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OBJECTIVES

OBJECTIVES. 1. Without a calculator; ADD , SUBTRACT , MULTIPLY , and DIVIDE decimal fractions. 2. With an approved calculator; ADD , SUBTRACT , MULTIPLY , and DIVIDE decimal fractions. 3. Without a calculator, CALCULATE the average of a series of numbers.

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OBJECTIVES

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  1. OBJECTIVES 1. Without a calculator; ADD, SUBTRACT, MULTIPLY, and DIVIDE decimal fractions. 2. With an approved calculator; ADD, SUBTRACT, MULTIPLY, and DIVIDE decimal fractions. 3. Without a calculator, CALCULATE the average of a series of numbers. 4. With an approved calculator, CALCULATE the average of a series of numbers. 5. Without a calculator, EXPRESS the solution of addition, subtraction, multiplication, and division operations using the appropriate number of significant digits.

  2. DECIMAL PLACE RELATIONSHIPS Fig 3-1

  3. EXAMPLE The magnitude of 7361.298 is: • Digit Place Value • 7  1,000 = 7,000 • 3  100 = 300 •  10 = 60 •  1 = 1 • 2  1/10 = 2/10 = 200/1,000 • 9  1/100 = 9/100 = 90/1,000 • 8  1/1,000 = 8/1,000 = 8/1,000 Ex 3-1

  4. EXAMPLE Sum = 7,361 + = 7,361 = 7,361.298 Ex 3-1

  5. EXAMPLE Find the decimal equivalent of Ex 3-2

  6. EXAMPLE is: The decimal equivalent of = 2  3 = 0.6666….. where the ….. indicates successive 6’s. Ex 3-3

  7. EXAMPLE Write the common fraction that is equivalent to the decimal fraction 0.375. Recall from the previous chapter to reduce the fraction to lowest terms, factor each term into its smallest component. Ex 3-4

  8. EXAMPLE Then cancel out each factor that occurs in each term. Then multiply the factors in each term together. Thus simplified, 0.375 in a fractional form is Ex 3-4

  9. EXAMPLE Find the sum of 39.62, 41.093, and 0.0327. 39.62 41.093 + 0.0327 80.7457 Ex 3-5

  10. EXAMPLE Subtraction is done in the same manner as with whole numbers. 32.100 – 16.379 15.721 Ex 3-6

  11. EXAMPLE Multiply 16.2 and 1.15. Multiply without concern for the decimal. 162  115 810 162 162 18630 Ex 3-7

  12. EXAMPLE Multiply 16.2 and 1.15. Ex 3-8

  13. EXAMPLE Divide 41.05 by 2.5. Divide without concern for the decimal. 25 160 150 105 100 50 50 0 Ex 3-9

  14. EXAMPLE Divide 41.05 by 2.5. Ex 3-10

  15. EXAMPLE Example 3-12 Convert 0.25 to a percentage. 0.25  100% = 25% Example 3-13 Convert 2 to a percentage. 2  100% = 200% Example 3-14 Convert 1.25 to a percentage. 1.25  100% = 125% Ex 3-12 Ex 3-13 Ex 3-14

  16. EXAMPLE Find the average of the following recorded temperatures: 600° F, 596° F, 597° F, 603° F. Step 1. After making sure that the individual quantities to be averaged have the same units, add the individual numbers of quantities to be averaged. 600 + 596 + 597 + 603 = 2,396 Step 2. Count the number of numbers or quantities to be averaged. The number of items is 4. Step 3. Divide the sum found in Step 1 by the number counted in Step 2. 2,396 ÷ 4 = 599° F Ex 3-36

  17. EXAMPLE The level in a tank is recorded once a day. The recorded level in the tank since the last addition over the last several days has been 500 gals, 490 gals, 487 gals, 485 gals, and 480 gals. Calculate the average tank level. Step 1. All levels have the same units (gals) so the individual numbers can be averaged. ADD the individual numbers. 500 + 490 + 487 + 485 + 480 = 2,442 Step 2. Count the quantities to be averaged. The number of items is 5. Ex 3-37

  18. EXAMPLE Step 3. Divide the sum found in Step 1 by the quantities in Step 2. 2,442 ÷ 5 = 488.4 The average tank level for the recorded days is 488.4 gals. Ex 3-37

  19. EXAMPLE Given the following price list of replacement pumps, find the average cost. $10,200; $11,300; $9,900; $12,000; $18,000; $7,600 Step 1. After making sure that the individual quantities to be averaged have the same units, add the individual numbers or quantities to be averaged. 10,200 + 11,300 + 9,900 + 12,000 + 18,000 + 7,600 = 69,000 Step 2. Count the numbers or quantities to be averaged. Total number of prices is 6. Ex 3-38

