北京交通大学-901-2011-真题

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2011材料力学真题

北京交通大学2011年硕士研究生入学考试试卷
科目代码：950科目名称：材料力学（共3页）
注意事项：答案一律写在答题纸上，写在试卷上不予装订和评分。
第1页
一、画图题与填空题（每小题10分，共30分）
1、画出图示单元体的最大剪应力的作用平面。
2、矩形截面梁如图所示，已知力，梁的尺寸，材料的弹性模量E。则梁上层纤维AB的伸长量为；B截面形心C点的铅垂位移为。
3、实验研究表明，材料的破坏有以下结论，其中错误的是（）
（A）不同的材料，只要应力状态相同，破坏形式一定相同。
（B）相同的材料，在不同的应力状态下，可有不同的破坏方式。
（C）不同的材料，在相同的应力状态下，可能有不同的破坏方式。
（D）相同材料，应力状态相同，破坏形式一定相同。
二、ห้องสมุดไป่ตู้20分）做图示梁的剪力图和弯矩图。
五、实验题（20分）
图示圆截面拐轴，已知轴受力，尺寸，a和直径d，试在m截面处确定合理的测点，布置应变片，设计一个电测桥路方案，通过该桥路测定材料的剪切模量G。
六、（20分）已知图示刚架的抗弯刚度EI，载荷集度q及尺寸l，试画出刚架的弯矩图。
七、计算题（20分）
圆截面折杆ABCD如图所示，已知材料常数，尺寸，杆AB和杆CD的直径均为d，若重物自高度l处以速度下落至D点。试求梁的第三相当应力的最大值。
三、计算题（15分）
跨长的铸铁梁受力如图所示，已知铸铁的许用拉应力，许用压应力。试根据截面最为合理的要求，确定T字形梁横截面的尺寸，并校核梁的强度。
四、计算题（25分）
图示结构中AB为刚性梁，CD,FG杆均为直径d=74mm的钢圆杆，杆长分别为，材料的弹性模量，比例极限许用应力，稳定安全系数，试求该结构的许可荷载。

北京交通大学管理学真题11

北京交通大学2011年硕士研究生入学考试803管理学真题一、单项选择题（每题1分，20个小题，共20分）1.管理者对某一情况进行分析，从而提出行动方案。

因此，他需要做一下工作：（1）分析评价各方案；（2）确定决策目标；（3）选择满意方案并实施；（4）认识和分析问题：（5）拟定备选方案。

正确的分析思路和程序应该是：（）A．（5）-（3）-（4）-（1）-（2）B．（4）-（5）-（1）-（2）-（3）C．（5）-（4）-（2）-（1）-（3）D．（4）-（2）-（5）-（1）-（3）2.在领导者的权力机构中，专家权在性质上属于：（）A．正式权力，具有持久性的影响和作用。

B．非正式权力，具有持久性的影响和作用。

C．正式权力，不具有持久性的影响和作用。

D．非正式权力，不具有持久性的影响和作用。

3.计划制定中的滚动计划法是动态的和灵活的，它的主要特点是：（）A．按前期计划执行情况和内外环境变化，定期修订已有计划。

B．不断逐期向前推移，时短、中期考虑有机结合。

C．按近细远粗的原则来制定，避免对不确定性远期的过早过死安排。

D．以上三方面都是。

4.近年来，许多组织都努力使自身的结构扁平化，因为扁平化组织结构具有许多优点。

请问下列哪一条不是扁平式组织结构的有点？（）A．降低管理费用B．有利于缩短上下级距离C．上下级易于协调D．有利于选择和培训下属5.在现代管理理论中，决策理论学派的代表人物是：（）A．西蒙B．孔茨C．巴纳德D．韦伯6.计划工作中强调抓关键问题、抓主要矛盾的原理是：（）A．改变航道原理B．许诺原理C．灵活性原理D．限定因素原理7.集权的缺点是：（）A．各部门统一协调困难B．易出现上级部门职权失控C．不能得到规模经济的好处D．限制了中基层管理者主动性和积极性的发挥8.关于群体形成的原因有很多，如①趋同、②外部压力、③安全需要、④自尊、⑤社会和情感需要、⑥领导需要，下列哪个选项正确？（）A．①②③④B．②③④⑤C．③④⑤⑥D．①③④⑤9.某人因获得上司的赏识而被提拔为部门主管，但上任之后却得不到同事们的支持，与同级之间的关系也远不如从前。

北京交通大学2011年硕士研究生入学考试试卷

北京交通大学2011年硕士研究生入学考试试卷（学术型）一、填空题（共15分，每题3分）1．电路如图所示，其端口Ｕ和Ｉ的伏安特性方程是＿＿＿２．图示对称三相电路，电源线电压为３８０V，负载吸收的总有功功率为３００Ｗ，当Ｃ相电源断路后负载吸收的总有功功率为＿＿＿Ｗ。

３．ＲＣ一阶电路全响应V，当初始状态不变而输入增加一倍，则全响应变为＿＿＿Ｖ。

４．已知某无源二端网络的端口电压和电流分别为,则此网络吸收的功率为＿＿＿W。

5.设非线性电感的韦安特性为：。

已知其静态工作点的电流为：，则其静态电感为＿＿＿H，动态电感为＿＿＿H。

二、（15分）图示电路中I=0，试求值。

三、（15分）在图示电感耦合电路中，已知，问负载N的（复）阻抗Z为何值时能获得最大功率？并求此最大功率。

四、（15分）图示电路已达稳态，t=0时，合上开关Ｓ，用拉氏变换求电流i(t≥0).（注：i 的方向为向下）五、（15分）图示电路，开关K在t=0时刻打开。

开关动作前电路已处稳态。

求t≥0时的六、（15分）如图所示对称三相电路中，电源线电压，求。

七、（15分）激励包含两个频率分量，要求响应含有的全部分量，但不含分量，则应选取何种无源元件，元件和参数如何选取？八、（15分）非线性电路如图，非线性电阻为电压控制型，可表示为：i=g(u)=.直流电流源，小信号。

