问题
解答题
已知x1,x2,…,xn的平均数
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答案
=. 2x+3y
[(2x1+3y1)+(2x2+3y2)+…+(2xn+3yn)]1 n
=
[(2x1+2x2+…+2xn)+(3y1+3y2+…+3yn)]1 n
=
×2(x1+x2+…+xn)+1 n
×3(y1+y2+…+yn)1 n
=2×
+3×x1+x2+…+xn n y1+y2+…+yn n
=2
+3. x . y
=2a+3b.