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1. May measures that a box of pin (with 100 pieces inside) weighs 168g, and the relative error is 1/84. Find the possible range of weight of each piece of pin.>2. It is known that the length of a cubic metal box is 8.5cm, cor. to the nearest 0.5 cm. Find the smallest possible volume of the box cor. to 5... 顯示更多 1. May measures that a box of pin (with 100 pieces inside) weighs 168g, and the relative error is 1/84. Find the possible range of weight of each piece of pin. > 2. It is known that the length of a cubic metal box is 8.5cm, cor. to the nearest 0.5 cm. Find the smallest possible volume of the box cor. to 5 sig. fig. > 3. Steven measures the width of a rectanglar playground to be 16.8m and the relative error is 1/840. Then he measures the length of the playground to be 25.58m and the % error is 25/1279%. Find the range of the actual width, length and area of the playground > Thz!
最佳解答:
1)maximum relative error = maximum absoulte error / measured value maximum absoulte error = maximum relative error X measured value ∴the absolute error of weight of each piece of pin = [168 X (1/84)] / 100 = 0.02g the weight of each piece of pin = 168 / 100 = 1.68g upper limit of weight of each piece of pin = measured value + maximum absoulte error = (1.68 + 0.02)g = 1.70g lower limit of weight of each piece of pin = measured value﹣maximum absoulte error = (1.68﹣0.02)g = 1.66g ∴the possible range of weight of each piece of pin is between 1.66g and 1.70g. 2)the maximum absoulte error of the length = (1/2)(0.5) = 0.25 cm the lower limit of the length = (8.5﹣0.25)cm = 8.25cm ∴the smallest possible volume of the box = (8.25)3 cm3 = 561.52 cm3 3)maximum relative error = maximum absoulte error / measured value maximum absoulte error = maximum relative error X measured value ∴the maximum absolute error of width = (16.8)(1/840) = 0.02cm the upper limit of the width = (16.8 + 0.02)cm = 16.82cm the lower limit of the width = (16.8﹣0.02)cm = 16.78cm ∴the range of the actual width is between 16.78cm and 16.82cm. the maximum absolute error of length = (25.58)(25/1279%) = 0.005cm the upper limit of the length = (25.58 + 0.005)cm = 25.585cm the lower limit of the length = (25.58﹣0.005)cm = 25.575cm ∴the range of the actual length is between 25.575cm and 25.585cm. the upper limit of the area = (16.82)(25.585)cm2 = 430.3397cm2 the lower limit of the area = (16.78)(25.575)cm2 = 429.1485cm2 ∴the range of the actual area is between 429.1485cm2 and 430.3397cm2
其他解答:31C9A75CB3B14398
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Maths Question (Approximation & Errors)發問:
1. May measures that a box of pin (with 100 pieces inside) weighs 168g, and the relative error is 1/84. Find the possible range of weight of each piece of pin.>2. It is known that the length of a cubic metal box is 8.5cm, cor. to the nearest 0.5 cm. Find the smallest possible volume of the box cor. to 5... 顯示更多 1. May measures that a box of pin (with 100 pieces inside) weighs 168g, and the relative error is 1/84. Find the possible range of weight of each piece of pin. > 2. It is known that the length of a cubic metal box is 8.5cm, cor. to the nearest 0.5 cm. Find the smallest possible volume of the box cor. to 5 sig. fig. > 3. Steven measures the width of a rectanglar playground to be 16.8m and the relative error is 1/840. Then he measures the length of the playground to be 25.58m and the % error is 25/1279%. Find the range of the actual width, length and area of the playground > Thz!
最佳解答:
1)maximum relative error = maximum absoulte error / measured value maximum absoulte error = maximum relative error X measured value ∴the absolute error of weight of each piece of pin = [168 X (1/84)] / 100 = 0.02g the weight of each piece of pin = 168 / 100 = 1.68g upper limit of weight of each piece of pin = measured value + maximum absoulte error = (1.68 + 0.02)g = 1.70g lower limit of weight of each piece of pin = measured value﹣maximum absoulte error = (1.68﹣0.02)g = 1.66g ∴the possible range of weight of each piece of pin is between 1.66g and 1.70g. 2)the maximum absoulte error of the length = (1/2)(0.5) = 0.25 cm the lower limit of the length = (8.5﹣0.25)cm = 8.25cm ∴the smallest possible volume of the box = (8.25)3 cm3 = 561.52 cm3 3)maximum relative error = maximum absoulte error / measured value maximum absoulte error = maximum relative error X measured value ∴the maximum absolute error of width = (16.8)(1/840) = 0.02cm the upper limit of the width = (16.8 + 0.02)cm = 16.82cm the lower limit of the width = (16.8﹣0.02)cm = 16.78cm ∴the range of the actual width is between 16.78cm and 16.82cm. the maximum absolute error of length = (25.58)(25/1279%) = 0.005cm the upper limit of the length = (25.58 + 0.005)cm = 25.585cm the lower limit of the length = (25.58﹣0.005)cm = 25.575cm ∴the range of the actual length is between 25.575cm and 25.585cm. the upper limit of the area = (16.82)(25.585)cm2 = 430.3397cm2 the lower limit of the area = (16.78)(25.575)cm2 = 429.1485cm2 ∴the range of the actual area is between 429.1485cm2 and 430.3397cm2
其他解答:31C9A75CB3B14398
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