已知。

求：(1)电路静态工作点；(2)非线性电阻上的电压u(t)、电流i(t)。

九、（15分）设电源频率为，电路如图(a)所示，其有向图如图(b)。

(1)写出关联矩阵A；(2)以支路3、4、5为树支，写出基本回路矩阵；(3)写出支路阻抗矩阵Z（以支路1-6为排列顺序）；(4)写出支路导纳矩阵Y（以支路1-6为排列顺序）。

十、（15分）电路如图所示，Ｎ为无源电阻二端口网络，其Ｔ参数为.已知电感无初始能量，且。

求电感上的电流、电压.这些图片都是我自己从拍出来的照片上截的，可能不是特别清楚。

这些题目也都是用键盘输入的，有的不是特别规范，你可以另存为07版本的看得会清楚一点！祝你考研顺利！。

2011年普通高等学校招生全国统一考试数学理试题(全国卷,含答案).doc

2011 年普通高等学校招生全国统一考试数学理试题（全国卷，含答案）本试卷分第Ⅰ卷 ( 选择题 ) 和第Ⅱ卷 ( 非选择题 ) 两部分。

第Ⅰ卷 1 至 2 页。

第Ⅱ卷 3 至 4 页。

考试结束后，将本试卷和答题卡一并交回。

第Ⅰ卷注意事项：1．答题前，考生在答题卡上务必用直径0.5 毫米黑色墨水签字笔将自己的姓名、准考证号填写清楚，并贴好条形码。

请认真核准条形码上的准考证号、姓名和科目。

2．每小题选出答案后，用 2B 铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号，在试题卷上作答无效。

．．．．．．．．．．3．第Ⅰ卷共 l2 小题，每小题 5 分，共 60 分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

一、选择题(1) 复数 z 1i ， z 为 z 的共轭复数，则 zz z 1（ A ） 2i（ B ） i（ C ） i（ D ） 2i【答案】 B(2) 函数 y 2 x( x 0) 的反函数为（ A ） yx 2( x R)（ B ）4（ C ）y 4x 2( x R)（）Dyx 2( x 0)4y 4x 2 ( x 0) 【答案】 B(3) 下面四个条件中，使 a b 成立的充分而不必要的条件是（ A ） a ＞b 1（ B ） a ＞b 1（C ） a 2＞ b 2（ D ） a 3＞ b 3【答案】 A(4) 设 S n 为等差数列a n 的前 n 项和，若 a 1 1，公差 d2 ， S k 2 S k 24 ，则 k（ A ） 8 （B ） 7（ C ） 6（ D ） 5【答案】 D(5) 设函数 f ( x) cos x(0) ，将 yf ( x) 的图像向右平移个单位长度后，所得的图3像与原图像重合，则的最小值等于（ A ）1（B ） 3（C ） 6（ D ） 93【答案】 C(6) 已知直二面角l , 点 A ， AC l , C 为垂足 , B ， BD l , D 为垂足．若 AB2, AC BD 1，则 D 到平面 ABC 的距离等于2 (B) 36 (D) 1(A)3 (C)33【答案】 CA(7) 某同学有同样的画册 2 本，同样的集邮册 3 本，从中取出 4 本赠送给 4 位朋友每位朋友 1 本，则不同的赠送方法共有(A) 4 种(B)10 种(C)18 种(D)20 种lD【答案】 BCB E(8) 曲线 y e 2 x1在点 (0,2) 处的切线与直线 y 0 和 y x 围成的三角形的面积为(A)1(B)1 (C)2 (D)1323【答案】 A(9) 设 f ( x) 是周期为 2 的奇函数，当 0x 1 时， f (x)2x(1 x) , 则 f (5 )11112(A) -(B)(C)(D)2442【答案】 A(10) 已知抛物线C ： y 24x 的焦点为 F ，直线 y2x 4 与 C 交于 A ， B 两点．则cos AFB(A)4(B)3 (C)3 (D)4 5555【答案】 D(11) 已知平面 α截一球面得圆 M ，过圆心 M 且与 α 成 600 二面角的平面 β 截该球面得圆 N .若该球面的半径为 4，圆 M 的面积为 4 ，则圆 N 的面积为(A) 7 (B) 9(C)11(D)13【答案】 D(12) r r rr rr r 1 rr r rr设向量 a ， b ， c 满足 | a | | b |1， agb， ac,bc60 ，则 | c | 的最大值2等于(A) 2 (B)3(c)2(D) 1【答案】 AB绝密★启用前2011 年普通高等学校招生全国统一考试ACD理科数学 ( 必修 +选修 II)第Ⅱ卷注意事项：1 答题前，考生先在答题卡上用直径0． 5 毫米黑色墨水签字笔将自己的姓名、准考证号填写清楚，然后贴好条形码。

