(D) (hat {mu )}_(4)=dfrac (1)(3)(X)_(1)+dfrac (3)(4)(X)_(2)-dfrac (1)(12)(X)_(3) .

x1，x2是取自总体N(μ，1)(μ未知)的样本。hat (mu )_(1)=dfrac (2)(3)(X)_(1)+dfrac (1)(3)(X)_(2) ;

设sim N(mu ,1), X1、X2、X来自总体样本, hat (mu )=dfrac (2)(5)(X)_(1)+dfrac (2)(5)(X)_(2)+

4.设总体 sim N(mu ,(sigma )^2), x1,x2,x3为来自X的样本 hat (mu )=dfrac (1)(4)(x)_(1)+b(x)_

dfrac (1)(2)x+dfrac (1)(3)=dfrac (1)(4)x-dfrac (1)(5)

lim _(xarrow 4)dfrac (sqrt {1+2x)-3}(x-4)= (-|||-A dfrac (2)(3) .-|||-B 2-|||-C

已知(x-dfrac (1)(x))=dfrac ({x)^3-x}(1+{x)^4}-|||-__，求(x-dfrac (1)(x))=dfrac ({x)^

... +({X)_(n)}^2)-|||-;(5) (mu )^2+dfrac (1)(3)((X)_(1)+(X)_(2)+(X)_(3))-|||-;(6

f(x)= ({e)^4-dfrac (1)(3))-|||-dfrac (1)(2)(e)^4-|||-dfrac (1)(2)((e)^2-dfrac

dfrac (1)(5)(X)_(1)+dfrac (2)(5)(X)_(2)+dfrac (2)(5)(X)_(3)-|||-D .dfrac (1)(7)(

dfrac (1)(x) C. -dfrac (1)({x)^2} D. dfrac (2)({x)^3}