How to solve problems with Chebyshev's theorem by using plain and understandable English?
99922
It might be helpful to indicate which theorem you are talking about.  After all, Chebyshev derived hundreds of results and theorems.
997626
http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP?Res=150\n\nhttp://www.math.sfu.ca/~cbm/aands/subj.htm\n\nhttp://clusterfie.epn.edu.ec/ibernal/html/CURSOS/AbrilAgosto06/Metodos/Libros/Fundamentals%20Of%20Numerical%20Computing%20-%201997%20-Wiley%20%20(By%20Laxxuss).pdf
175327
Try this, maybe?\n\nhttp://mathforum.org/library/drmath/drmath.elem.html
448669
Do you mean Chebychev's theorem of proof of Bertrand's postulate on prime number distributions, or Chebychev's theorem concerning standard deviations? \n\nEdit your question to clarify and I'll answer more...\n\nOk, Chebychev's theorem states that at least 1-(1/k^2) of a distribution lies within k standard deviations of the mean.  The reason k must be more than 1 is that at that low of a standard deviation exactly what type of distribution you have becomes important.  Chebychev's theorem works for any distribution, so it can be used to assess general population size, at least within a fairly wide range, with less analysis using just the sample standard deviation.  So, it works for sparse data, non-normal distributions, and so on, within its limits of only holding for calculations of what lies within a range larger than one standard deviation.  \n\nSo, for example, you can take a small subset from a set of data with a mean of 50, and a *sample* standard deviation  of 8, and from Chebychev's theorem know that if you were to take a larger sample, at least 75% (1-1/2^2) would lie within the range of 44 to 66, that is, 50+/-16.
951948
When I do word problems, I break it down into manageable chunks.  After reading the entire problem once to get a sense of what I'm dealing with, I re-read each sentence one at a time and try to write down in mathematical terms what each sentence is saying.  Often each sentence is only a portion of the overall equation, but I find if I tackle it like this, I can piece together the overall equation by looking at all the partial equations I got from each sentence.\n\nHope this helps a little.  Good luck!\n\n~ ♥ ~
920579
use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
797929
1 -> b\n2 -> e\n3 -> c\n4 -> d\n5 -> a\n6 -> f
976659
Start here:\nhttp://www.math.pitt.edu/~sussmanm/2071Spring05/lab07/lab07.html#TheGramSchmidtMethod
432979
Try these (math dictionaries):\n\nhttp://www.teachers.ash.org.au/jeather/maths/dictionary.html\n\nhttp://www.mathwords.com/\n\nhttp://www.intermath-uga.gatech.edu/dictnary/homepg.asp\n\nhttp://www.harcourtschool.com/glossary/math2/index_temp.html\n\nThere are many more, but these should get you started.
311264
Well. Do you know how to work rational expressions in general? The first link below is a useful general resource and has a section on rational expressions and polynomial long division; the second is specifically on rational expressions at college intro-level.\n\nI used to tutor math and I found that most people's difficulty with word problems was not the actual math, it was extracting the appropriate equations from the words. Once they had the equations, they could solve them (assuming they'd learned the basics). \n\nI'd find it pretty difficult to give you general guidelines on "how to solve word problems with rational expressions" but the main trick with word problems is to figure out a) what variables are involved, and assign each of them a letter (e.g., speed of first train is x, time it has been travelling is y, speed of second train is z...), b) sort out the relationship between them (e.g. x = 2z/5), and c) make sure you know what *exactly* is being asked. It is not at all uncommon for word problems to include extraneous information (colour of train, age of driver) to muddle the issue.\n\nYour best bet is almost certainly to go ask your teacher, professor, TA, tutor, study group, or classmate-who-understands-it for help. They will be better able to figure out where your problem is if they can watch you work through an actual example.
702610
I'll try to help... \n\nFirst, I'm going to leave out all the x-terms, but just imagine they are there:\n\n.................. ........ ......4/5 56/25 -156/125 -1028/625\n5 ... -4 ... 9 ) 4 ... 8 ... -8 ... 4 ... 1 ... -8\n................... 4 .-16/5 36/5\n...................------------------\n................... ......56/5 -76/5 4\n................... ......56/5 -224/25 504/25\n................... ......-------------- -----------\n................... ............. -156/25 -404/25 1\n................... ............. -156/25 624/125 -1404/125\n................... ............. ------------------- ----------------\n................... ............. ........... -1028/125 1529/125 -8\n................... ............. ........... -1028/125 4112/625 -9252/625\n................... ............. ........... ------------- --------------- --------------\n................... ............. ................. .......... 3553/625 4252/625\n\nIf I did it correctly (which there is no guarantee), I get the answer as:\n\n(4/5)x^3 + (56/25)x^2 - (156/125)x - 1028/625)\nwith a remainder of:\n[ (3553/625)x + 4252/625 ] / (5x^2 -4x+9)\n\nHowever, with all those fractions I probably made a mistake somewhere... your math teacher is a masochist!
923039
email me, calcu_lust@yahoo.com, i'll write out a nice solution.\n\nEdit: Go to my website, and aforementioned nice solution is typed out as a .pdf for you to download, with all the correct symbols and such:\nhttp://geocities.com/calcu_lust/yahooans.pdf
837082
Hint: Let C=X&#92;O and let f(x)=dist(x,C). The other way is trivial.
862589
maybe you could try practising. For maths, the phrase practice makes perfect is very true. hmm...short cuts in multiplication and division...maybe you could try learning the abaccus (a mathematical instrument used by the chinese => functions like a calculator, but manually powered, no batteries)...I only know how to add and subtract using it...
914601
http://mathworld.wolfram.com/PoincareConjecture.html
567661
the Chebyshev's inequality goes Pr(|X-mean| > or = k.sd)< or = 1/k^2\n\ni m assuming u meant standard deviation when u said 4.3818, which is 2 sd.\n\nThe prob of points outside 2sd at both sides is not more than 1/k^2.. or 1/2^2=1/4.\n\nSo, the probability of points within 2sd will be at least 1-1/4=3/4.\n\nhow many data points? if u have 100 data points in total, there will be at least 100x3/4=75 data points within the mean+/-4.3818 range.\n\nNote: The exactly probability depends on the type of distribution the data follows, for example, for Normal/Gaussian, 95% lies within +/- 1.96sd. Chebyshev's inequality states the minimum possible for any type of distribution.
898860
First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D