北京交通大学-学年概率论与数理统计期末考试试卷(A卷)答案.doc

北京交通大学2009~2010学年第一学期概率论与数理统计期末考试试卷（A 卷）答案一．（本题满分8分）某城市有汽车100000辆，牌照编号从00000到99999．一人进城，偶然遇到一辆车，求该车牌照号中含有数字8的概率．解：设事件{}8汽车牌照号中含有数字=A ，所求概率为()A P ．…………….2分()()40951.01091155=-=-=A P A P ．…………….6分二．（本题满分8分）设随机事件A ，B ，C 满足：()()()41===C P B P A P ，()0=AB P ，()()161==BC P AC P ．求随机事件A ，B ，C 都不发生的概率．解：由于AB ABC ⊂，所以由概率的非负性以及题设，得()()00=≤≤AB P ABC P ，因此有()0=ABC P ．…………….2分所求概率为()C B A P ．注意到C B A C B A ⋃⋃=，因此有…………….2分 ()()C B A P C B A P ⋃⋃-=1…………….2分()()()()()()()ABC P BC P AC P AB P C P B P A P -+++---=1 83016116104141411=-+++---=．…………….2分三．（本题满分8分）某人向同一目标进行独立重复射击，每次射击时命中目标的概率均为p ，()10<<p ．求此人第6次射击时恰好第2次命中目标的概率．解：{}次命中目标次射击时恰好第第26P{}次射击时命中目标次目标，第次射击中命中前615P =…………….2分 {}{}次射击时命中目标第次目标次射击中命中前615P P ⋅=…………….2分()()424115151p p p p p C -=⋅-=．…………….4分四．（本题满分8分）某种型号的电子元件的使用寿命X （单位：小时）具有以下的密度函数：()⎪⎩⎪⎨⎧≤>=1000100010002x x x x p ．⑴ 求某只电子元件的使用寿命大于1500小时的概率（4分）；⑵ 已知某只电子元件的使用寿命大于1500小时，求该元件的使用寿命大于2000小时的概率（4分）．解：⑴ 设{}小时于电子元件的使用寿命大1500=A ，则(){}()321000100015001500150021500=-===>=+∞+∞+∞⎰⎰x dx x dx x p X P A P ．…………….4分 ⑵ 设{}小时于电子元件的使用寿命大0002=B ，则所求概率为()A B P ． ()()(){}(){}()A P X P A P X X P A P AB P A B P 20002000,1500>=>>==．…………….2分而 {}()211000100020002000200022000=-===>+∞+∞+∞⎰⎰x dx x dx x p X P ，所以， (){}()4332212000==>=A P X P A B P ．…………….2分五．（本题满分8分）设随机变量X 服从区间[]2,1-上的均匀分布，而随机变量⎩⎨⎧≤->=0101X X Y ．求数学期望()Y E ．解：(){}(){}1111-=⨯-+=⨯=Y P Y P Y E …………….2分 {}(){}0101≤⨯-+>⨯=X P X P …………….2分()()⎰⎰⎰⎰-∞-+∞-=-=0120003131dx dx dx x p dx x p X X313132=-=．…………….4分六．（本题满分8分）设在时间t （分钟）内，通过某路口的汽车数()t X 服从参数为t λ的Poisson （泊松）分布，其中0>λ为常数．已知在1分钟内没有汽车通过的概率为2.0，求在2分钟内至少有1辆汽车通过的概率．解：()t X 的分布列为(){}()tk e k t k t X P λλ-==!，()Λ,2,1,0=k ．…………….2分因此在1=t 分钟内，通过的汽车数为 (){}λλ-==e k k X P k!1，()Λ,2,1,0=k ．由题设，(){}2.001===-λe X P ，所以5ln =λ．…………….3分因此，(){}(){}()252425111!0521021125ln 220=-=-=⋅-==-=≥--e e X P X P λ．…………….3分七．（本题满分8分）设二维随机变量()Y X ,的联合密度函数为()⎩⎨⎧<<<<=其它020,101,xy x y x f 求：⑴ 随机变量Y 边缘密度函数()y f Y （4分）；⑵ 方差()Y D （4分）．解：⑴ ()()⎰+∞∞-=dx y x f y f Y ,．因此，当0≤y 或者2≥y 时，()0=y f Y ．…………….1分当20<<y 时，()()2,2y dx dx y x f y f y Y ===⎰⎰∞+∞-．所以， ()⎪⎩⎪⎨⎧<<=其它202y y y f Y ．…………….3分⑵ ()()34621203202====⎰⎰+∞∞-y dy y dy y yf Y E Y ． ()()2821242322====⎰⎰∞+∞-ydy y dy y f y Y E Y …………….2分所以， ()()()()929162342222=-=⎪⎭⎫⎝⎛-=-=Y E Y E Y D ．…………….2分八．（本题满分8分）现有奖券10000张，其中一等奖一张，奖金1000元；二等奖10张，每张奖金200元；三等奖100张，每张奖金10元；四等奖1000张，每张奖金2元．而购买每张奖券2元，试计算买一张奖券的平均收益．解：设X ：购买一张奖券所得的奖金．则X 的分布律为所以，…………….2分 ()531000010002100001001010000102001000011000=⨯+⨯+⨯+⨯=X E …………….4分再令Y 表示购买一张奖券的收益，则2-=X Y ，因此 ()()572532-=-=-=X E Y E （元）．…………….2分九．（本题满分8分）两家电影院竞争1000名观众，假设每位观众等可能地选择两个电影院中的一个，而且互不影响．试用中心极限定理近似计算：甲电影院应设多少个座位，才能保证“因缺少座位而使观众离去”的概率不超过1%？附：标准正态分布()1,0N 的分布函数()x Φ的某些数值表解：设甲电影院应设N 个座位才符合要求．设1000名观众中有X 名选择甲电影院，则⎪⎭⎫⎝⎛21,1000~B X ．…………….1分由题意，{}99.0≥≤N X P ．而 ()500211000=⨯=X E ，()25021211000=⨯⨯=X D ．…………….2分所以，{}()()()()⎭⎬⎫⎩⎨⎧-≤-=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧-≤-=≤250500250500N X P X D X E N X D X E X P N X P99.0250500≥⎪⎭⎫⎝⎛-Φ≈N …………….3分查表得33.2250500≥-N ，所以有 84.53625033.2500=⨯+≥N ．