1. Approximation and Errors 3 3.1Significant Figures 3.2Scientific Notation 3.3Approximation and Errors Case Study Chapter Summary

2. P. 2 Suppose that there are actually 116 paper clips on the plate. Case Study John’s estimate  Actual number  120  116 Did John or Linda make a better estimation? I guess there are 120 paper clips. What’s your guess, Linda? I guess there are 110 clips, John. We need to find the difference between each estimate and the actual number. The smaller the difference is, the better is the estimate.  4 Actual number  Linda’s estimate  116  110  6 Since 4 is less than 6, John made a better estimation.

3. P. 3 3.1 Significant Figures A. Basic Concepts The population of Hong Kong at the end of 2007 is  7 000 000. (correct to the nearest million)  6 950 000. (correct to the nearest ten thousand)  6 953 000. (correct to the nearest thousand) In most cases, the far left non-zero digit, having the largest place value in a number, is the first significant figure. Subsequent important digits are called the second significant figure, the third significant figure, and so on. The important digits of a number are called significant figures.  7  6 95  6 953 This is also the most important figure.

4. P. 4 1 0 4 9 1st significant figure 2nd significant figure 3rd significant figure 4th significant figure For all numbers, any zeros between 2 non-zero digits are significant figures. For example, the number 1049 has 4 significant figures. 3.1 Significant Figures A. Basic Concepts We also study significant figures in decimals. For example, the decimal 5.307 has 4 significant figures.

5. P. 5 3.1 Significant Figures A. Basic Concepts 0. 0 3 8 0 1st significant figure 2nd significant figure 3rd significant figure For all decimals, any zeros after the last non-zero digit are significant figures. For decimals less than 1, all zeros in front of the first non-zero digit are not significant figures. These zeros are called place holders. For example, the number 0.0380 has only 3 significant figures.

6. P. 6 44008 Number of Significant FiguresNumbers The following table shows some examples. All the significant figures are marked with green colour. 3.1 Significant Figures A. Basic Concepts 3900 (correct to the nearest integer) has 4 significant figures. But 3900 (correct to the nearest ten) has 3 significant figures. 46.092 10.7 30.002 04 49.000 30.130

7. P. 7 Using the technique of rounding off for estimation, the number 47 054 can be expressed as: (a)47 100 (cor. to the nearest hundred) (b)47 000 (cor. to the nearest thousand) (c)50 000 (cor. to the nearest ten thousand) Alternatively, we can round off a number to a certain number of significant figures. 3.1 Significant Figures B. Round off to the Required Significant Figures We can express 47 054 as: (a)47 100 (cor. to 3 sig. fig.) (b)47 000 (cor. to 2 sig. fig.) (c)50 000 (cor. to 1 sig. fig.) The phrase ‘cor. to 3 sig. fig.’ is the short form of ‘correct to 3 significant figures’.

8. P. 8 3.1 Significant Figures B. Round off to the Required Significant Figures The following steps show how to round off 51 672, correct to 3 significant figures. Step 1:Determine which digit is to be rounded off. 51 672 3rd significant figure Step 2:Round off according to the next significant figure. 51 672  51 700 (cor. to 3 sig. fig.) Since 7 is larger than 5, we add 1 to the 3rd significant figure.

9. P. 9 Round off 472 780 correct to (a)2 significant figures, (b)3 significant figures, (c)4 significant figures. (a)470 000 (cor. to 2 sig. fig.) (b)473 000 (cor. to 3 sig. fig.) (c)472 800 (cor. to 4 sig. fig.) 3.1 Significant Figures B. Round off to the Required Significant Figures Example 3.1T Solution:

10. P. 10 3.1 Significant Figures B. Round off to the Required Significant Figures Example 3.2T Round off 0.300 649 correct to (a)1 significant figure, (b)2 significant figures, (c)5 significant figures. Solution: (a)0.3 (cor. to 1 sig. fig.) (b)0.30 (cor. to 2 sig. fig.) (c)0.300 65 (cor. to 5 sig. fig.)

11. P. 11 3.1 Significant Figures B. Round off to the Required Significant Figures Example 3.3T Round off 9995 correct to (a)1 significant figure, (b)2 significant figures, (c)3 significant figures. Solution: (a)10 000 (cor. to 1 sig. fig.) (b)10 000 (cor. to 2 sig. fig.) (c)10 000 (cor. to 3 sig. fig.)

