practice Question on sigma notation?
115539
Check out wikipedia.org for info on sigma notation.
755601
26. The pattern has to start 1, 2, 4, 8, 15. I see the differences are 1, 2, 4, 7, and the differences of the differences are 1, 2, 3. The next difference of the difference has to be 4, so, working my way back up, I figure the next difference has to be 11, making the next number 26.\n\nAs for why this makes sense: Consider the question in two dimensions, i.e. into how many parts will many random lines diivide a plane?  That pattern starts 1, 2, 4, 7, 11 - the same numbers on the differences line above.  If you want to get really abstruse, I would submit that the number parts random spaces divide the fourth dimension is 1, 2, 4, 8, 16, 31, 57.  :)\n\n* * * * * *\n\nHey, you really do ask an interesting question, but it appears I'm the only one willing to answer it.  (Yes, I see you asked the same question a few days ago.)  OK, then, let's look at it more systematically:\n\n\nOne-dimensional question: Into how many parts will random points divide a line?\n\nEasy question, obviously:  Zero points leave one part (i.e. the whole line); one point breaks it into two parts, two points into three parts, etc.  The sequence is 1, 2, 3, 4, 5, etc.  Generally, n random points divide a line into n + 1 parts. \n\n\nTwo-dimensional question: Into how many parts will random lines divide a plane?\n\nHarder question, but you can work this one out on paper.  Zero lines leave one part (the whole plane); one line breaks it in two; two lines breaks it in four.  Then it gets tricky - the third line can't go thru all four parts, but it can go through three of them, so you end up with 7 parts.  The fourth line can go through 4 of the 7 parts.  Now, how you'd prove this rigorously I don't know, but the sequence is 1, 2, 4, 7, 11, 16, etc. and is clearly derived from the one-dimensional sequence: start with 1, then add 1 to get 2, then add 2 to get 4, then add 3 to get 7, then add 4 to get 11, etc.  To put this into a formula, n random lines divide a plane into (n^2 + n + 2)/2 parts.        \n\n\nNow for the three-dimensional question: Into how many parts will random planes divide a three-dimensional space?\n\nTo visualize this one, look at a corner of your room.  Obviously, zero planes leave one part (the whole space), one plane  breaks it in two (visualize the floor of your room as the plane, for example), two planes breaks it in four (visualize the floor plus one wall), and three planes breaks it in eight (visualize the floor plus two adjoining walls).  Now, try to imagine the fourth plane - it can't go through all eight parts, but it can go through seven of them, making a total of 15 parts.  The three-dimensional  sequence starts 1, 2, 4, 8, 15, and again we see it is derived from the sequence for the previous dimension: start with 1, then add 1 to get 2, then add 2 to get 4, then add 4 to get 8, then add 7 to get 15, etc.  To get the number for 5 planes, add 11 to get 26.  To put this into a formula, n random planes divide space into (n^3 + 5n + 6)/6 parts.   \n\nBased on what we are seeing, what the heck - why not carry it forward to the fourth dimension.  Using the numbers from the three-dimensional sequence, the sequence becomes 1, 2, 4, 8, 16, 31, 57, etc.  We can conclude, for example, that 6 random three-dimensional spaces divide the fourth dimension in 57 parts.  Now that should really impress your instructor!
124951
try this site: http://tutorial.math.lamar.edu/AllBrowsers/1314/1314.asp
623638
Remaining length of string\n= 7m - 2.56m\n= 4.44m\n= 444 cm\n\nHow did I do it, below is the working.\n7m is also same as 7.00m.\n\n  7.00\n- 2.56\n------\n  4.44\n------\nCancel the last 0 to 10.\nCancel the middle zero to 9.\nCancel the 7 to 6.\n\n1m = 100cm\n4.44 m = 100 cm * 4.44 = 444cm
319006
http://www.sigmaaldrich.com/catalog/search/ProductDetail/SIGMA/L1376
205352
divide 68 by 4 \n17 for each space in between\n\nso,  0, 17, 34, 51, and 68
165476
How about (a+c)/(b+d)  ? more interesting!!\n\nbtw this only work when a/b != c/d
469250
Discrete Case:\nE[X] = Sigma_{i=1}^n x_i*p(X = x_i) \n\nwhere Sigma_{i=1}^n p(X = x_i) = 1\n\nContinuous Case:\nE[X] = int(x*p(X = x)dx,-infty,infty)\n\nThink of it as a weighted *average*.