所以，应至少设537个座位，才符合要求．…………….2分十．（本题满分8分）设总体X 的密度函数为()⎩⎨⎧<<=其它0102x x x f ， ()n X X X ,,,21Λ是从总体X 中抽取的一个简单随机样本．令()()n n X X X X ,,,max 21Λ=，试求()n X 的密度函数()()x f n ．解：总体X 的分布函数为()⎪⎩⎪⎨⎧≥<<≤=111002x x x x x F ．…………….3分因此()n X 的密度函数为()()()()()()⎪⎩⎪⎨⎧<<⋅==--其它102121x x x n x f x F n x f n n n …………….4分⎩⎨⎧<<=-其它010212x nx n ．…………….1分十一．（本题满分12分）设总体X 的密度函数为()⎪⎩⎪⎨⎧≤>=+ααβαβαββx x x x f 01，；，其中1,0>>βα为参数，()n X X X ,,,21Λ是从总体X 中抽取的一个简单随机样本．⑴ 当1=α时，求未知参数β的矩估计量M βˆ（6分）；⑵ 当1=α时，求未知参数β的最大似然估计量Lβˆ（6分）．解：⑴ 当1=α时，密度函数为()⎩⎨⎧≤>=--10111x x x x f βββ，；，所以，()()1111-==⋅==⎰⎰⎰+∞-+∞--+∞∞-βββββαββdx x dx xx dx x xf X E ，；．…………….2分解方程：()1-=ββX E ，得解：()()1-=X E X E β．…………….2分将()X E 替换成X ，得未知参数β的矩估计量为1ˆ-=X X Mβ．…………….2分 ⑵ 当1=α时，密度函数为()⎩⎨⎧≤>=--10111x x x x f βββ，；，所以，似然函数为()()()111+-===∏ββββi n ni i x x f L ，；，()()n i x i ,,1,1Λ=>．…………….2分所以，()()()n x x x n L Λ21ln 1ln ln +-=βββ．对β求导，得()n x x x nL Λ21ln ln -=∂∂ββ．…………….2分令0ln =∂∂βL ，得方程()0ln 21=-n x x x nΛβ．解得 ()n x x x nΛ21ln =β．因此，β的最大似然估计量为 ()n X X X nΛ21ln ˆ=β．…………….2分十二．（本题满分8分）设总体()2,~σμN X ，()n X X X ,,,21Λ是从总体X 中抽取的一个简单随机样本．X 与2S 分别表示样本均值与样本方差．令nS X T 22-=，求()T E ，并指出统计量T 是否为2μ的无偏估计量．解：()μ=X E ，()nX D 2σ=，…………….2分由 ()()()()22X E X E X D -=，得 ()()()()2222μσ+=+=nX E X D XE ．…………….2分又 ()22σ=S E ，所以有…………….1分()()⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛-=n S E X E n S X E T E 2222()2222μμσ=-⎪⎪⎭⎫ ⎝⎛+=n S E n ．…………….2分这表明nS X T 22-=是2μ的无偏估计量．…………….1分北京交通大学2010~2011学年第二学期概率论与数理统计期末考试试卷（A 卷）参考答案一．（本题满分8分）在正方形(){}1,1,≤≤=q p q p D ：中任取一点()q p ,，求使得方程02=++q px x 有两个实根的概率．解：设=A “方程02=++q px x 有两个实根”，所求概率为()A P ．设所取的两个数分别为p 与q ，则有11<<-p ，11<<-q ．因此该试验的样本空间与二维平面点集(){}11,11,<<-<<-=q p q p D ：中的点一一对应．…………………………………2分随机事件A 与二维平面点集(){}04,2≥-=q p q p D A ：，即与点集()⎭⎬⎫⎩⎨⎧≥=q p q p D A 4,2：…………………2分中的点一一对应．所以， ()241312412214113112=⎪⎪⎭⎫ ⎝⎛+=⨯⎪⎪⎭⎫⎝⎛+==--⎰p p dp p D D A P A的面积的面积．…………………4分二．（本题满分8分）从以往的资料分析得知，在出口罐头导致索赔的事件中，有%50是质量问题；有%30是数量短缺问题；有%20是产品包装问题．又知在质量问题的争议中，经过协商解决的占%40；在数量短缺问题的争议中，经过协商解决的占%60；在产品包装问题的争议中，经过协商解决的占%75．如果在发生的索赔事件中，经过协商解决了，问这一事件不属于质量问题的概率是多少？解：设=1A “事件属于质量问题”，=2A “事件属于数量短缺问题”， =3A “事件属于产品包装问题”．=B “事件经过协商解决”．所求概率为()B A P 1．…………………2分由Bayes 公式，得 ()()()()()()()()()332211111A B P A P A B P A P A B P A P A B P A P B A P ++=…………………2分37735849.075.02.060.03.040.05.040.05.0=⨯+⨯+⨯⨯=．…………………2分所以，()()62264151.037735849.01111=-=-=B A P B A P ．…………………2分三．（本题满分8分）设随机事件A 满足：()1=A P ．证明：对任意随机事件B ，有()()B P AB P =．解：因为()1=A P ，所以，()()0111=-=-=A P A P ．…………………2分所以，对任意的随机事件B ，由A B A ⊂，以及概率的单调性及非负性，有 ()()00=≤≤A P B A P ，因此有()0=B A P ．…………………2分所以，对任意的随机事件B ，由B A AB B ⋃=，以及AB 与B A 的互不相容性，得 ()()()()()()AB P AB P B A P AB P B A AB P B P =+=+=⋃=0．………………4分四．（本题满分8分）设随机变量X 的密度函数为()⎩⎨⎧<<+=其它0102x bx ax x p ，并且已知()21=X E ，试求方差()X D ．解：由()1=⎰+∞∞-dx x p 及()()21==⎰+∞∞-dx x xp X E ，得()()32112ba dx bx ax dx x p +=+==⎰⎰+∞∞-，…………………2分 ()()432112ba dx bx ax x dx x xp +=+==⎰⎰+∞∞-．…………………2分由此得线性方程组 ⎪⎩⎪⎨⎧=+=+2143132b a ba ．解此线性方程组，得6,6-==b a ．