【费费数学】第六部分（1-5）

hjnh

xie xie !    zhong  yu  ming bai  liao!

steven-2004

不对啊，5题说的的是五个不同整数，楼上说的是相同的整数啊，不符合条件

kekegrace

以下是引用耳朵在2004-2-10 3:16:00的发言：
第一题，题目第一句话说的是：“each policy has a paper record or an electric record”，就是两者只居其一即可，有只有P的也有只有E的，也有两个都有的。
但是问的是P和E都正确的概率。可是不知道P和E单独的比例，因此这个算法不完整。

以下是引用耳朵在2004-2-10 3:49:00的发言：
（见32楼）画了个图，表示得更清楚。
题目中要求的是黄色的部分，即B。
已知A+B+C+D+E+F=1，而现在E等于（D+E）的60%，也等于（E+F）的75%，但是算不出来A和C，所以B占的比例无法求出。

以下是引用bunnier在2004-6-20 0:06:00的发言：

Once again, if the answer must be right, then its question must be wrong...  for the record: 94% is the right answer for the following questions:

• what is the probability that a policy isn't recorded incorrectly? (~P & ~E)


• what is the probability that it is one having correct paper [or] correct electric record, [or both]? (P | E)

As for the question "what is the probability that it is one having both correct paper and correct electric record?" (P & E), agree with 耳朵gg that the problem doesn't offer enough information to solve it. However, if the problem states that each policy has a paper record and an electric record, the question is solvable... a revised #1:

In an insurance company, each policy has a paper record and an electric record. 60 percent of the policies having incorrect paper record have incorrect electric record and 75 percent of the policies having incorrect electric record have incorrect paper record. 3 percent of all the policies have both incorrect paper and incorrect electric records. If we randomly pick out one policy, what is the probability that it is one having both correct paper and correct electric record?

偶也是这样想的，偶同意耳朵和bunnier的结论和分析！

严重感谢耳朵和bunnier！！！

建议版主在方便的时候可以在费费详解版里注明一下此题目本身有争议，请大家进入论坛看具体讨论。这样后来的XDJM可以看到完整的分析哦。

horsefish

4、xyz<>0, 问x(y+z)>0?
（1）|x+y|=|x|+|y|
（2）|y+z|=|y|+|z|
【答案】C
【思路】由（1）可知X、Y 是同号，由（2）可知Y、Z 也是同号，则X、Y、Z 都是同号，所以 x(y+z)>0

质疑highlight部分, 倒底能不能推出来x,y,z同号, 请NN解答. (因为我见以前的jj类似这样的问题, 答案都选的是E)

horsefish

2、If for all x, x^2+ax+b = (x+c)^2, then a =?
（1）b=0；
（2）c=-3；
【答案】B
【思路】x^2+ax+b = (x+c)^2= x^2+2cx+c^2，则A=2C，只要知道C 的值就可以知道A 值了

由上式b=c^2,a=2C,那么条件1是不是也可以呢?

流沙

以下是引用horsefish在2004-9-18 12:07:00的发言：

2、If for all x, x^2+ax+b = (x+c)^2, then a =?
（1）b=0；
（2）c=-3；
【答案】B
【思路】x^2+ax+b = (x+c)^2= x^2+2cx+c^2，则A=2C，只要知道C 的值就可以知道A 值了

由上式b=c^2,a=2C,那么条件1是不是也可以呢?

同意

tiger1234

马雨姐:

2、If for all x, x^2+ax+b = (x+c)^2, then a =?
（1）b=0；
（2）c=-3；

so we want to know the value of a while as know that as for all x, x^2 + ax + b = (x + c)^2 = x^2 + 2xc + c^2

1) b = 0, now we have x^2 + ax + 0 = x^2 + 2xc + c^2 => ax = 2xc + c^2 since for all x, ax = 2xc + c^2 then ax = 2xc(the reason why this is true is because x here is a varible not a number, because this works for all x) => a = 2c, and since ax = 2xc + c^2 => c^2 = 0 => c = 0 => a = 0, thus 1) by itself is sufficient to know the answer

2) c = -3, now we have x^2 + ax + b = x^2 -6x + 9 => ax = -6x => a = -6, thus 2) by itself is sufficient to know the answer

thus the answer is D

[此贴子已经被作者于2004-9-19 8:40:40编辑过]

tiger1234

4、xyz<>0, 问x(y+z)>0?
（1）|x+y|=|x|+|y|
（2）|y+z|=|y|+|z|

so if we know xyz != 0, we want to know whether x(y + z) > 0

1) |x + y| = |x| + |y| => x*y >=0(meaning x and y have the same sign), since we know xyz != 0, that means none of the x, y, and z is 0, thus we know that x*y >0, now, even though we know x*y > 0, but x and y could be both positive and negative, thus x + y could be greater than 0 and less than 0, since we have no further information about z other than the fact that it's not 0, 1) by itself is not sufficient to establish the answer

2) |y+z|=|y|+|z| => y*z >=0, because of  the same reason that applies to 1), 2) is not sufficient to know the answer

now with 1) and 2) together we know that x*y > 0 => x and y have the same sign, y*z > 0 => y and z have the same sign => x, y, and z all have the same sign, since xyz != 0, it is sufficient to establish that x*(y + z) > 0, thus 1) and 2) together is sufficient to know the answer, thus the answer is C

[此贴子已经被作者于2004-9-19 8:39:57编辑过]

tiger1234

1、In an insurance company, each policy has a paper record or an electric record, or both of them. 60 percent of the policies having incorrect paper record have incorrect electric record and 75 percent of the policies having incorrect electric record have incorrect paper record. 3 percent of all the policies have both incorrect paper and incorrect electric records. If we randomly pick out one policy, what is the probability that it is one having both correct paper and correct electric record?

we have no information about the ratio among policies that only have papre record, policies that only have electric record, and ones that have both of them; if the question states: "each policy has a paper record and an electric record" then the answer is 94%

[此贴子已经被作者于2004-9-19 8:45:11编辑过]

horsefish

以下是引用tiger1234在2004-9-19 8:27:00的发言：

4、xyz<>0, 问x(y+z)>0?
（1）|x+y|=|x|+|y|
（2）|y+z|=|y|+|z|

so if we know xyz != 0, we want to know whether x(y + z) > 0

1) |x + y| = |x| + |y| => x*y >=0(meaning x and y have the same sign), since we know xyz != 0, that means none of the x, y, and z is 0, thus we know that x*y >0, now, even though we know x*y > 0, but x and y could be both positive and negative, thus x + y could be greater than 0 and less than 0, since we have no further information about z other than the fact that it's not 0, 1) by itself is not sufficient to establish the answer

2) |y+z|=|y|+|z| => y*z >=0, because of  the same reason that applies to 1), 2) is not sufficient to know the answer

now with 1) and 2) together we know that x*y > 0 => x and y have the same sign, y*z > 0 => y and z have the same sign => x, y, and z all have the same sign, since xyz != 0, it is sufficient to establish that x*(y + z) > 0, thus 1) and 2) together is sufficient to know the answer, thus the answer is C

我还是质疑:当 XY>0; YZ>0联合后能否推出xyz同号?