= B) 1 2 1 2C) ( X X1 2)2  ( X X1 2)2 4 X X1 2 S -4P2 = 122 –...

35 .

=

b)

1 2 1 2

c) ( x x

1

2

)

2

( x x

1

2

)

2

4 x x

1 2

S -4P

2

= 12

2

– 4.35 = 4.

d) x

13

x

23

( x x

1

2

) 3

3

x x x x

1 2

(

1

2

) = S

3

– 3PS = 12

3

– 3.35.12 = 468.