…………………2分所以，()()()1035164166612222=⋅-⋅=-==⎰⎰+∞∞-dx x x x dx x p x XE ，所以，()()()()20121103222=⎪⎭⎫ ⎝⎛-=-=X E X E X D ．…………………2分五．（本题满分8分）经验表明，预定餐厅座位而不来就餐的顾客比例为%20．某餐厅有50个座位，但预定给了52位顾客，问到时顾客来到该餐厅而没有座位的概率是多少？解：设X 表示52位预订了座位的顾客中来就餐的顾客数，则()8.0,52~B X ．…………1分则所求概率为()50>X P ．…………………2分 ()()()525150=+==>X P X P X P …………………2分052525215151522.08.02.08.0⋅⋅+⋅⋅=C C 9330001278813.0=．…………………3分六．（本题满分10分）将一颗均匀的骰子独立地掷10次，令X 表示这10次出现的点数之和，求()X E （5分）与()X D （5分）．解：设k X 表示第k 次出现的点数，()10,,2,1Λ=k ．则1021,,,X X X Λ相互独立，而且∑==101k k X X ．而k X 的分布列为 ()61==j X P k ，()6,,2,1Λ=j ．…………………2分所以，()()∑∑==⋅==⋅=616161j j k k j j X P j X E2721616161=⨯==∑=j j ， ()10,,2,1Λ=k ．…………………2分所以，由数学期望的性质，得()()35102727101101101=⨯===⎪⎭⎫ ⎝⎛=∑∑∑===k k k k k X E X E X E ．…………………2分()()∑∑==⋅==⋅=612612261j j k kj j X P jXE691916161612=⨯==∑=j j ， ()10,,2,1Λ=k ．…………………2分所以，由1021,,,X X X Λ的相互独立性，及数学期望的性质，得()()345510691691101101101=⨯===⎪⎭⎫ ⎝⎛=∑∑∑===k k k k k X D X D X D ．…………………2分七．（本题满分10分）设随机变量()1,0~N X ，求随机变量122+=X Y 的密度函数．解：由题意，随机变量X 的密度函数为()2221x X e x p -=π，()+∞<<∞-x ．………1分设随机变量122+=X Y 的分布函数为()y F Y ，则有()()()⎪⎭⎫ ⎝⎛-≤=≤+=≤=211222y X P y X P y Y P y F Y ，…………………2分所以，当1≤y 时，()0=y F Y ；…………………1分当1>y 时，()⎪⎪⎭⎫⎝⎛-≤≤--=⎪⎭⎫⎝⎛-≤=2121212y X y P y X P y F Y⎰⎰------==210221212222221y x y y x dx edx eππ…………………2分因此有 ()⎪⎪⎩⎪⎪⎨⎧≤>=⎰--112221022y y dxey F y x Y π ，…………………2分所以，随机变量122+=X Y 的密度函数为()()⎪⎪⎩⎪⎪⎨⎧≤>⎪⎭⎫⎝⎛-⋅='=-⎪⎪⎭⎫⎝⎛--1121212122212212y y y ey F y p y Y Y π ()⎪⎩⎪⎨⎧≤>-=--10112141y y e y y π ．…………………2分八．（本题满分10分）设二维随机变量()Y X ,的联合密度函数为()⎩⎨⎧<<<=其它0103,x y x y x p ，求X 与Y 的相关系数Y X ,ρ．解：()()4333,13102====⎰⎰⎰⎰⎰+∞∞-+∞∞-dx x dy x dx dxdy y x xp X E x ， ()()83233,103100====⎰⎰⎰⎰⎰+∞∞-+∞∞-dx x ydy xdx dxdy y x yp Y E x，…………………2分()()5333,141322====⎰⎰⎰⎰⎰+∞∞-+∞∞-dx x dy x dx dxdy y x p x X E x，()()513,1410222====⎰⎰⎰⎰⎰+∞∞-+∞∞-dx x dy y xdx dxdy y x p y Y E x ，…………………2分()()103233,1041002====⎰⎰⎰⎰⎰+∞∞-+∞∞-dx x ydy dx x dxdy y x xyp XY E x ，所以有 ()()()()16038343103,cov =⨯-=-=Y E X E XY E Y X ，…………………2分 ()()()()8034353222=⎪⎭⎫ ⎝⎛-=-=X E X E X D ， ()()()()320198351222=⎪⎭⎫ ⎝⎛-=-=Y E Y E Y D ，…………………2分因此，有()()()573320198031603,cov ,=⋅==Y D X D Y X Y X ρ．…………………2分九．（本题满分10分）一生产线生产的产品成箱包装，假设每箱平均重kg 50，标准差为kg 5．若用最大载重量为kg 5000的汽车来承运，试用中心极限定理计算每辆车最多装多少箱，才能保证汽车不超载的概率大于977.0（设()977.02=Φ，其中()x Φ是标准正态分布()1,0N 的分布函数）．解：若记i X 表示第i 箱的重量，()n i ,,2,1Λ=．则n X X X ,,,21Λ独立同分布，且()()25,50==i i X D X E ， ()n i ,,2,1Λ=．…………………2分再设n Y 表示一辆汽车最多可装n 箱货物时的重量，则有 ∑==ni i n X Y 1．由题意，得 ()977.010100055050005505000>⎪⎭⎫ ⎝⎛-Φ≈⎪⎭⎫ ⎝⎛-≤-=≤n n n n n n Y P Y P n n ．…………4分查正态分布表，得 2101000>-=nnx ，…………………2分当99=n 时，2005.1<=x ；98=n 时，202.2>=x ，故取98=n ，即每辆汽车最多装98箱货物．…………………2分十．（本题满分8分）设总体()1,0~N X ，()621,,,X X X Λ是取自该总体中的一个样本．令()()26542321X X X X X X Y +++++=，试确定常数c ，使得随机变量cY 服从2χ分布．解：因为()1,0~N X i ，()6,,1Λ=i ，而且61,,X X Λ相互独立，所以()3,0~321N X X X ++，()3,0~654N X X X ++．…………………2分因此()1,0~3321N X X X ++，()1,0~3654N X X X ++．…………………2分而且3321X X X ++与3654X X X ++相互独立．因此由2χ分布的定义，知 ()2~33226542321χ⎪⎭⎫ ⎝⎛+++⎪⎭⎫ ⎝⎛++X X X X X X ，…………………2分即()()()2~3226542321χX X X X X X +++++．