Let the average value be x, then x1+x2+...+x10 =10x.

Disassemble (x1-2) 2 = x12-4x1+4 here.

(x 1-2)2+(X2 2)2+...+(x 10-2)2 = 1 10

So (x12+x22+...+x102)-4 (x1+x2+...+x10)+10× 4 =/kloc-.

(X 12+X22+...+X 102)=40X+70？ The same form

Because the variance (x1-x) 2+(x2-x) 2+...+(x10-x) 2 = 2.

(X 12+X22+...+X 102)-2X(X 1+X2+...+X 10)+ 10X2=2

Substitute into the formula? 40X+70-2X2+ 10X2=2，

4X2+20X- 1=0

The solution is X=(-5)/2.

2. As shown in the figure, the vertex of the parabola is (0,0), which is the closest point to m (a a,0) on the parabola. M is on the X axis, and M must be to the left of (0,0), so A.

Personally, remember to adopt!