# Question #3790f

Aug 26, 2017

3 sig figs in the case of the addition of the numbers given.

#### Explanation:

When counting significant figures, two simple rules to remember ...
1. If the number has a decimal point in the number, count digits from left to right. When 1st non-zero digit is reached, start counting => number of sig. figs.

1. If the number does not have a decimal point in the number, count digits from right to left. When 1st non-zero digit is reached, start counting => number of sig. figs.

For multiplication and division operations, the final value is rounded to the data value having the least number of sig. figs.

For addition and subtraction operations, the final value is rounded to the number having the least number of digits to the right of the decimal point.

For the posted question, values given should be expressed in non-exponential form and added. That is ...

$3.25 \times {10}^{4} = 32 , 500$
$7.4 \times {10}^{3} = 7 , 400$

32,500 + 7400 = 39,900 => number value with assumed decimal point placement. Count right to left until reaching non-zero digit and start counting digits. In the case of 39,900 => 3 sig. figs. and should be reported as $3.99 \times {10}^{4}$ with 3 sig. figs.

Aug 26, 2017

Here's what's going on here.

#### Explanation:

When you need to add or subtract numbers written in scientific notation, the first thing that you need to do is to make sure that both numbers have the same order of magnitude, i.e. the same exponent.

Numbers written in scientific notation take the form

$\textcolor{w h i t e}{a a} \textcolor{b l u e}{m} \times {10}^{\textcolor{p u r p \le}{n} \textcolor{w h i t e}{a} \stackrel{\textcolor{w h i t e}{a a a a a a}}{\leftarrow}} \textcolor{w h i t e}{a \textcolor{b l a c k}{\text{the")acolor(purple)("exponent}} a a}$
$\textcolor{w h i t e}{\frac{a}{a} \textcolor{b l a c k}{\uparrow} a a a a}$
$\textcolor{w h i t e}{\textcolor{b l a c k}{\text{the")acolor(blue)("mantissa}} a}$

$\textcolor{b l u e}{3.25} \cdot {10}^{\textcolor{p u r p \le}{4}} \text{ " and " } \textcolor{b l u e}{7.4} \cdot {10}^{\textcolor{p u r p \le}{3}}$

In order to be able to add these two numbers, you must pick one and convert its exponent to the exponent of the second number.

Let's pick $3.25 \cdot {10}^{4}$. You can rewrite this number as

$3.25 \cdot {10}^{4}$

$= 3.25 \cdot 10 \cdot {10}^{3}$

$= \textcolor{b l u e}{32.5} \cdot {10}^{\textcolor{p u r p \le}{3}}$

Now you have two numbers written in scientific notation that can be added or subtracted because they have the same exponent, $\textcolor{p u r p \le}{3}$.

You will thus have

$\textcolor{b l u e}{32.5} \cdot {10}^{\textcolor{p u r p \le}{3}} + \textcolor{b l u e}{7.4} \cdot {10}^{\textcolor{p u r p \le}{3}}$

$= \left(\textcolor{b l u e}{32.5} + \textcolor{b l u e}{7.4}\right) \cdot {10}^{\textcolor{p u r p \le}{3}}$

Now, when you're working with addition or subtraction, you know that the result must have the same number of decimal places as the measurement with the least number of decimal places.

In this case, you have

• $32.5 \text{ " -> " 1 decimal place}$
• $7.4 \textcolor{w h i t e}{1} \text{ " -> " 1 decimal place}$

This means that the result of the addition of the two mantissae will have $1$ decimal place.

$\textcolor{b l u e}{32.5} + \textcolor{b l u e}{7.4} = \textcolor{b l u e}{39.9}$

You can thus say that you have

$\textcolor{b l u e}{32.5} \cdot {10}^{\textcolor{p u r p \le}{3}} + \textcolor{b l u e}{7.4} \cdot {10}^{\textcolor{p u r p \le}{3}}$

$= \left(\textcolor{b l u e}{32.5} + \textcolor{b l u e}{7.4}\right) \cdot {10}^{\textcolor{p u r p \le}{3}}$

$= \textcolor{b l u e}{39.9} \cdot {10}^{\textcolor{p u r p \le}{3}}$

Now, in normalized scientific notation, you need

$1 \le | \textcolor{b l u e}{m} | < 10$

Since

$1 \le 39.9 \textcolor{red}{\cancel{\textcolor{b l a c k}{<}}} 10$

you must rewrite the result as

$\textcolor{b l u e}{39.9} \cdot {10}^{\textcolor{p u r p \le}{3}}$

$= 3.99 \cdot 10 \cdot {10}^{3}$

$= \textcolor{b l u e}{3.99} \cdot {10}^{\textcolor{p u r p \le}{4}}$

Therefore, you can say that

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{3.25 \cdot {10}^{4} + 7.4 \cdot {10}^{3} = 3.99 \cdot {10}^{4}}}}$

So remember, when adding or subtracting two numbers written in scientific notation, the number of sig figs is given by the result of addition or the subtraction of the two mantissae after both numbers have been converted to the same order of magnitude.

In this case, the answer must be reported to three sig figs.

For example, you can redo the calculation using

$7.4 \cdot {10}^{3} = 0.74 \cdot {10}^{4}$

This time, you will have

$3.25 \cdot {10}^{4} + 0.74 \cdot {10}^{4}$

$= \left(3.25 + 0.74\right) \cdot {10}^{4}$

In this case, you're adding two numbers that have $2$ decimal places, so the result will have $2$ decimal places as well.

$3.25 + 0.74 = 3.99$

Since

$1 \le 3.99 < 10$

you don't have to convert the result to match the normalized scientific notation, so once again, you will have

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{3.25 \cdot {10}^{4} + 7.4 \cdot {10}^{3} = 3.99 \cdot {10}^{4}}}}$