取31=c ，则有()2~2χcY ．…………………2分十一．（本题满分12分）设总体X 的密度函数为()⎪⎩⎪⎨⎧<<=-其它；0101x xx f θθθ ，其中0>θ为参数，()n X X X ,,,21Λ是从总体X 中抽取的一个简单随机样本．⑴ 求参数θ的矩估计量Mθˆ（6分）；⑵ 求参数θ的最大似然估计量L θˆ（6分）．证明：⑴ ()()11101+==⋅==⎰⎰⎰-+∞∞-θθθθθθθdx x dx xx dx x xf X E ；，…………………3分因此，得方程 ()1+=θθX E ，解方程，得 ()()21⎪⎪⎭⎫⎝⎛-=X E X E θ，将()X E 替换成X ，得参数θ的矩估计量为21ˆ⎪⎪⎭⎫ ⎝⎛-=X X M θ．…………………3分 ⑵ 似然函数为 ()()∏∏=-===ni i n ni i x x f L 1121θθθθ；，…………………2分取对数，得 ()()∑=-+=ni ix nL 1ln 1ln 2ln θθθ，对θ求导，得 ()⎪⎭⎫⎝⎛+=+=∑∑==ni i ni i x n x n L d d 11ln 21ln 212ln θθθθθθ，所以，得似然方程 0ln 211=⎪⎭⎫⎝⎛+∑=ni i x n θθ，…………………2分解似然方程，得21ln ⎪⎪⎪⎪⎭⎫ ⎝⎛=∑=ni i x n θ，因此，参数θ的最大似然估计量为 21ln ˆ⎪⎪⎪⎪⎭⎫⎝⎛=∑=ni i L X n θ．…………………2分北京交通大学2010~2011学年第一学期概率论与数理统计期末考试试卷（A 卷）答案一．（本题满分8分）一间宿舍内住有6位同学，求这6位同学中至少有2位的生日在同一个月份（不考虑出生所在的年份）的概率．解：设=A “6位同学中至少有2位的生日在同一个月份”．所求概率为()A P ．…………………………..1分考虑事件A 的逆事件：=A “6位同学的生日各在不同的月份”．…………………………..1分()()777199074.02985984665280112116612=-=-=-=P A P A P ． ……..2分 …..2分 …………..2分二．（本题满分8分）有朋友自远方来访，他乘火车、轮船、汽车、飞机来的概率分别是3.0，1.0，4.0和2.0．如果他乘火车、轮船、汽车、飞机来的话，迟到的概率分别为31、72、52、61，结果他未迟到，试问他乘火车来的概率是多少？解：设=B “朋友来访迟到”，=1A “朋友乘火车来访”， =2A “朋友乘轮船来访”，=3A “朋友乘汽车来访”， =4A “朋友乘飞机来访”．……..1分所求概率为()B A P 1，由Bayes 公式得 ……..1分 ()()()()()()()()()()()44332211111A B P A P A B P A P A B P A P A B P A P A B P A P B A P +++=…..2分⎪⎭⎫ ⎝⎛-⨯+⎪⎭⎫ ⎝⎛-⨯+⎪⎭⎫ ⎝⎛-⨯+⎪⎭⎫ ⎝⎛-⨯⎪⎭⎫ ⎝⎛-⨯=6112.05214.07211.03113.03113.0 ……..2分652.0534.0751.0323.0323.0⨯+⨯+⨯+⨯⨯=1050.29494382356==． ……………..2分三．（本题满分8分）设随机变量X 的密度函数为()⎪⎪⎪⎩⎪⎪⎪⎨⎧<≤-<≤=其它010525525025x x x xx f试求随机变量X 的分布函数()x F ．解：当0<x 时， ()()00===⎰⎰∞-∞-xx dt dt t f x F ； …….1分当50<≤x 时，()()50250200x dt t dt dt t f x F xx=+==⎰⎰⎰∞-∞-；……..2分当105<≤x 时，()()255055015212552250x x dt t dt t dt dt t f x F xx -+-=⎪⎭⎫⎝⎛-++==⎰⎰⎰⎰∞-∞-；……..2分当10≥x 时，()()102552250105505=+⎪⎭⎫⎝⎛-++==⎰⎰⎰⎰⎰∞-∞-xx x dt dt t dt t dt dt t f x F ．……..2分因此，随机变量X 的分布函数为()⎪⎪⎪⎩⎪⎪⎪⎨⎧≥<≤-+-<≤<=10110550152150500022x x xx x x x x F ．……..1分四．（本题满分8分）试决定常数C ，使得!k C p kk λ=，()Λ,2,1=k 为某一离散型随机变量X 的分布列，其中0>λ为参数．解：若使!k Cp kk λ=，()Λ,2,1=k 是某一随机变量X 的分布列，当且仅当0!≥=k Cp kk λ，()Λ,2,1=k ，而且11=∑∞=k k p ， ……..2分因此有()11111!!kkk k k k p CC C e k k λλλ∞∞∞=======-∑∑∑，……..4分所以有 11C e λ=-．……..2分五．（本题满分8分）设U 与V 分别是掷一颗均匀的骰子两次先后出现的点数．试求一元二次方程02=++V Ux x有两个不相等的实数根的概率．解：一元二次方程02=++V Ux x 有两个不相等的实数根的充分必要条件是042>-V U ，或者V U 42>．……..2分又()V U ,的联合分布列为()361,===j V i U P ，()6,,2,1,Λ=j i ．……..2分所以，一元二次方程02=++V Ux x 有两个不相等的实数根的充分必要条件是()V U ,的取值应为下列情形之一：()1,3，()2,3，()1,4，()2,4，()3,4，()1,5，()2,5，()3,5，()4,5，()5,5，()6,5，()1,6，()2,6，()3,6，()4,6，()5,6，()6,6．……..2分()361702==++有两个不相等的实数根一元二次方程V Ux x P ．……..2分六．（本题满分8分）设随机变量X 服从区间()1,2-上的均匀分布，试求随机变量2X Y =的密度函数()y f Y ．解：随机变量X 的密度函数为()⎪⎩⎪⎨⎧<<-=其它01231x x p X ．……..1分设2X Y =的分布函数为()y F Y ，则有 ()()()y X P y Y P y F Y ≤=≤=2．……..1分当0≤y 时，()0=y F Y ；当40≤<y 时，()()()()()y F y F y X y P y X P y F XX Y --=≤≤-=≤=2；当4>y 时，()1=y F Y ．……..1分综上所述，得随机变量2X Y =的分布函数为()()()⎪⎩⎪⎨⎧≥<<--≤=11400y y y F y F y y F XXY ．……..1分因此，随机变量2X Y =的密度函数为()()()()()⎪⎩⎪⎨⎧<<-+='=其它04021y y p y p y y F y p XXY Y ．