question 3 of 10\na student measured the density of gold to be 18.5 g/cm³. the accepted value of the density…

question 3 of 10\na student measured the density of gold to be 18.5 g/cm³. the accepted value of the density of gold is 19.3 g/cm³. what is the percent error of the students measurement?\na. 5.92%\nb. 4.15%\nc. 0.60%\nd. 2.11%
Answer
Explanation:
Step1: Calculate the error
Error = |Measured value - Accepted value| = |18.5 - 19.3| = 0.8
Step2: Calculate percent - error
Percent error = $\frac{\text{Error}}{\text{Accepted value}}\times100%=\frac{0.8}{19.3}\times100% \approx 4.15%$
Answer:
B. 4.15%