it's DN60; d56 = DN56 and d50 = DN50
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I think that you're doing the same thing that I did for IB. Here's what I think you're asking for : 0, 2, 6, 12, 20, 30, 42... c₁ = 0 c₂ = 2 c₃ = 6 (c₂ + 4 = 2 + 4 = 6) c₄ = 12 (c₃ + 6 = 2 + 4 + 6 = 12) c₅ = 20 (c₄ + 8 = 2 + 4 + 6 + 8 = 20) c₆ = 2 + 4 + 6 + 8 + 10 etc... dn = (n/2) <2c₁ + (n-1) 2> dn = (n/2) <2 (2) + (n-1) 2> dn = (n/2) (4 + 2n - 2) dn = (n/2) (2 + 2n) dn = (2n/2) + (2n²/2) dn = n + n²
Your answer is not complete, as you need something after the last "plus". To give an answer that's still correct though, let's call it n: If: y = x2/(x2 + n) dy/dx = [2x(x2 + n) - (2x + dn/dx) ] / (x2 + n)2 = (2x3 + 2nx - 2x - dn/dx) / (x4 + 2nx2 + n2) = (2x3 + [2n - 2]x - dn/dx) / (x4 + 2nx2 + n2) If n is a constant, then you can remove the term dn/dx, and simplify the term [2n - 2]x
D=E1+((dN/10)-Cf)(I)/F The formula of deciles is different for grouped and ungrouped data.