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Hi, we said we wanted to quote a systematics from the slope observed in the MM2 scan close to the cut. http://www.slac.stanford.edu/~ursl/talks/070202/Q0BALLMM-scan.eps Note that although this scan does not show any significant effect we decided to quote something because of the nasty slope in the B0 sample: http://www.slac.stanford.edu/~ursl/talks/070202/Q0B0MM-scan.eps We need now to agree on what to quote. The numbers in the individual values of the MM2 cut are (cut in GeV): 0.1 BRBR = 0.0194121 +- 0.00500246(stat) +- 0.00143125(MC stat) 0.2 BRBR = 0.0179828 +- 0.00449651(stat) +- 0.00129028(MC stat) 0.3 BRBR = 0.0178606 +- 0.00431818(stat) +- 0.00126132(MC stat) 0.4 BRBR = 0.0172839 +- 0.0042426(stat) +- 0.00123396(MC stat) 0.5 BRBR = 0.0169568 +- 0.00429284(stat) +- 0.00124551(MC stat) <- nominal 0.6 BRBR = 0.0147281 +- 0.00441028(stat) +- 0.00124826(MC stat) 0.7 BRBR = 0.0164152 +- 0.00455604(stat) +- 0.00130385(MC stat) 0.8 BRBR = 0.0159495 +- 0.00465518(stat) +- 0.00133811(MC stat) 0.9 BRBR = 0.0152921 +- 0.00473141(stat) +- 0.00136396(MC stat) 1.0 BRBR = 0.0163479 +- 0.00487447(stat) +- 0.00142404(MC stat) The bin at 0.6 is ~2.5 sigmas off, but then the measurement goes back, I would not resonate on this small fluctuation. Ignoring it 0.0010 seems to me to be fair, since it comprises results (from 0.2 to 1.0) with errors up to 0.0049 (as opposed to the 0.0043 of the default). If you like an algorithm rather than some algorithm instead of eye-balling, I would suggest to take the whole spread (including the point at 0.6) and dividing by sqrt(12) under the assumption of flat distribution. This would yield 0.0014. I will add the 0.0014 in the BAD that added up with the present 0.0027 gives 0.0030) unless somebody has a better proposal. bye Ric