A student reports the density of an article to be 1.92 g/mL. The accepted value of density is 1.89 g/mL. Calculate the student's percent error.

Respuesta :

Given :

Actual value observed , [tex]v_a=1.92\ g/ml[/tex].

Expected value , [tex]v_e=1.89\ g/ml[/tex].

To Find :

The student's percent error.

Solution :

Percentage error is given by :

[tex]P=|\dfrac{v_a-v_e}{v_e}|\times 100\ \%[/tex]

Putting given values in above equation, we get :

[tex]P=|\dfrac{1.92-1.89}{1.89}|\times 100\ \%\\\\P=1.587\ \%[/tex]

Therefore, percentage error is 1.587 %.

Hence, this is the required solution.