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x²+y²≥2x+2y-2
x²-2x+1+y²-2y+1≥0
(x-1)²+(y-1)²≥0
Ta nierówność jest prawdziwa dla dow.liczb rzeczywistych x,y
zad.25
zad.28
xs=(1+3+4+7)/4=15/4=3 3/4
β²=[(1-15/4)²+(3-15/4)²+(4-15/4)²+(7-15/4)²]/4=[(-11/4)²+(-3/4)²+(1/4)²+(13/4)²]/4=[121/16+9/16+1/16+169/16]/4=300/64
β=√(300/64)=10/8√3=(5/4)√3≈2,17 -odchylenie standardowe