# How many significant figures are there in the result of the following calculation? (4.321/28) x (6.9234 x 10)

May 15, 2017

One significant figure.

#### Explanation:

You're dealing with multiplication and division, so you know that the answer must have as many significant figures as the measurement with the least number of sig figs.

• $\textcolor{w h i t e}{1} 4.321 \text{ "->" }$ four sig figs
• $\textcolor{w h i t e}{12.1} 28 \text{ "->" }$ two sig figs
• $6.9234 \text{ "->" }$ five sig figs
• $\textcolor{w h i t e}{12.1} 10 \text{ "->" }$ one sig fig

So, you know that the answer must be rounded to one significant figure because that is how many sig figs you have for your least precise measurement.

Since your expression contains a couple of parentheses, focus on solving these first. Keep in mind that you don't really need parentheses for same-level operations, you can solve these left-to-right.

$\left(\frac{4.321}{28}\right) \times \left(6.9234 \times 10\right)$

You're going to be doing a multi-step calculation, so make sure that you keep enough sig figs for each step to avoid the loss of significance.

As a general rule, it's better to have too many sig figs for each step than too few, so I'll go with $5$ sig figs for each calculation step.

$\frac{4.321}{28} = 0.15432$

$6.9234 \times 10 = 69.234$

Put this together to get

$0.15432 \times 69.234 = 10.684$

Finally, round the answer to one significant figure to get

$\left(\frac{4.321}{28}\right) \times \left(6.9234 \times 10\right) = 10$