Chapter 5.3, Problem 27WE

To determine

To Find: One significant digit estimate of each of the following.

Answer to Problem 27WE

  $\frac{73.1\times {\left(0.493\right)}^2}{0.620\times {\left(32.6\right)}^2}=0.03$

Explanation of Solution

Given information:

  $\frac{73.1\times {\left(0.493\right)}^2}{0.620\times {\left(32.6\right)}^2}$

Concept:

There are three rules on determining how many significant figures are in a number.

They are

1.Non-zero digits are always Significant.

2.Any zeros between two significant digits are Significant.

3.A final zero or trailing zeros in the decimal portion ONLY are Significant.

  $\frac{73.1\times {\left(0.493\right)}^2}{0.620\times {\left(32.6\right)}^2}$

The following rule applies for Multiplication and Division:

The LEAST number of significant figures in any number of the problem determines the number of Significant figures in the answer.

Using Above rule

We get

  $\begin{array}{l}\Rightarrow \frac{73.1\times {\left(0.493\right)}^2}{0.620\times {\left(32.6\right)}^2}=\frac{73.1\times 0.243}{0.620\times 1060}\\ \Rightarrow \frac{73.1\times {\left(0.493\right)}^2}{0.620\times {\left(32.6\right)}^2}=\frac{17.8}{657}\\ \Rightarrow \frac{73.1\times {\left(0.493\right)}^2}{0.620\times {\left(32.6\right)}^2}=0.0271\end{array}$

But we need only one Significant figure.

Therefore,

  $\Rightarrow \frac{73.1\times {\left(0.493\right)}^2}{0.620\times {\left(32.6\right)}^2}=0.03$