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标题: 费费数学第七章几题不明！ [打印本页]

作者: hedonism555 时间: 2003-9-10 23:19
标题: 费费数学第七章几题不明！
38、K is the square of an integal N,which of the choice below could be the number of factors of K?
Ⅰ奇数  Ⅱ偶数  Ⅲ质数

该题并没问有多少“不同”的因子个数，所以答案是否应是2。偶数。

44、The possibility that the value of stock A will increase is 0.34 and the possibility that stock B will increase is 0.68.What is the biggest possibility that neither will happen？

Why the answer is 0.32?  My way is 1-0.34*0.68

48、一位男士有5套衣服，7条领带；其中有一套衣服只能配一条领带，其余衣服和领带都可搭配，问一共有几种搭配方法。

题目说“有一套衣服”是否是指具体的某一套，如果按C1,5+4*6的算法是不是违背题意？

49、一个平行四边形的边长分别为6、7，问以下那个可能是平行四边形的面积？
(1)21  (2)42  (3)84

Why 21 ?

64、x,y为整数，y能否被3整除？
(1)y=2x^3+9x^2-8x (2)x=3

No clue why A.

72、四封信装入四个已写好正确地址的信封(每封信只装一个信封)，仅有一封信装对地址的概率是多少？

The question is similar to 48. Since the question refers to only one single letter can be correctly enveloped, then why C1,4*2. Its hidden meaning is that each letter can be fitted into a correct envelop in each one time. This is not what the question indicates.  My answer is

3,3-1(the remaining 3 all correct)-3(2 out of 3 wrong)/p4/4

Any strong mind can kindly pierce a few ray of inspirational light into my confusing mental fog ? Thank you very much !

作者: hedonism555 时间: 2003-9-11 15:47
Nobody can do me a favor? help!!!!!!!!!!!!!!!!!!!!!!

作者: sftd 时间: 2003-9-11 17:26
49、一个平行四边形的边长分别为6、7，问以下那个可能是平行四边形的面积？
(1)21 (2)42 (3)84

Why 21 ?

64、x,y为整数，y能否被3整除？
(1)y=2x^3+9x^2-8x (2)x=3

No clue why A.

49. s=absinA=42sinA, A为a,b夹角

64. y=2x^3+9x^2-8x, 9x^2能被3整除, 只要考虑2x^3-8x能否被3整除,
2x^3-8x=2x(x+2)(x-2)能被3整除

作者: hedonism555 时间: 2003-9-11 18:37
Thank you! 49 题还是不太明白。能详细讲一下吗？

作者: sftd 时间: 2003-9-11 18:56
49. 平行四边形, 如果a为底边, bsinA为底边上的高啊, 画个图就知道了.
同理,三角形面积为1/2absinA

作者: hedonism555 时间: 2003-9-11 19:22
Okay, i roughly grasp the idea. This area is my weak point. Thanks again!

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