12. P. 12 The total price of 120 oranges is $250. Round off the average price of each orange correct to 2 significant figures. 3.1 Significant Figures B. Round off to the Required Significant Figures Example 3.4T The average price of each orange  $(250  120)  $2.0833 Solution:  $2.1 (cor. to 2 sig. fig.)

13. P. 13 3.2 Scientific Notation A. Introduction  Our body produces about 173 000 000 000 red blood cells each day.  The mass of 1 water molecule is about 0.000 000 000 000 000 000 000 03 g. It is convenient to rewrite these numbers as follows:  173 000 000 000  1.73  100 000 000 000  1.73  10 11 This kind of representation of numbers is called scientific notation. In science, we often deal with numbers which are very large or small:  0.000 000 000 000 000 000 000 03 g  3  10  23 g A positive number is expressed in scientific notation if it is in the form a  10 n, where 1  a  10 and n is an integer.

14. P. 14 Express each of the following numbers in scientific notation. (a)22 000(b)  0.000 000 7 (c)95 000  10 4 (d)0.008  10 2 3.2 Scientific Notation A. Introduction Example 3.5T Solution: (a)22 000  2.2  10 000 (b)  0.000 000 7   7  0.000 000 1 (c) 95 000  10 4  (9.5  10 4 )  10 4 (d)0.008  10  2  (8  10  3 )  10  2  2.2  10 4   7  10  7  9.5  10 8  8  10  5 Move the decimal point 4 places to the left and then times 10 4 Move the decimal point 7 places to the right and then times 10  7 a m  a n  a m  n

15. P. 15 3.2 Scientific Notation A. Introduction Example 3.6T Convert the following numbers into integers or decimals. (a)1.002  10 7 (b)6  10  5 Solution: (a) 1.002  10 7  1.002  10 000 000 (b) 6  10  5  6  0.000 01  10 020 000  0.000 06

16. P. 16 3.2 Scientific Notation A. Introduction Example 3.7T Without using a calculator, evaluate each of the following expressions. Express the answers in scientific notation. (a)(b)(4  10 3 )(3.3  10 5 ) Solution:

17. P. 17 3.2 Scientific Notation A. Introduction In calculator, we should key in ‘1.32 9’ to represent 1.32  10 9. EXP The calculator would display ‘ 1.32 E 9 ’ before execution. If we key in ‘4  11 ’, then the calculator would display ‘ 4  10  11 ’, which means 4  10  11. EXPEXE Students should note that the key-in sequences of different calculators may be different.

18. P. 18 B. Applications of Scientific Notation For example: 1 228 550 000 m 3 of water was consumed in Hong Kong in 2005  2006. 1 228 550 000 m 3  1 230 000 000 m 3 (cor. to 3 sig. fig.)  1.23  10 9 m 3 3.2 Scientific Notation When presenting estimates, we can express numbers in scientific notation and round off the value correct to a certain number of significant figures.

19. P. 19 Example 3.8T B. Applications of Scientific Notation 3.2 Scientific Notation 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  10 4

20. P. 20 Example 3.9T B. Applications of Scientific Notation 3.2 Scientific Notation 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.)

21. P. 21 3.3 Approximation and Errors They are only approximations. Hence, every measurement or estimation has errors. In Book 1A Chapter 2, we learnt that no measurements give exact values.

22. P. 22 A. Absolute Error 3.3 Approximation and Errors The difference between a measured value (or an estimated value) and the actual value is called the absolute error. The absolute error is always positive. If the actual value is larger than a measured value, then absolute error  actual value  measured value If a measured value is larger than the actual value, then absolute error  measured value  actual value Example: A story book has 117 pages. Sara guessed that the book has 100 pages. ∴ The absolute error of Sara’s estimation is 17 pages.

