跪求 数学2道解析 来自Prep 不胜感激

九天仙女

1. For a finite sequence of nonzero numbers, the number of variation in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of two consecutive terms is negative. What is the number of variation in sign for the sequence 1, -3, 2,5,-4,-6?

A 1 B 2 C 3 D4 E5

题读了很久，实在太绕了，然后也没做出来，答案是C3.

2. Last month 15 homes were sold in Town X, the averange (atithmetic mean ) sale price of the homes was$150,000 and the median sale price was $130,000, which of the following statement must be ture?

（1）. At least one of the homes was sold for more than $165,000

（2）.At least one of the homes was sold more than $130,000 and less than $150,000

（3）.At least one of the homes was sold less than 130,000

A. (1) only B.(2)only C. (3)only D. (1)&(2)E.(1)&(3)

答案是忘了，一会儿回去看看再补充上来。

a10285

1、题意为the number of variation in sign定义为排列中有多少对（连续的两个数相乘的积为负数），则 1*（-3）=-3      （-3）*2=-6    5*（-4）=-20  共三对数，选C

a10285

2、我用的极限法，因为中数为$130,000，则前8个数最大的时候为前8个数全为$130,000，此时后7个 数最小，此时后7个数平均数为（150000*15-130000*8）/7≈172000，
  此时（1）正确，排除（2）（3）选A
  我说的有些概括，不知能看懂不~

九天仙女

第二个题算法看懂了，只是逻辑上有点没转过来

总觉得everything is possible

1 2 3 都挺有可能的。

ps：为什么要用极限法呢？为什么中间的那个数130，000就要分别往两边算呢？150，000是所有数的平均数，不是后7个的呀？

嘿嘿~问题有点多。脑子转不过来了

麻烦你了

a10285

因为（1）（2）（3）问的都是至少有balabala，因此只要反例出现就能排除，此时用极限法最煎蛋。

因为第8个数是130000，则前七个数都≤130000，（即130000只是往前算）想得出后7个数中最小数规律，则令前8个数最大（130000）就行~

150，000的确是所有数的平均数，才有算式中的150000*15

九天仙女

哦~好难滴说

万分感谢呵