5 Approximation and Errors 5.1 Significant Figures 5.2 Scientific Notation 5.3 Errors

5.1 Significant Figures A. Basic Concepts There were 73 074 candidates sitting for the HKDSE Examination in 2012. (a) If we are only interested in the number of candidates correct to the nearest ten thousand, i.e., 73 074  70 000, then:  73 074  70 000 (cor. to the nearest 10 000)  73 074  70 000 (cor. to 1 sig. fig.) The phrase ‘cor. to 1 sig. fig.’ is the short form of ‘correct to 1 significant figure’. (b) If we want to have more information on the number of candidates in thousands, i.e., 73 074  73 000, then:  73 074  73 000 (cor. to the nearest 1000)  7 3 074  73 000 (cor. to 2 sig. fig.)

5.1 Significant Figures A. Basic Concepts In most cases, the far left non-zero digit, having the largest place value in a number, is the first significant figure (also called the most important figure). Subsequent important digits are called the second significant figure, the third significant figure, and so on. The concept of significant figures also applies to decimals. For example, the number 5.307 has 4 significant figures.

5.1 Significant Figures A. Basic Concepts In general, the same quantity expressed in different units should have the same number of significant figures, such as 0.038 km and 38 m.

5.1 Significant Figures A. Basic Concepts

5.1 Significant Figures B. Rounding off to the Required Significant Figures

5.1 Significant Figures B. Rounding off to the Required Significant Figures

Example 1T 5 Approximation and Errors Solution: Round off 472 780 correct to (a) 2 significant figures, (b) 3 significant figures, (c) 4 significant figures. Solution: (a) 472 780 = (cor. to 2 sig. fig.) (b) 472 780 = (cor. to 3 sig. fig.) (c) 472 780 = (cor. to 4 sig. fig.)

Example 2T 5 Approximation and Errors Solution: Round off 0.300 649 correct to (a) 1 significant figure, (b) 2 significant figures, (c) 5 significant figures. Solution: (a) 0.300 649 = (cor. to 1 sig. fig.) (b) 0.300 649 = (cor. to 2 sig. fig.) (c) 0.300 649 = (cor. to 5 sig. fig.)

Example 3T 5 Approximation and Errors Solution: Round off 9995 correct to (a) 1 significant figure, (b) 2 significant figures, (c) 3 significant figures. Solution: (a) 9995 = (cor. to 1 sig. fig.) (b) 9995 = (cor. to 2 sig. fig.) (c) 9995 = (cor. to 3 sig. fig.)

Example 4T 5 Approximation and Errors Solution: The total price of 120 oranges is $250. Round off the average price of each orange correct to 2 significant figures. Solution: The average price of each orange (cor. to 2 sig. fig.)

5.2 Scientific Notation A. Introduction

5.2 Scientific Notation A. Introduction You may also determine the power of 10 by counting. For example, the decimal point is moved to the left by 11 places:

5.2 Scientific Notation A. Introduction

Example 5T 5 Approximation and Errors Solution: Express each of the following numbers in scientific notation. (a) 22 000 (b) 0.000 000 7 (c) 95 000  104 (d) 0.008  10–2 Solution: (a) 22 000  2.2  10 000  2.2  104 (b) 0.000 000 7  7  0.000 000 1  7  107 (c) 95 000  104  (9.5  104)  104  9.5  108 (d) 0.008  102  (8  103)  102  8  105

Example 6T 5 Approximation and Errors Solution: Convert the following numbers into integers or decimals. (a) 1.002  107 (b) 6  105 Solution: (a) 1.002  107  1.002  10 000 000  10 020 000 (b) 6  105  6  0.000 01  0.000 06

Example 7T 5 Approximation and Errors Solution: Without using a calculator, evaluate each of the following expressions. Express the answers in scientific notation. (a) (b) (4  103)(3.3  105) Solution:

5.2 Scientific Notation B. Using a Calculator to Perform Operations Involving Scientific Notation Different calculators have different key-in sequence. In this book, we use the model of CASIO fx-3650P for discussion.

5.2 Scientific Notation C. Applications of Scientific Notation

Example 8T 5 Approximation and Errors Solution: 36 084 people were injured in road accidents last year. Round off the number correct to 3 significant figures and express the answer in scientific notation. Solution: 36 084  36 100 (cor. to 3 sig. fig.)  3.61  104

Example 9T 5 Approximation and Errors Solution: The atomic radii of a helium atom and a gold atom are 3.1  10-11 m and 1.35  10-10 m respectively. How many times is the atomic radius of a gold atom to that of a helium atom? Give the answer correct to 2 significant figures. Solution: The required ratio (cor. to 2 sig. fig.)

