# If you measured the volume of something that had a length of 5.0 m and a width of 3.456 m, how many significant figures would be in your answer?

Sep 27, 2017

Your final answer should consist of two significant figures to reflect the number of significant figures in your least accurate given data.

#### Explanation:

In this example, the length was not measured with the same degree of accuracy as the width, possibly due to different measuring equipment, different method, or that other person being sloppy.

Regardless, the resulting answer can only be returned with precision to two significant figures. To maintain the accuracy as much as possible, calculations should use all the significant figures given, with the rounding-off occurring only at the answer.

Example: $V = l w h = 5.0 m \times 3.456 m \times h m = 17.28 \left(h\right) {m}^{3}$

This would need to round-off to $17 \left(h\right) {m}^{3} \to$ [ANS]

So volume in this case could not be $17.2 \left(h\right) {m}^{3}$ nor $17.28 \left(h\right) {m}^{3}$