and 8 or 9- because they suck (not as bad now that we have 2 pt conversions though)\n\nstay away from the edges??? hmm, I'll have to remember that. hehe
540362
goe here and find what you seek\nhttp://shareware.search.com/search?cat=237&tag=ex.sa.fd.srch.sa_win&q=draw
881790
No formula can predict that as with every table being a different size the calculated relief angles would bouce from different points along the rails, therfore making it impossible for any equations to accurately measure or predict which pocket the ball would fall.\n\nBut if you were only using one size table all of the time, then yes your simple calculated equation would work fine.
383543
For a moment, I thought that Gabrielwyl had the correct pattern, then I noticed a mistake in for n=14. If there are 14 people, you should sit in the 13th spot.\n\nThen I checked gulabo's answer, and he did a good job of trying to find an expression to the pattern, except that it doesn't work. He wrote, "So, if n is even just SIT AT FIRST POSITION." Not true. You can check it yourself. His expression for if n = an odd integer is also incorrect.\n\n\n\nAnyways, the pattern for n people (yourself included) is to sit at the listed position. It's similiar to Gabrielwyl's except with the mistake corrected, and the pattern defined.\n\nN.Pos\n------\n1.1\n2.1\n3.3\n4.1\n5.3\n6.5\n7.7\n8.1\n9.3\n10.5\n...\n14.13\n15.15\n16.1\n...\n30.29\n31.31\n32.1\n...\n63.63\n64.1\n...\nn=2^n+x : 1+2x\nn=2^k : 1\n\n\nSo here's the strategy. It's simple:\n\nLet:\n\nn = # of knights including you\nk = an integer >=0\nx = an integer such that 1 <= x < 2^k\n\n1. For every n=2^k where k is an interger>=0, you must sit in the first seat to win.\n\n2. For every interger, n=2^n+x, you must sit in the seat # (1+2x).
427607
yeah.. but you might want to call your bank and ask.\nmy bank won't take coins in sleeves. they have to be loose.
246896
Close, but no cigar...\n\nABCDE FGHIJ KLMNO\nCJDMG OEKHB INALF\nGFBKL NAOJC DHIEM\nEKNCI LMGAH JOFBD\n\nConstraint: Fifteen riders yields (15 * 14 / 2) = 105 unordered pairs.  A single five rider heat yields (5 * 4 / 2) = 10 unordered pairs.  Over twelve heats that makes 120 unordered pairs -- a little bit more than the 105 that we want to cover.  So we want to have each pair of riders be together in at least one heat, with no pair of riders being together in more than two heats.\n\nResult: Not quite -- 21 "head-to-head" matchups do not take place.  But no pair of riders are together more than twice.\n\nAK BI BM BN CF CH CL\nDK DL DN EF EG EJ EL\nFM GN GO HN IO JK JL\n\nConstraint: There are four sets and five positions.  So we want to have each rider in a different position in each heat.\n\nResult: Constraint is met!\n\nConstraint: There are three heats per set and fours sets. So we want to have each rider be the first seat of a set, the second heat of a set, and the third heat of a set (obviously the fourth set will double up on one of the heats).\n\nResult: Not quite -- 5 riders miss one of the first/second/third heats in a set.\n\nC:310 D:202 G:220 H:031 O:022\n\nAlternative:\n\nABCDE FGHIJ KLMNO\nJDLGM HOEBK INFCA\nLFOAH DEGKN MIBJC\nEHNMI OCJFD GKALB\n\nThe same on the first two constraints.  On the third constraint there are now 7 riders (up from 5) missing one of the first/second/third heats in a set.  But they are all 2-2-0 with none of the 3-1-0's of the first schedule.\n\nAJ AM BF BN CG CH CK\nCL DH DI EF EJ EL FK\nFM GO IK IL IO JK JN\n\nA:202 D:220 E:220 H:220 K:022 L:202 M:202
263161
This is a good probability problem -- I LOVE probability problems.  This is really asking 4 different questions.  First, what are the chances of the first pull being black.  (9 total socks, 5 are black ...).  Then what are the chances of the second one being blue (Keep in mind that now you only have EIGHT socks in the drawer, unless you put the last one back - so 1 blue and eight total socks).  Then chances of a brown sock with only 7 socks left (3 brown, seven total socks).  \n\nWhen you get all those probabilities, in order to see what the chances of this AND this AND this happening, you have to multiply the fractions (if it was "What are the chances of this OR this happening, you'd add them).\n\nGood luck.  This stuff is FUN!  (Especially if you ever want to go to VEGAS!)