23. P. 23 Example 3.10T A. Absolute Error 3.3 Approximation and Errors The actual weight of a pack of potato chips is 183.4 g. Find the absolute error if Julie corrects the weight 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

24. P. 24 Therefore, errors of measurements are related to the scale intervals of the tools. In the figure, the scale interval of the ruler is 1 cm. B. Maximum Absolute Error 3.3 Approximation and Errors Measuring tools with smaller scale intervals can give estimations with higher accuracy. ∴ The minimum and the maximum actual lengths are 14.5 cm and 15.5 cm respectively. The measured length is 15 cm, correct to the nearest cm. The absolute error of this measurement does not exceed 0.5 cm. This figure is called the maximum absolute error.

25. P. 25  Lower limit (  14.5 cm) Minimum possible value of measurement  Upper limit (  15.5 cm) Maximum possible value of measurement They can be found by the following formulas. B. Maximum Absolute Error 3.3 Approximation and Errors maximum absolute error   scale interval of the measuring tool In general, In the figure, the scale interval of the ruler is 1 cm. Lower limit  Measured value  Maximum absolute error Upper limit  Measured value  Maximum absolute error

26. P. 26 Example 3.11T B. Maximum Absolute Error 3.3 Approximation and Errors The capacity of a bottle is 5.38 L, correct to 3 significant figures. What is the lower limit of the capacity? Solution: The maximum absolute error   0.01 L The lower limit of the capacity  (5.38  0.005) L  5.375 L  0.005 L The scale interval of measurement  0.01 L The 3rd significant figure of the number 5.38 is 8. Hence the capacity is correct to the nearest 0.01 L.

27. P. 27 Lower limit  (8  0.5) cm  7.5 cm Upper limit  (8  0.5) cm  8.5 cm Lower limit  10.5 cm Upper limit  11.5 cm Lower limit of the area Upper limit of the area ∴ The actual area lies between 39.375 cm 2 and 48.875 cm 2. Example 3.12T B. Maximum Absolute Error 3.3 Approximation and Errors 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 area of the triangle? Solution: The maximum absolute error of measurement   1 cm  0.5 cm Base length:Height:

28. P. 28 AnimalActual weight (kg)Measured weight (kg)Absolute error (kg) Pig63.42630.42 Dog5.194.770.42 Both of the absolute errors of the 2 measurements are the same, but they do not reflect the degree of accuracy. To compare the accuracy of 2 estimations, we also have to determine the relative error. C. Relative Error 3.3 Approximation and Errors Consider the following set of data:

29. P. 29 The absolute error of the pig’s weight is small (insignificant) when compared to its actual weight. On the other hand, the absolute error of the dog’s weight is significant because it is relatively large when compared to its actual weight. In general, we use the relative error to determine the degree of accuracy of a measurement. C. Relative Error 3.3 Approximation and Errors 0.08090.425.19Dog 0.006 620.4263.42Pig Absolute error (kg)Actual weight (kg)Animal Relative error 

30. P. 30 For example, the degree of accuracy of the measurement of the pig’s weight is higher than that of the dog’s weight. Sometimes it is impossible to find the actual values in some real-life situations. So we can use the maximum absolute error and the measured value instead to find the relative error. C. Relative Error 3.3 Approximation and Errors The smaller the relative error is, the higher is the accuracy of a measurement. Relative error 

31. P. 31 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, (c)the relative error of the length, correct to 3 significant figures. (b)Maximum absolute error (c)Relative error (cor. to 3 sig. fig.) Example 3.13T C. Relative Error 3.3 Approximation and Errors Solution: (a)The length of the string  75 cm

32. P. 32 Let x g be the weight of a package of sugar. x  2.5  72  180 ∴ The weight of a package of sugar is 180 g. Example 3.14T C. Relative Error 3.3 Approximation and Errors 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: ∴ Lower limit of the actual weight  (180 – 2.5) g  177.5 g First find out the measured weight of the package of sugar.