974437
hi,\ndo loop is first excecute the task and then its check the condition.\n\ntime 1:\n---------\nd=0, i=20.\n\nd=d+(1.0/i)\nd=0+(1.0/20)\nso new d=0.05\n\nand at the same time,\ni=i/(-2)\nso new i=20/(-2)\nnew i=(-10)\n\nnow the d=0.05\nand i=(-10)\n\nand then its check the condition.\ni<5\n(-10) < 5 is true\nso it wil run the loop again,\n\nso nw come to the 2nd level...\n\nTime 2:\n----------\nd=d+(1.0/i)\nso d=0.05+ (1.0 /(-10)\nnew d =0.05+ (-0.1)\nnew d=0.05-0.1\nd= (-0.05)\n\nand concider abt i,\ni=i/(-2)\ni=(-10)/(-2)\nso new i = (+5)\n\nnow lets checks the condition,\ni < 5\n+5 < 5 is fals, \nso the program s terminated.\n\nso now the value of d= (-0.05)\n\nr u clear?\nthis is a gud egsample for tha do while loop,\nHav fun bro,\ntc.
690206
Just slide the decimal over for each multiple:\n\ngiga(G)=1,000,000,000\nmega(M)=1,000,000\nkilo(k)=1,000\nhecto(h)=100\ndeca(D)=10\n1=1 (duh!)\ndeci(d)=.1\ncenti(c)=.01\nmilli(m)=.001\nmicro(that "u" looking thing with a tail)=.000001\nnano(n)=.000000001
492690
S(I)+3f (g)--> SF (g)\n..........2...........6
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1/rt3 *1/rt2-1/rt3\n[1/rt3](1/rt2-1)\n[1/rt3](1-rt2/rt2)\nrationalising\n[(1-rt2)*rt6]/6
775240
Just apply the formulas you gave and the definitions.\nNote that the mean of the measurements (usually mu) is L here.\n\nThe expected value of S(1) is, expanding and using the independence of L(1) and L(2), \n\nES(1) = E[0.5(L(1)+ L(2))]^2\n= (1/4){E[L(1)^2] + 2E[L(1)L(2)] + E[L(2)^2]}\n= (1/4){...............+ 2E[L(1)]E[L(2)] +  ........}\n= (1/4)[(sigma^2+L^2) + 2L^2 + (sigma^2+L^2)]\n= L^2 + sigma^2/2 = A + (1/2)sigma^2\n\nso S(1) is biased upwards (as an estimator of A).\n\nSimilarly \n\nES(2) = E[0.5(L^2(1) + L^2(2))] = (1/2){E[L(1)^2] + E[L(2)^2]}\n= (1/2)[(sigma^2+L^2) + (sigma^2+L^2)] = L^2 + sigma^2\n= A + sigma^2\n\nwhich is more biased tahn S(1).\nHow about S(3) = L(1)L(2)? Using the independence of L(1) and L(2), \n\nE[L(1)L(2)] = E[L(1)] E[L(2)] = L^2\n\nso S(3) is unbiased.
734849
x+78 [greater than or equal to] 120\nx [greater than or equal to] 120-78\nx [greater than or equal to] 42
900946
S.D. = ((Sigma xi^2 - (Sigma xi / n)) / (n-1))^1/2
464863
greek letters?\nChi Xi and Sigma
200234
if this series goes upto infinity then just find the nth term which will be 2n-4.4 \nthen let E represent sigma notation\n∞\nE 2n-4.4 \nn=1\n\ni hope i was help