5.3 Errors

5.3 Errors A. Absolute Error The absolute error is always positive. Depending on the situation, it may or may not have a unit.

Example 10T 5 Approximation and Errors Solution: The actual weight of a pack of potato chips is 183.4 g. Find the absolute error if the weight is measured correct to 1 significant figure. Solution: 183.4 g  200 g (cor. to 1 sig. fig.) The absolute error  (200  183.4) g  16.6 g

5.3 Errors B. Maximum Absolute Error

5.3 Errors B. Maximum Absolute Error

5.3 Errors B. Maximum Absolute Error

Example 11T 5 Approximation and Errors Solution: The capacity of a bottle is 5.38 L, correct to 3 significant figures. What is the lower limit of the actual capacity? Solution: The scale interval of measurement  0.01 L The maximum absolute error   0.01 L  0.005 L The lower limit of the actual capacity  (5.38  0.005) L  5.375 L

Example 12T 5 Approximation and Errors Solution: Gordon measured the base length and height of a triangle to be 8 cm and 11 cm respectively, both correct to the nearest 1 cm. What is the range of the actual area of the triangle? Solution: The maximum absolute error of measurement   1 cm  0.5 cm Lower limit of the base length  (8  0.5) cm  7.5 cm Lower limit of the height  (11  0.5) cm = 10.5 cm Upper limit of the base length  (8  0.5) cm  8.5 cm Upper limit of the height  (11 + 0.5) cm = 11.5 cm

Example 12T 5 Approximation and Errors Solution: Gordon measured the base length and height of a triangle to be 8 cm and 11 cm respectively, both correct to the nearest 1 cm. What is the range of the actual area of the triangle? Solution: Lower limit of the area Upper limit of the area ∴ The actual area lies between 39.375 cm2 and 48.875 cm2.

5.3 Errors C. Relative Error 0.006 62 0.0823

5.3 Errors C. Relative Error We use this alternative approach to find the relative error when an actual value cannot be obtained.

Example 13T 5 Approximation and Errors Solution: The figure below shows the length of a piece of string. Find (a) the length of the string, (b) the maximum absolute error of the length, the relative error of the length, correct to 3 significant figures. Solution: (a) The length of the string  75 cm (b) Maximum absolute error (c) Relative error (cor. to 3 sig. fig.)

Example 14T 5 Approximation and Errors Solution: Louis used a balance to measure the weight of a package of sugar. The maximum absolute error of the balance is 2.5 g. If the relative error of the result is , find the lower limit of the actual result. Solution: Let x g be the measured weight of a package of sugar. x  2.5  72  180 ∴ The measured weight of a package of sugar is 180 g. ∴ Lower limit of the actual weight  (180 – 2.5) g  177.5 g

5.3 Errors D. Percentage Error

Example 15T 5 Approximation and Errors Solution: A university bought 1258 new computers last year. If the number of computers bought is correct to the nearest hundred, find the percentage error of the estimation. (Give the answer correct to 3 significant figures.) Solution: 1258  1300 (cor. to the nearest hundred) Absolute error  1300 – 1258  42 Percentage error   3.34% (cor. to 3 sig. fig.)

Example 16T 5 Approximation and Errors Solution: Sally and Christine measured the capacity of 2 boxes separately. Sally’s result was 300 mL correct to 2 significant figures. Christine’s result was 1250 mL correct to the nearest 50 mL. (a) Find the percentage errors of their measurements. (Give the answer correct to 3 significant figures if necessary.) (b) Hence, determine whose measurement is more accurate. Solution: (a) For Sally’s measurement: For Christine’s measurement: Maximum absolute error  10 mL  2  5 mL Maximum absolute error  50 mL  2  25 mL Percentage error Percentage error (cor. to 3 sig. fig.)

Example 16T 5 Approximation and Errors Solution: Sally and Christine measured the capacity of 2 boxes separately. Sally’s result was 300 mL correct to 2 significant figures. Christine’s result was 1250 mL correct to the nearest 50 mL. (a) Find the percentage errors of their measurements. (Give the answer correct to 3 significant figures if necessary.) (b) Hence, determine whose measurement is more accurate. Solution: (b) 1.67% is less than 2%. ∴ Sally measurement is more accurate.