Yahoo 知識+ 將於 2021 年 5 月 4 日 (美國東岸時間) 停止服務，而 Yahoo 知識+ 網站現已轉為僅限瀏覽模式。其他 Yahoo 資產或服務，或你的 Yahoo 帳戶將不會有任何變更。你可以在此服務中心網頁進一步了解 Yahoo 知識+ 停止服務的事宜，以及了解如何下載你的資料。
Maths (measurement)
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!
2 個解答
- 9 年前最愛解答
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.
- ?Lv 79 年前
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
