APPENDIX B
Calculations and Significant Figures
Most of the applied examples and exercises in this book involve approximate values.
For example, one exercise states that the moon has a radius of 1074 miles. This does
not mean that the moon’s radius is exactly 1074 miles; it simply means that this is the
radius rounded to the nearest mile.
One simple method for specifying the accuracy of a number is to state how many
significant digits it has. The significant digits in a number are the ones from the first
nonzero digit to the last nonzero digit (reading from left to right). Thus 1074 has four
significant digits, 1070 has three, 1100 has two, and 1000 has one significant digit. This
rule may sometimes lead to ambiguities. For example, if a distance is 200 km to the
nearest kilometer, then the number 200 really has three significant digits, not just one.
This ambiguity is avoided if we use scientific notation—that is, if we express the number as a multiple of a power of 10:
2.00 10 2
When working with approximate values, students often make the mistake of giving a
final answer with more significant digits than the original data. This is incorrect because
you cannot “create” precision by using a calculator. The final result can be no more accurate than the measurements given in the problem. For example, suppose we are told that
the two shorter sides of a right triangle are measured to be 1.25 in. and 2.33 in. long. By
the Pythagorean Theorem we find, using a calculator, that the hypotenuse has length
"1.252 1 2.332  2.644125564 in.
But since the given lengths were expressed to three significant digits, the answer cannot
be any more accurate. We can therefore say only that the hypotenuse is 2.64 in. long,
rounding to the nearest hundredth.
In general, the final answer should be expressed with the same accuracy as the leastaccurate measurement given in the statement of the problem. The following rules make
this principle more precise.
Rules for Working with Approximate Data
1. When multiplying or dividing, round off the final result so that it has as
many significant digits as the given value with the fewest number of significant digits.
2. When adding or subtracting, round off the final result so that it has its last
significant digit in the decimal place in which the least-accurate given value
has its last significant digit.
3. When taking powers or roots, round off the final result so that it has the
same number of significant digits as the given value.
Example 1 ■ Working with Approximate Data
A rectangular table top is measured to be 122.64 in. by 37.3 in. Find the area and
perimeter.
Solution Using the formulas for area and perimeter, we get the following.
Area 5 length 3 width 5 122.64 3 37.3 < 4570 in2
Three significant
digits
Perimeter 5 21 length 1 width2 5 21 122.54 1 37.32 < 319.9 in. Tenths digit
So the area is approximately 4570 in2, and the perimeter is approximately 319.9 in.
■
B-1
B-2 APPENDIX B ■ Calculations and Significant Figures
Note that in the formula for the perimeter the value 2 is an exact value, not an approximate measurement. It therefore does not affect the accuracy of the final result. In
general, if a problem involves only exact values, we may express the final answer with
as many significant digits as we wish.
Note also that to make the final result as accurate as possible, you should wait until
the last step to round off your answer. If necessary, use the memory feature of your
calculator to retain the results of intermediate calculations.
B Exercises
1–10 ■ Significant Figures Evaluate the expression. Round
your final answer to the appropriate number of decimal places or
significant figures.
1. 3.27 2 0.1834 2. 102.68 1 26.7
3. 28.36 3 501.375 4.
5. 1 1.362 3 6. !427.3
7. 3.31 642.75 1 66.787 2 201,186
5238
701
8.
1.27 2 10.5
9. 1 5.10 3 1023 2 1 12.4 3 107 2 1 6.007 3 1026 2
7
10.
1 1.361 3 10 2 1 4.7717 3 10 2
25
1.281876
11–12 ■ Significant Figures in Geometry Use the geometric
formulas on the inside front cover of the book to solve these
problems.
11. Find the circumference and area of a circle whose radius is
5.27 ft.
12. Find the volume of a cone whose height is 52.3 cm and
whose radius is 4.267 cm.
13–14 ■ Newton’s Law of Gravity The gravitational force F (in
newtons) between two objects with masses m1 and m2 (in kg),
separated by a distance r (in meters), is given by Newton’s Law
of Gravity: F5G
m1m2
r2
where G < 6.67428 3 10211 Nm2/kg2 .
13. Find the gravitational force between two satellites in stationary earth orbit, 57.2 km apart, each with a mass of 11,426 kg.
14. The sun and the earth are 1.50 3 1011 m apart, with masses
1.9891 3 1030 kg and 5.972 3 1024 kg, respectively.
(a) Find the gravitational force between the sun and the
earth.
(b)Convert your answer in part (a) from newtons to pounds,
using the fact that 1 N < 0.225 lb .
ANSWERS to Appendix B Exercises
1. 3.09 2. 129.4 3. 14,220 4. 38.41 5. 2.52 6. 20.67 7. 2300 8. 75.9 9. 3.80 10. 506.6 11. 33.1 ft, 87.3 ft2 12. 997 cm3 13. 2.66  1012 N 14. (a) 3.52  1022 N (b) 7.93  1021 lb
B-3