  20. EXAMPLE Step 3:. Divide the sum found in Step 1 by the number counted in Step 2. 69,000 ÷ 6 = 11,500 Thus, the average price of the replacement pump is $11,500. Ex 3-38

  21. EXAMPLE The Purchasing department ordered replacement valves for the upcoming outage • 4 valves cost $145.25 each • 2 valves cost $137.85 each • 3 valves cost $150.00 each $ 145.25 $ 145.25 $ 145.25 $ 145.25 $ 137.85 $ 137.85 $ 150.00 $ 150.00 $ 150.00 $ 1,306.70 Step 1. The valves have the same units ($). Add the quantities to be averaged. Ex 3-39

  22. EXAMPLE Step 2. Count the quantities to be averaged. 4 + 2 + 3 = 9 Step 3. Divide the sum found in Step 1 ($1,306.70) by the number counted in Step 2 (9). $1,306.70 ÷ 9 = $145.19 The average price for the valves ordered is $145.19. Alternately, Step 1 could have been performed as (4)($145.25) + (2)($137.85) + (3)($150.00) = ($581.00) +($275.70) + ($450.00) = $1,306.70 Ex 3-39

  23. EXAMPLE Calculate the average of the following lengths. 2 ft, 30 in, 1.5 ft, 18 in Step 1. The items to be averaged have different units (ft and in). Determine which unit would be preferred. (In this case it doesn’t matter.) 2 ft = 24 in 30 in = 2.5 ft 1.5 ft =18 in 18 in = 1.5 ft Ex 3-40

  24. EXAMPLE You could convert all to feet, and sum and average. Step 1. 2 ft + 2.5 ft + 1.5 ft + 1.5 ft = 7.5 ft Step 2 4 items Step 3 7.5 ft ÷ 4 = 1.875 ft Ex 3-40

  25. EXAMPLE Alternately, you could convert to inches, sum and average. Step 1. 24 in + 30 in + 18 in + 18 in = 90 in. Step 2 4 items Step 3 90 in ÷ 4 = 22.5 in Ex 3-40

  26. THREE TEMPERATURE SCALES Fig 3-2

  27. EXAMPLE Measured Number Most Significant Digits (left) Least Significant Digits Number of Significant Digits 12,345 1 5 5 123.45 1 5 5 1,986 1 6 4 37.806 3 6 5 201 2 1 3 300.7 3 7 4 500.08 5 8 5 0.00875 8 5 3 Ex 3-41

  28. EXAMPLE Measured Number Most Significant Digits (left) Least Significant Digits Number of Significant Digits 900.030 9 right‑most 0 6 0.090 9 right‑most 0 2 200 2 2 1 200. 2 right‑most 0 3 200.0 2 right‑most 0 4 200.00 2 right‑most 0 5 Ex 3-41

  29. EXAMPLE Yard Stick Length 8.0” Width 5.0” Calculated Area 40 sq. in. Significant Digits 2 Correct Answer 40. sq. in. Ex 3-42

  30. EXAMPLE Ruler Length 8.2” Width 4.8” Calculated Area 39.36”. Significant Digits 2 Correct Answer 39 sq. in. Ex 3-43

  31. EXAMPLE Micrometer Length 8.164” Width 4.795” Calculated Area 39.14638 sq. in. Significant Digits 4 Correct Answer 39.15 sq. in. Ex 3-44

  32. EXAMPLE Round to two figures 0.193 0.19 Round to five figures 157,632 157,630 Round to three figures 7,591 7,590 Round to four figures 0.98764 0.9876 Ex 3-45

  33. EXAMPLE Round to one figure 0.193 0.2 Round to three figures 157,632 158,000 Round to two figures 7,591 7,600 Round to four figures 0.98764 0.9877 Ex 3-46

  34. EXAMPLE Round to two figures 0.1853 0.19 Round to four figures 195,753 195,800 Round to one figure 7,591 8,000 Round to three figures 0.98751 0.988 Round to two figures 18,501 19,000 Round to four figures 19,555,005 19,560,000 Ex 3-47

  35. EXAMPLE Round to two figures 0.18500 0.18 Round to four figures 195,750 195,800 Round to one figure 7,500 8,000 Round to one figure 6,500 6,000 Round to three figures 0.98450 0.984 Ex 3-48

  36. EXAMPLE Add 0.0146, 0.0950, and 0.43 Step 1 Set up to do the math as normal. 0.0146 0.0950 + 0.043 Step 2 Identify the least significant digits in each term to be added or subtracted. 0.0146 the six is the least significant digit 0.0950 the zero is the least significant digit + 0.043 the three is the least significant digit Ex 3-49

  37. EXAMPLE Step 3 Draw a vertical line to the right of the term with the least accuracy (the least significant digit that is farthest to the left). 0.0146 0.0950 + 0.043 Step 4 Do the math as normal. 0.0146 0.0950 + 0.043 0.1526 Ex 3-49

