數學一問～

1)
根據題目, (a+b+c)/3=6, (c+d+e)/3=7, (a+b+c+d+e)/5=6.4
所以a+b+c=18, c+d+e=21, a+b+c+d+e=32
c=(a+b+c)+(c+d+e)-(a+b+c+d+e)=18+21-32=7

2)
設原本有x項數據,
則這些數據的和為:5x
故加入13這項數據後的平均值為:(5x+13)/(x+1)
即(5x+13)/(x+1)=6
5x+13=6(x+1)
5x+13=6x+6
13-6=6x-5x
x=7
所以原本有7項數據

1.a+b+c=18...(1)
c+d+e=21...(2)
a+b+c+d+e=32...(3)
(1) put into (2)
a+b+2c+d+e=39...(4)
(4)-(3)
(a+b+2c+d+e)-(a+b+c+d+e)=39-32
c=7

if 有2個數據 (5+13)/2=9,so wrong
if 有3個數據 (10+13)/3=7.66.......... , so wrong
if 有4個數據 (15+13)/4=7,so wrong
if 有5個數據 (20+13)/5=6.6,so wrong
if 有6個數據 (25+13)/6=6.33............ ,so wrong
if 有7個數據 (30+13)/7=6+1/7,so wrong
if 有8個數據 (35+13)/8=6,so the answer is 8個數據

2007-07-25 17:00:08 補充：
the numbers in these sentenses(if 有2個數據,if 有3個數據,if 有4個數據,if 有5個數據,if 有6個數據,if 有7個數據,if 有8個數據)are wrong, they should be 1,2,3,4,5,6,7, not 2,3,4,5,6,7,8

1)
a,b,c的平均數為６-> a + b + c = 6x3 = 18 -- (1)
c,d,e平均數為７-> c + d + e = 7x3 = 21 -- (2)
a,b,c,d,e的平均數為6.4 -> a + b + c + d + e = 6.4x5 = 32 -- (3)
(1) + (2) - (3) -> c = 7

2) Let n be the total number of figures,
Total = nx5 -- (1)
Total + 13 = (n+1)x6 -- (2)
(2) - (1) -> 13 = n + 6; n = 7