33. P. 33 For example, since the relative errors of the pig’s and the dog’s weight are 0.006 62 and 0.0809 respectively. The percentage errors of the pig’s and the dog’s weight are 0.662% and 8.09% respectively. The smaller the percentage error is, the higher is the accuracy of a measurement. D. Percentage Error 3.3 Approximation and Errors When the relative error is expressed as a percentage, it is called the percentage error. Percentage error  Relative error  100%

34. P. 34 A university bought 1258 new computers last year. If the number of computers bought is now correct to the nearest hundred, find the percentage error of the estimation. (Give the answer correct to 3 significant figures.) Absolute error  1300 – 1258  42 Percentage error  (cor. to 3 sig. fig.) Example 3.15T D. Percentage Error 3.3 Approximation and Errors Solution: 1258  1300 (cor. to the nearest hundred)  3.34%

35. P. 35 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 who measured more accurately. (cor. to 3 sig. fig.) (a)For Sally’s measurement: Example 3.16T D. Percentage Error 3.3 Approximation and Errors Solution: Maximum absolute error  10 mL  2  5 mL Percentage error For Christine’s measurement: Maximum absolute error  50 mL  2  25 mL Percentage error (b)Since 1.67% is less than 2%, Sally measured more accurately.

36. P. 36 Chapter Summary 3.1 Significant Figures 1.For all numbers, any zeros between 2 non-zero digits are significant figures. 2.For all decimals, any zeros after the last non-zero digit are significant figures.

37. P. 37 3.2 Scientific Notation All positive numbers can be expressed in the form a  10 n, where 1  a  10 and n is an integer. Chapter Summary

38. P. 38 3.3 Approximation and Errors 1. Absolute error (a)If the actual value is larger than a measured value, then the absolute error  actual value  measured value. (b)If a measured value is larger than the actual value, then the absolute error  measured value  actual value. Chapter Summary (b)Lower limit  Measured value  Maximum absolute error 2. Maximum absolute error (a) Maximum absolute error   Scale interval of the measuring tool (c)Upper limit  Measured value  Maximum absolute error

39. P. 39 3.3 Approximation and Errors Chapter Summary 4.Percentage error Percentage error  Relative error  100% 3. Relative error

40. B. Round off to the Required Significant Figures 3.1 Significant Figures Follow-up 3.1 Round off 94 506 correct to (a)1 significant figure, (b)2 significant figures, (c)3 significant figures. (a)90 000 (cor. to 1 sig. fig.) (b)95 000 (cor. to 2 sig. fig.) (c)94 500 (cor. to 3 sig. fig.) Solution:

41. B. Round off to the Required Significant Figures 3.1 Significant Figures Follow-up 3.2 Round off 0.203 617 correct to (a)2 significant figures, (b)3 significant figures, (c)4 significant figures. (a)0.20 (cor. to 2 sig. fig.) (b)0.204 (cor. to 3 sig. fig.) (c)0.2036 (cor. to 4 sig. fig.) Solution:

42. B. Round off to the Required Significant Figures 3.1 Significant Figures Follow-up 3.3 Round off 89 989 correct to (a)1 significant figure, (b)2 significant figures, (c)4 significant figures. (a)90 000 (cor. to 1 sig. fig.) (b)90 000 (cor. to 2 sig. fig.) (c)89 990 (cor. to 4 sig. fig.) Solution:

43. Total weight of the mobile phones  (85.53  3) g  256.59 g B. Round off to the Required Significant Figures 3.1 Significant Figures Follow-up 3.4 The weight of each of 3 mobile phones is 85.53 g. Round off the total weight of the mobile phones correct to 3 significant figures. Solution:  257 g (cor. to 3 sig. fig.)

44. Follow-up 3.5 3.2 Scientific Notation A. Introduction Express each of the following numbers in scientific notation. (a)1 475 000(b)0.000 000 061 (c)  28 600 000(d)1300  10 4 Solution: (a)1 475 000  1.475  1 000 000 (b)0.000 000 061  6.1  0.000 000 01 (c)  28 600 000   2.86  10 000 000 (d)1300  10 4  (1.3  10 3 )  10 4  1.475  10 6  6.1  10  8   2.86  10 7  1.3  10 7

45. Follow-up 3.6 3.2 Scientific Notation A. Introduction Convert the following numbers into integers or decimals. (a)4.6  10 9 (b)5.02  10  4 Solution: (a) 4.6  10 9  4.6  1 000 000 000 (b) 5.02  10  4  5.02  0.0001  4 600 000 000  0.000 502

46. Follow-up 3.7 3.2 Scientific Notation A. Introduction Without using a calculator, evaluate each of the following expressions. Express the answers in scientific notation. (a)3.7  10 8  8.9  10 7 (b) Solution: the same as 2  10  4

