Question

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.

Answer 1

1.59%.

Step-by-step explanation:

The following data were obtained from the question:

Measured density = 1.92 g/mL

Accepted density = 1.89 g/mL

Percentage error =...?

The percentage error error of the student can be obtained as follow:

Percentage error = |Measured density – Accepted density |/Accepted density × 100

Percentage error = |1.92 – 1.89|/1.89 × 100

Percentage error = 0.03/1.89 × 100

Percentage error = 1.59%

Therefore, the student percentage error is 1.59%