maths (只有一題)

2007-11-22 2:10 am
1.若某(2n+1)個連續數的平均數是12, , 求該(2n+1)個數的中位數。

回答 (2)

2007-11-23 12:38 am
✔ 最佳答案
設首個數是(2n+1)

[(2n+1)+(2n+2)+(2n+3)] /3=12

2n+1+2n+2+2n+3=36

6n+6=36

6(n+1)=36

n=5

(2n+1)個數的中位數=(2n+2)
=(2 x 5 +2)
=12
2007-11-22 2:30 am
答案應該係5.5


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