中三數 triangle and probability

2011-03-24 2:28 am
急急急急急
如:
在2001至3000的數字卡中
(A)抽到4的倍數的機會率
(B)抽到7的倍數的機會率
(C)抽到4及7的倍數的機會率
有冇式可以計?考試一個一個數太麻煩!!
點PROVE triangle既median同altitude??
更新1:

可唔可以解埋條式點解呀??

回答 (4)

2011-03-24 4:03 am
✔ 最佳答案
(a)在2001至3000的數字有(3000-2000)/4=250個4的倍數
P(抽到4的倍數)
250/1000=1/4
(B)在2001至3000的數字有(2996-1995)/7=143個7的倍數
P(抽到7的倍數)
143/100
(C)在2001至3000的數字中有(2996-1988)/28=36個有28的倍數
P(抽到4和7的倍數)
=250/1000+143/1000-36/1000=357/1000

2011-03-23 20:03:51 補充:
慢左D TIM >.
2011-03-24 4:01 am
在2001至3000的數字卡中
(A)抽到4的倍數的機會率
Sol
2001<=4x<=3000
500.25<=x<=750
501<=x<=750
750-501+1=250
p=250/1000=1/4
(B)抽到7的倍數的機會率
2001<=7y<=3000
285.85<=y<=428.58
286<=yx<=428
428-286+1=143p=143/1000
(C)抽到4及7的倍數的機會率
Sol
[4,7]=28
2001<=28z<=3000
71.46<=z<=107.2
72<=z<=107
107-72=1=36
250+143-36=357
p=357/1000


2011-03-24 2:46 am
2001至3000的數字中有:
(3000-2004)/4
=996/4
=249(個4的倍數)
---------------------------------------
P(4的倍數)
=249/1000
====================2001至3000的數字中有:
(3000-2006)/7
=994/7
=142(個4的倍數)
---------------------------------------
P(7的倍數)
=142/1000
=71/500
====================2001至3000的數字中有:
(3000-2020)/(4x7)
=980/28
=35(個4及7的倍數)
---------------------------------------
P(4及7的倍數)
=35/1000
=7/200
====================
參考: Hope I can help you!
2011-03-24 2:31 am


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