1、ulate the wages.4.3 The Personnel Administration is responsible for supervising the implementation of this plan. And examine the assessment result according to the statistics they get.4.4 If this plan should cause the problem of the changes and optimization of the organization structure and work flo
2、w, the Warehousing Department is responsible for following up.5. The monthly wages of piece workers and related terms interpretation 5.1 The monthly wages of piece workers in the Warehousing Department are divided into Class A and Class B. Class A includes piecework staff like: warehouse keeper and
3、forklift operator. Their monthly income as following formula: Class A monthly income= basic salary + composite allowance A + piecework pay +assessment bonus deductionClass B piecework positions include group leader and shift leader whose monthly salaries formulas are as follows:Group leaders income
4、= basic salary + composite allowance B+ assessment bonus - deductionShift leaders income = basic salary+ composite allowance B+ assessment bonus - deduction5.2 Related terms interpretation5.2.1 Composite allowance A includes company welfare bonus (double salary, bonuses, holiday benefits etc.), paid
5、 holiday, average shift, night shift and rotating shift allowances.5.2.2 Assessment bonus: It is divided in to Lvl.A, B, and C according to the monthly operation efficiency and quality. The bonus ranges from 0-200 Yuan RMB. (See the detail in The assessment bonus standards of individuals and teams)5
6、.2.3 Composite allowance B includes integrated management commission, company welfare benefits, paid vacation allowance, average shift, night shift and rotating shift allowances. (shift leader on duty only)5.2.4 Basic salary: Team leaders salaries are calculated according to their administration in
7、time assessment.5.2.5 Piece rate wage: Under the premise of guaranteeing quality, the piece rate wage is calculated according to workers actual operation procedure and quantity.Piece rate wage = piecework unit salary x personal piecework quantity5.2.6 Deduction: It includes operation accident, forkl
8、ift accident, loss due to prohibited operation as well as the assessment of related leaders.5.2.7 Quality of piecework sees also Piecework methods and assessment standards6. The calculation methods about piece rate wages6.1 Piecework contentClass: Including all sectors of delivery and receiving of g
9、oods,replenishment,all sectors of packing operation, packing material delivery and receiving and batch purchasing.Class : No separate calculation, receipt confirmation/QA release (Document staff operation), stocktaking (10 Yuan per hour, not included in piecework)Class : Team tasks including sorting
10、 the pallets for the shift duty and pallets recycling 6.2 Piecework unit salary: 6.2.1 For the basic operation unit salary, see The piecework labor hour cal天下教育网 http:/ 中小学ppt课件、教案、学案、试题、等教育资源网！（考试时间120分钟 满分150分）本试卷分为选择题（共40分）和非选择题（共110分）两部分【试题总体说明】本套试卷的题型分布与2011年北京高考题没有区别，延续了北京的8、6、6分布。6道大题的考点与以往也没
11、有什么不同，分别涉及了解三角形、立体几何、概率、导数、解析几何、集合新题型。所以可见，命题人在命题过程中是有考虑的，在试题整体上，没有过多变化，力求平稳。1命题覆盖面广，琐碎知识考察力度加大。这套前14道小题，几乎没有高中同一章节的内容，考察内容十分分散。其实，这是新课标的一个重要特点。新课标的理科教材与原大纲相比，内容有增无减，增加了算法、三视图、积分、几何概型、平面几何、参数方程极坐标等许多内容，而这些内容一定要体现在高考试卷中。本套试题的小题1,2,3,4,5,6，9,10等试题难度较低，考查学生的基础知识掌握情况.2.中档题注重综合，难题注重新颖。这次试题中的8、14题都是综合问题，第
12、8题是线性规划与集合综合、第14题是新概念的题目，考察学生综合运用知识的能力，稍有失误就会失分。这套试卷的小题有很鲜明的特色，活而不难。3.解答题构思巧妙，体现知识的综合性，考查学生的素质和能力.这次解答题的命题点与以往是没有变化的，变化的只是具体的题目。第17题立体几何，考查探索性问题。15解三角形和向量结合，试题比单独考查三角函数便增加了难度. 18题的背景较为新颖，需读懂题意，考查基古典概率问题。第20题，以数列为背景考查学生的综合素质,难度较大。第一部分（选择题 共40分）注意事项：考生务必将答案答在答题卡上，在试卷上答无效。一、选择题：本大题共8小题，每小题5分，共40分在每小题给出
13、的四个选项中，选出符合题目要求的一项.1.已知集合，则等于（ ）ABC D【答案】D【解析】.2.已知平面向量，且，则实数的值为 （ ）A B C D【答案】B【解析】因，可得3. 函数的图象大致是 （ ）来源:学,科,网【答案】B【解析】当函数的图象是抛物线；当只需把函数的图象在y轴右侧部分向下平移1个单位即可，故图象大致为B.4. 设数列是公差不为0的等差数列，且成等比数列，则的前项和等于 （ ）A B C D【答案】A【解析】因成等比数列，故5.执行如图所示的程序框图，输出的值为（ ）A B C D 【答案】D【解析】程序运行一次：T=1，S=0;运行两次：T=1，S=-1;运行三次：T
14、=0，S=-1; 运行四次：T=-1，S=0，输出S=0，程序结束.6. 函数的一个零点在区间内，则实数的取值范围是（ ）A B C D 【答案】C【解析】由条件可知7. 已知函数，设，则的大小关系是 （ ）A. B. C. D. 【答案】B【解析】因为函数在上单调递增，所以8. 已知集合, .若存在实数使得成立,称点为“”点，则“”点在平面区域内的个数是 ( ) A. 0 B. 1 C. 2 D. 无数个【答案】A【解析】要使成立,首先函数的图象与函数的图象必须有公共点，由可得若点在区域C内，则必有代入（1）可得方程无整数解，故满足条件的点不存在，选A.第二部分（非选择题 共110分）二、填空题：本大题共6小题，每小题5分，共30分.把答案填在答题卡上.9. 若变量，满足约束条件 则的最大值为 . 【答案】【解析】画出约束条件所表示的平面区域如图所示：且A(1,1), 在点处取得最大值.10. 已知有若干辆汽车通过某一段公路，从中抽取辆汽车进行测速分析，其时速的频率分布直方图如图所示，则时速在区间上的汽车大约有 辆.来源:学科网ZXXK【答案】80【解析】在上的数据的频率为则时速在的汽车大约有辆.11.