  38. EXAMPLE Step 5 Round the digit to the left of the line following the rounding rules. 0.1526 rounds to 0.153 (Rounding Rule 1) Ex 3-49

  39. EXAMPLE Example 3-50 Add 850 and 5.90 850 + 5.90 855.90 855.90 rounds to 860 (Rounding Rule 3) Example 3-51 Subtract 5.6 from 875 875 – 5.6 869.4 869.4 rounds to 869 (Rounding Rule 1) Ex 3-50 EX 3-51

  40. EXAMPLE Example 3-52 Subtract 85 from 1,000,000 1,000,000 – 85 999,915 999,915 rounds to 1,000,000 (Rounding Rule 2) Example 3-53 Subtract 0.375 from 0.5 0.5 – 0.375 0.125 0.125 rounds to 0.1 (Rounding Rule 1) Ex 3-52 EX 3-53

  41. EXAMPLE The altimeter in an airplane reads to the nearest 100 ft. A cargo plane is cruising at 35,500 ft. Inside the cargo plane are crates. The crates are each 4 ft tall. These crates are stacked five high. On top of the highest crate is a ball bearing which measures 0.350 inches in diameter. How far is the top of the ball bearing from the ground? Add 35,500 ft + (5 × 4 ft) + 0.350 in. 35,500 ft 20 ft + 0.029166 ft 35,520.029166 ft If we convert 0.350 in. to ft the problem becomes: Ex 3-54

  42. EXAMPLE The altimeter is the least accurate measurement, and it controls the accuracy of the answer. Since the plane’s altitude is only accurate to within 100 ft, this controls the accuracy of the addition problem. By rule 2 above, the sum has the same accuracy as the least accurate measurement. The altimeter cannot distinguish between 35,500 ft and 35,520.029 ft. Therefore the answer is 35,500 ft, not 35,520.029 ft. Ex 3-54

  43. EXAMPLE A scale used to weigh trucks is marked in tons (T). In an hour, the following weights are recorded: 18T, 22T, 17T, 19T, 25T, 30T, 11T, 8T. • Calculate the total weight of the trucks weighed. • The total is 150 T. • b. Calculate the average weight of all the trucks weighed. • The average is 18.75 T = 19 T Since the scale can only measure to the nearest ton, the average cannot be more accurate than the scale. Ex 3-55

  44. EXAMPLE Multiply 3.3 and 0.025. • Step 1 • 3.3 has two significant digits (rule 3). • 0.025 has three significant digits (rule 3). • 3.3  0.025 = 0.0825 • Step 3 • The answer contains two significant digits. • Step 4 • The most significant digit is 8. • Step 5 • 0.0825 rounds to 0.082 (rounding rule 3b) Ex 3-56

  45. EXAMPLE Multiply 1,780.0 and 0.050. • Step 1 • 1,780.0 has five significant digits (rule 3). • 0.050 has three significant digits (rule 3). • Step 2 • (1,780)(0.050) = 89 • Step 3 • Answer has three significant digits. • Step 4 • The most significant digit is the 8 followed by two other significant digits. 89.0 • Step 5 • 89 rounds to 89.0 (rounding rule 1) • (Note decimal and zero) Ex 3-57

  46. EXAMPLE Divide 23.5 into 180,000 • Step 1 • 23.5 has three significant digits (rule 3) • 180,000 has two significant digits (rule 2) • Step 2 • 180,000 ÷ 23.5 = 7,659.5745 • Step 3 • Answer will contain two significant digits. • Step 4 • The most significant digit is the first 7 followed by one other significant digit. • Step 5 • 7,659.5745 rounds to 7,700 (rounding rule 3a). • (Note no decimal.) Ex 3-58

  47. EXAMPLE Divide 888 by 464. • Step 1 • 888 has three significant digits (rule 2). • 464 has three significant digits (rule 2). • Step 2 • 888 ÷ 464 = 1.9137931 • Step 3 • Answer will have three significant digits. • Step 4 • The most significant digit is one and is followed by two more significant digits. • Step 5 • 1.9137931 rounds to 1.91 (rounding rule 2). Ex 3-59

  48. EXAMPLE Calculate the area of a rectangle measuring 4.473 in by 6.238 in. • 4.473 in 6.238 in • = 27.902574 in2 • = 27.90 in2 • Both measurements have the same accuracy, four significant digits. • . Ex 3-60

  49. EXAMPLE Calculate the area of a rectangle measuring 9.825 in by 3.0 in. 9.825 in 3.0 in = 29.475 in2 = 29 in2 The measurement with the least accuracy has two significant digits. Therefore the answer must have two significant digits. Ex 3-61

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