47. Follow-up 3.8 B. Applications of Scientific Notation 3.2 Scientific Notation A flower shop made $764 280 in profit last year. Round off the number correct to 3 significant figures and express the answer in scientific notation. Solution: $764 280  $764 000 (cor. to 3 sig. fig.)  $7.64  10 5

48. Follow-up 3.9 B. Applications of Scientific Notation 3.2 Scientific Notation The mass of the Moon is about 7.35  10 22 kg. It is 1.23  10  2 times as heavy as the Earth. Find the mass of the Earth. Express the answer in scientific notation and correct to 3 significant figures. Solution: The mass of the Earth (cor. to 3 sig. fig.)

49. Follow-up 3.10 A. Absolute Error 3.3 Approximation and Errors The actual capacity of a cup is 751.5 mL. Find the absolute error if Jane corrects the value of the capacity to 3 significant figures. Solution: 751.5 mL  752 mL (cor. to 3 sig. fig.) The absolute error  (752  751.5) mL  0.5 mL

50. Follow-up 3.11 B. Maximum Absolute Error 3.3 Approximation and Errors The weight of a ring is measured to be 45.7 g, correct to 3 significant figures. What is the lower limit of the weight? Solution: The maximum absolute error   0.1 g The lower limit of the weight  (45.7  0.05) g  45.65 g  0.05 g The 3rd significant figure of the number 45.7 is 7. Hence the weight 45.7 g is correct to the nearest 0.1 g. The scale interval of measurement  0.1 g

51. Follow-up 3.12 B. Maximum Absolute Error 3.3 Approximation and Errors Tiffany measured the length of the sides of an isosceles triangle. The results are shown in the figure. If the measurements were correct to the nearest 2 cm, find the range of the perimeter of the triangle. Solution: Lower limit  (20  1) cm  19 cm Upper limit  (20  1) cm  21 cm Lower limit  23 cm Upper limit  25 cm Lower limit of the perimeter  (19 + 23 + 23) cm  65 cm Upper limit of the perimeter  (21 + 25 + 25) cm  71 cm ∴ The range of the perimeter is 65 cm  71 cm. The maximum absolute error of measurement   2 cm  1 cm Base length:Length of the sides ’24 cm’:

52. Follow-up 3.13 C. Relative Error 3.3 Approximation and Errors The figure shows the thermometer in a classroom. Find (a)the temperature in the classroom, (b)the maximum absolute error of the temperature, (c)the relative error of the temperature, correct to 3 significant figures. Solution: (b)Maximum absolute error  1  C (a)The temperature in the classroom  24  C  0.5  C (cor. to 3 sig. fig.) (c)Relative error  

53. Follow-up 3.14 C. Relative Error 3.3 Approximation and Errors Let x s be the actual result.  62.5 ∴ The actual result is 62.5 s. Jimmy recorded the time for him to finish a 400 m run with a maximum absolute error of 0.01 s. If the relative error of the result is 1.6  10  4, find the lower limit of the actual result. ∴ Lower limit of the actual result  (62.5 – 0.01) s  62.49 s Solution:

54. Follow-up 3.15 D. Percentage Error 3.3 Approximation and Errors 364 students joined a charity event. If the number of students is now correct to the nearest ten, find the percentage error of the estimation. (Give the answer correct to 2 significant figures.) Solution: Absolute error  364 – 360  4 Percentage error  (cor. to 2 sig. fig.) 364 students  360 students (cor. to the nearest ten)  1.1%

55. Follow-up 3.16 D. Percentage Error 3.3 Approximation and Errors Kelvin and Florence measured the durations of 2 fire drills separately. Kelvin’s result was 210 seconds correct to the nearest 10 seconds. Florence’s result was 360 seconds correct to the nearest 20 seconds. (a)Find the percentage errors of their measurements. (Give the answers correct to 3 significant figures.) (b)Hence determine who measured more accurately. Solution: (cor. to 3 sig. fig.) (a)For Kelvin’s measurement: Maximum absolute error  10 s  2  5 s Percentage error For Florence’s measurement: Maximum absolute error  20 s  2  10 s Percentage error (b)Since 2.38% is less than 2.78%, Kelvin measured more accurately. (cor. to 3 sig. fig.)