標題:
Maths (measurement)
發問:
There is a rectangle with length 10cm and width 6cm, where all measurements are corrected to the nearest cm. If the actual value of the perimeter is p cm, find the range of values of p. Ans.:30<=p<34 Why the answer is not 30<=p<=34 ??? Please explain, thz!
最佳解答:
Because the data are to be corrected. If the length >10.4 e.g 10.45<--this will be corrected to 10.5 but not 10 the width >6.4 e.g 6.45<--also will be corrected to 6.5 not 6 So,we can know that ,the length and the width must be <=10.4 and <=6.4 respectively. The greatest value of p =(10.4+6.4)x2 p =33.6. Therefore the value of p must<34,it can not be equal to 34.
The range of length 9.5 <= x < 10.5 The range of width 5.5 <= y < 6.5 As p = 2(x + y) Min. of p = 2(9.5 + 5.5) = 30 Max. of p < 2(10.5 + 6.5) = 34 So, 30 <= p < 34
Maths (measurement)
發問:
There is a rectangle with length 10cm and width 6cm, where all measurements are corrected to the nearest cm. If the actual value of the perimeter is p cm, find the range of values of p. Ans.:30<=p<34 Why the answer is not 30<=p<=34 ??? Please explain, thz!
最佳解答:
Because the data are to be corrected. If the length >10.4 e.g 10.45<--this will be corrected to 10.5 but not 10 the width >6.4 e.g 6.45<--also will be corrected to 6.5 not 6 So,we can know that ,the length and the width must be <=10.4 and <=6.4 respectively. The greatest value of p =(10.4+6.4)x2 p =33.6. Therefore the value of p must<34,it can not be equal to 34.
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其他解答:The range of length 9.5 <= x < 10.5 The range of width 5.5 <= y < 6.5 As p = 2(x + y) Min. of p = 2(9.5 + 5.5) = 30 Max. of p < 2(10.5 + 6.5) = 34 So, 30 <= p < 34
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