……..1分当10<<y 时，10<<y ，01<-<-y ，于是有()31=y p X，()31=-y p X，因此有()()()()yy y p y p y y p XXY 3131312121=⎪⎭⎫ ⎝⎛+=-+=；当41<<y 时，21<<y ，12-<-<-y ，于是有()0=y p X，()31=-y p X，因此有()()()()yy y p y p y y p XXY 613102121=⎪⎭⎫ ⎝⎛+=-+=．……..2分因此，随机变量2X Y =的密度函数为()⎪⎪⎪⎩⎪⎪⎪⎨⎧<<≤<=其它41611031y y y y y p Y ．……..1分七．（本题满分8分）试解释“在大量独立重复试验中，小概率事件几乎必然发生”的确切意思．解：设A 是一随机事件，其概率()10<<A P ．……..1分现独立重复做试验，则在n 次独立重复试验中，事件A 至少发生一次的概率为()()nA P --11．……..2分令∞→n ，则有()()()()()11lim 111lim =--=--∞→∞→nn nn A P A P ．……..2分这表明，只要试验次数n 充分大，不管随机事件A 的概率多么小，随机事件A 在n 次独立重复试验中至少发生一次的概率与1可以任意接近，即随机事件A 在n 次独立重复试验中至少发生一次是几乎必然的．……..3分八．（本题满分8分）一公寓有200户住户，一户住户拥有汽车辆数X 的分布列为试用中心极限定理近似计算，至少要设多少车位，才能使每辆汽车都具有一个车位的概率至少为95.0？（设：()95.0645.1=Φ，其中()x Φ是()1,0N 的分布函数．）解：设需要的车位数为n ，i X 表示第i 个住户需要的车位数，()200,,2,1Λ=X ．则随机变量20021,,,X X X Λ独立同分布，而且()2.13.026.011.00=⨯+⨯+⨯=i X E ，()8.13.026.011.002222=⨯+⨯+⨯=i X E ，……..2分于是有()()()()36.02.18.1222=-=-=i i i X E X E X D ．……..1分由题意，得⎪⎪⎪⎪⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-≤⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛≤∑∑∑∑∑∑======200120012001200120012001i i i i i i i i i i i i X D X E n X D X E X P n X P ⎪⎪⎪⎪⎪⎭⎫⎝⎛⨯⨯-≤⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=∑∑∑===36.02002.1200200120012001n X D X E X P i i i i i i⎪⎭⎫⎝⎛-Φ≈72240n ．……..3分由题设，95.072240≥⎪⎭⎫⎝⎛-Φn ，因此得645.172240≥-n ，所以有 9583.25372645.1240=⨯+≥n ．因此至少需要254个车位，才能满足题设要求．……..2分九．（本题满分8分）设随机变量X 与Y 相互独立，而且都服从参数为λ的指数分布，令Y X V Y X U +=-=3,34，试求二维随机变量()V U ,的相关系数V U ,ρ．解：因为X 与Y 都服从参数为λ的指数分布，所以()()λ1==Y E X E ，()()21var var λ==Y X ．……..1分于是有()()()()λλλ113143434=⋅-⋅=-=-=Y E X E Y X E U E ，()()()()λλλ411333=+⋅=+=+=Y E X E Y X E V E ．再由X 与Y 的相互独立性，得()()()()2222519116var 9var 1634var var λλλ=⋅+⋅=+=-=Y X Y X U ，()()()()22210119var 93var var λλλ=+⋅=+=+=Y E X Y X V ． ……..3分()()()[]()223512334Y XY X E Y X Y X E UV E --=+-= ()()()223512Y E XY E X E --=()()()()()()()()()()22var 35var 12Y E Y Y E X E X E X +⋅-⋅-+⋅=⎪⎭⎫⎝⎛+⋅-⋅⋅-⎪⎭⎫ ⎝⎛+⋅=22221131151112λλλλλλ2222136524λλλλ=--=．……..2分所以有()()()()2294113,cov λλλλ=⋅-=-=V E U E UV E V U ．因此有()()()105910259var var ,cov 222,===λλλρV U V U VU ．……..2分十．（本题满分8分）设总体X 存在二阶矩，总体期望()μ=X E ，总体方差()2σ=X D ，()n X X X ,,,21Λ是从中抽取的一个样本，X 是样本均值，2S 是样本方差．⑴ 计算方差()X D （4分）；⑵ 如果()2,~σμN X ，计算方差()2S D （4分）．解：⑴ ()()n n n n X D n X n D X D n i n i i n i i 2221221211111σσσ=⋅===⎪⎭⎫ ⎝⎛=∑∑∑===．……..4分⑵ 因为总体()2,~σμN X ，()n X X X ,,,21Λ是取自总体X 中的一个样本，所以()()1~1222--n S n χσ．……..2分所以，()()()()()()12121111142422242222-=-⋅-=⎪⎪⎭⎫ ⎝⎛--=⎪⎪⎭⎫ ⎝⎛-⋅-=n n n S n D n S n n D S D σσσσσσ．……..2分十一．（本题满分10分）设()10<<B P ，证明：随机事件A 与B 相互独立的充分必要条件是()()1=+B A P B A P ．证明：必要性：设随机事件A 与B 相互独立，所以随机事件A 与B 也相互独立．因此有()()A P B A P =， ()()A P B A P =，……..3分因此有()()()()1=+=+A P A P B A P B A P ．……..2分充分性：由于 ()()1=+B A P B A P ，所以有 ()()()B A P B A P B A P =-=1．因此有()()()()()()()()()B P AB P A P B P AB A P B P B A P B P AB P --=--==11．……..3分由()10<<B P ，得()01>-B P ，因此有 ()()()()()()()AB P A P B P B P AB P -=-1．整理，得 ()()()()()()()B P AB P B P A P AB P B P AB P -=-．