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Analysis and Questions showa setup for each problem When reading a ruler, the precision is good to about +/-0.05 cm. Suppose
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Answer #1

Ans 1

If dia of sphere D1 = 1.20 cm

Volume of sphere V1 = (4/3)*3.14*(D1/2)^3

= (4/3)*3.14*(1.20/2)^3

= 0.904 cm3

Correct dia D2 = 1.15 cm

Volume of sphere V2 = (4/3)*3.14*(D2/2)^3

= (4/3)*3.14*(1.15/2)^3

= 0.7959 cm3

% error = (V1 - V2)*100/V2

= (0.904 - 0.7959)*100/0.7959

= 13.58%

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