The possible error involved in measuring each dimension of a rectangular box is ±0.02 inches. The dimensions of the box are 8 inches by 5 inches by 12 inches. Approximate the propagated error and the relative error in the calculated volume of the box.

Respuesta :

Answer:

Step-by-step explanation:

Given

Possible dimension error is [tex]\pm 0.02 in.[/tex]

length of box [tex]L=8 in.[/tex]

Width [tex]W=5 in.[/tex]

height [tex]h=12 in.[/tex]

Volume V is given by

[tex]V=LWh[/tex]

[tex]dV=LWdh+LhdW+WhdL[/tex]

and it is given

[tex]dL=dW=dh=\pm 0.02\ in.[/tex]

since error always add therefore

[tex]dV=(8\times 5+5\times 12+8\times 12)\cdot 0.02[/tex]

[tex]dV=3.92 in.^3[/tex]

Propagated error is [tex]3.92 in.^3[/tex]

relative error[tex]=\frac{dV}{V}[/tex]

[tex]=\frac{3.92}{480}[/tex]

[tex]=0.00816[/tex]

[tex]=0.816 \%[/tex]

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