166.
70, 75, 80, 85, 90, 105, 105, 130, 130, 130The list shown consists of the times, in seconds, that it took each of 10 schoolchildren to run a distance of 400 meters. If the standard deviation of the 10 running times is 22.4 seconds, rounded to the nearest tenth of a second, how many of the 10 running times are more than 1 standard deviation below the mean of the 10 running times?
(A) One
(B) Two
(C) Three
(D) Four
(E) Five
选B,看不懂,也不会做。。
3 A manufacturer produced x percent more video cameras in 1994 than in 1993 and y percent more video cameras in 1995 than in 1994. If the manufacturer produced 1,000 video cameras in 1993, how many video cameras did the manufacturer produce in 1995 ?
1 xy=20
2 X+Y+XY/100=9.2
答案选B
请问2为什么可以求出?
4. Working alone at its own constant rate, a machine seals k cartons in 8 hours, and working alone at its own constant rate, a second machine seals k cartons in 4 hours. If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate?
(A) 25%
(B)
(C) 50%
(D) 66*2/3%
(E) 75%
选D, 看不懂题目也不会解答
166 题目的意思:
70 75 80 85 90 105 105 130 130 130
上述列表所示为10个学生跑400米跑所用的时间。该10个数的标准差为22.4,四舍五入到10分位,有多少学生的跑步成绩比上述列表的平均数低出一个标准差以上?
平均数为: (70+75+80+85+90+105+105+130+130+130)/10 = 100
比平均数低1个标准差的数字 必须 小于 100 - 22.4 = 77.6
所有数字中 比77.6 小的只有 70和75 两个 所以B Tow
3 1993的产量为 1000
1994的产量为 1000(1+y%)
1995的产量为 1000(1+y%)(1+x%) = 1000(1+x%+y%+x%y%)
如果知道 X+Y+XY/100=9.2 则 1+x%+y%+x%y% = 9.2%
从而 1000(1+x%+y%+x%y%) 可以计算
所以条件2 是可以的
4 比较奇怪
自愧资质有限 难以领悟题目要旨
1,额..要找10个数中,在比MEAN小的数中,与MEAN的差比标准方差小的数...
只有70,75
2,X+Y+XY/100=9.2
看到这,我们准备求1+X/100)(1+Y/100)=1+x/100+y/100+xy/10000
到这,应该明白了吧..
3,V1=K/8
V2=K/4
面对K CARTONS
SET THE NEW ONE COST AS Y
1/Y=1/8+1/4
Y=8/3
V3=3K/8
额,后来看到问题,发现上面不要算的...随便写出来...权当偶的思路
问最快机器在相同时间内生产的产品在所有中的比率.
一看就知道.8小时,4小时.肯定是2/3的产品...所以是66.67%
... 我把第三个题目翻译一遍吧 我困惑的是 66*2/3%这个数字是什么意思
Working alone at its own constant rate, a machine seals k cartons in 8 hours, and working alone at its own constant rate, a second machine seals k cartons in 4 hours. If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate?
当单独工作于其自身的稳定工作率时,某机器密封k个纸箱需8小时;同样当单独工作于自身的稳定的工作频率时,另一机器密封k个纸箱需4个小时。如果两台机器(每个都工作在与上述相同的工作率)一起密封一定数量的纸箱时,那么工作率较高的那台机器密封了百分之多少的纸箱?
这个题目其实很容易理解,题目的第一句话描述了两个机器其中一个的工作率为另一个的两倍
所以一起工作的话 必然其中快的那个密封的纸箱书为另一个的两倍 所以占总量的66%
第一题:
十个小孩儿跑400米所需时间分别是以上十个数,这十个时间的标准方差是22.4,精确到小数点后第一位。这十个数中,有几个数比这十个数的算术平均数小一个以上(more than one)标准方差(22.4)?
答案:两个,分别是70,75。求出算数平均数为100,then100-22.4=77.6,因为是more than one standard deviation,所以结果肯定该比77.6小,那就是前两个喽。
第二题:
(1+X%)x1000x(1+Y%)=一九九五年的产量=(1+X%+Y%+XY%%)x1000
=1000+10x+10y+0.1xy
=1000+92
=1092
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