KADD 2016 Zadanie 5

From Łukasz Graczykowski

Revision as of 09:07, 31 March 2016 (view source) Lgraczyk (Talk | contribs) ← Older edit		Latest revision as of 15:19, 24 February 2017 (view source) Lgraczyk (Talk | contribs)
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		+	<!--
	{| align="right"		{| align="right"
	| __TOC__		| __TOC__
Line 41:		Line 42:
	Output:		Output:
-	X1=3.00691+/-0.495242	+	X1=3.00691, u(X1)=0.495242
-	X2=2.00581+/-0.40909	+	X2=2.00581, u(X1)=0.40909
-	X3=9.97287+/-1.98102	+	X3=9.97287, u(X3)=1.98102
	rho(X1,X2)=0.156132		rho(X1,X2)=0.156132
	rho(X1,X3)=-0.00698853		rho(X1,X3)=-0.00698853
	rho(X2,X3)=0.00875851		rho(X2,X3)=0.00875851
		+
		+	po zakragleniu do 2 cyfr znaczacych:
		+	X1=3.01, u(X1)=0.50
		+	X2=2.01, u(X1)=0.41
		+	X3=9.97, u(X3)=1.98
	To jest macierz kowariancji wielkosci X		To jest macierz kowariancji wielkosci X
Line 65:		Line 71:
	1 |      3.949      2.865		1 |      3.949      2.865
-	Y1=26.0157+/-9.76487	+	Y1=26.0157, u(Y1)=9.76487
-	Y2=12.5267+/-2.86537	+	Y2=12.5267, u(Y2)=2.86537
	rho(Y1,Y2)=0.746574		rho(Y1,Y2)=0.746574
	rho(Y2,Y1)=0.746574		rho(Y2,Y1)=0.746574
		+
		+	po zakragleniu do 2 cyfr znaczacych:
		+	Y1=26.02, u(Y1)=9.76
		+	Y2=12.53, u(Y2)=2.87
		+	-->

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Latest revision as of 15:19, 24 February 2017