即得 ()()()B P A P AB P =．这表明随机事件A 与B 相互独立．……..2分十二．（本题满分10分）⑴ 设总体X 等可能地取值1，2，3，Λ，N ，其中N 是未知的正整数．()n X X X ,,,21Λ是取自该总体中的一个样本．试求N 的最大似然估计量．（7分）⑵ 某单位的自行车棚内存放了N 辆自行车，其编号分别为1，2，3，…，N ，假定职工从车棚中取出自行车是等可能的．某人连续12天记录下他观察到的取走的第一辆自行车的编号为12， 203， 23， 7， 239， 45， 73， 189， 95， 112， 73， 159，试求在上述样本观测值下，N 的最大似然估计值．（3分）解：⑴ 总体X 的分布列为 {}Nx X P 1==， ()N x ,,2,1Λ=．所以似然函数为 (){}nni i i N x X P N L 11===∏=， ()()n i N x i ,,2,1,1Λ=≤≤．……..3分当N 越小时，似然函数()N L 越大；另一方面，N 还要满足：()n i N x i ,,2,1,1Λ=≤≤，即{}()n n x x x x N =≥,,,max 21Λ．所以，N 的最大似然估计量为()n X N =ˆ．……..4分 ⑵ 由上面的所求，可知N 的最大似然估计值为()239ˆ==n x N ．……..3分北京交通大学2012~2013学年第一学期概率论与数理统计期末考试试卷（A 卷）参考答案某些标准正态分布的数值其中()x Φ是标准正态分布的分布函数．一．（本题满分5分）口袋中有10个球，分别标有号码1到10，从中任意取出4个球．求最小号码是5的概率．解：设=A “取出4个球，最小号码是5”．10个球取出4个球，有取法410C 种．………….2分若最小号码是5，有取法35C 种，因此()2112101041035===C C A P ．………….3分二．（本题满分5分）一间宿舍住有5位同学，求他们之中至少有两位的生日在同一个月份的概率．解：设=A “5位同学至少有两位的生日在同一月份”．5位同学，每一位在12个月份中任意选择，共有512种可能．………….2分考虑A 的逆事件A ，它表示5位同学中，没有两位的生日是同一月份的．则 ()()6181.012115512=-=-=PA P A P ．………….3分三．（本题满分8分），已知男人中%5的是色盲患者，女人中色盲患者占%25.0，今从男女比例为21:22的人群中随机地挑选一人，发现是色盲患者，问此人是男性的概率是多少？解：设=A “任选一人为男性”，=B “任选一人是色盲患者”．所求概率为()B A P ．由Bayes 公式，得 ()()()()()()()A B P A P A B P A P A B P A P B A P +=………….3分9544.00025.0432105.0432205.04322=⨯+⨯⨯=．………….5分四．（本题满分8分）在一小时内，甲、乙、丙三台机床需要维修的概率分别是9.0，8.0和85.0，而且这三台机床是否需要维修是相互独立的．求在一小时内⑴ 至少有一台机床不需要维修的概率；（4分） ⑵ 至多只有一台机床需要维修的概率．（4分）解：设{}甲机床需要维修=A ，{}乙机床需要维修=B ，{}丙机床需要维修=C ．则 ⑴ {}()()C B A P C B A P P ⋃⋃-=⋃⋃=1维修至少有一台机床不需要…….2分 ()()()388.085.08.09.011=⨯⨯-=-=C P B P A P ．………….2分⑵ {}()C B A C B A C B A C B A P P ⋃⋃⋃=修至多有一台机床需要维………….2分 ()()()()C B A P C B A P C B A P C B A P +++=()()()()()()()()()()()()C P B P A P C P B P A P C P B P A P C P B P A P +++=059.085.02.01.015.08.01.015.02.09.015.02.01.0=⨯⨯+⨯⨯+⨯⨯+⨯⨯=．…….2分五．（本题满分8分）试确定常数a ，b ，c ，d 的值，使得函数()⎪⎩⎪⎨⎧>≤≤++<=e x d e x d cx x bx x ax F 1ln 1为一连续型随机变量的分布函数．解：因为连续型随机变量的分布函数()x F 是连续函数，因此函数()x F 在分段点1=x 及e x =处连续，所以有()()()10101F F F =+=-，即有d c a +=．………….2分 ()()()e F e F e F =+=-00，即有d d ce be =++．………….2分又分布函数()x F 必须满足：()0lim =-∞→x F x ，()1lim =+∞→x F x ．因而有()0lim ==-∞→x F a x ，()1lim ==+∞→x F d x ．………….2分由此得方程组 ⎩⎨⎧=++=+1101ce be c ，解此方程组，得1,1,1,0=-===d c b a ．………….2分六．（本题满分8分）某地区成年男子的体重X （以kg 计）服从正态分布()2,σμN ．若已知()5.070=≤X P ，()25.060=≤X P ，⑴ 求μ与σ的值；⑵ 如果在该地区随机抽取5名成年男子，求至少有两个人的体重超过kg 65的概率．解：⑴ 由已知()5.0707070=⎪⎭⎫⎝⎛-Φ=⎪⎭⎫ ⎝⎛-≤-=≤σμσμσμX P X P ，()25.0606060=⎪⎭⎫⎝⎛-Φ=⎪⎭⎫ ⎝⎛-≤-=≤σμσμσμX P X P ………….2分得⎪⎪⎩⎪⎪⎨⎧=-=⎪⎭⎫ ⎝⎛-Φ-=⎪⎭⎫ ⎝⎛-Φ75.025.016015.070σμσμ ．即⎪⎪⎩⎪⎪⎨⎧=⎪⎭⎫ ⎝⎛--Φ=⎪⎭⎫⎝⎛-Φ75.0605.070σμσμ ，查正态分布表，得⎪⎩⎪⎨⎧=--=-675.060070σμσμ ，解方程组，得70=μ，81.14=σ．………….2分⑵ 设=A “从该地区任意选取一名成年男子，其体重超过kg 65”．则()()⎪⎭⎫⎝⎛-≤--=⎪⎭⎫ ⎝⎛-≤--=≤-=>3376.081.1470181.14706581.1470165165X P X P X P X P ()()6631.03376.03376.01=Φ=-Φ-=．………….2分设X ：该地区随机抽取的5名成年男子中体重超过kg 65的人数．则 ()6631.0,5~B X ．设=B “5人中至少有两人的体重超过kg 65．则 ()()()()()101112===-=≤-=≥=X P X P X P X P B P9530.03369.06631.03369.06631.0141155005=⨯⨯-⨯⨯-C C ．（已知()75.0675.0=Φ，()6631.034.0=Φ）………….2分七．（本题满分8分）设二维随机变量()Y X ,的联合密度函数为()()⎪⎩⎪⎨⎧-<<+=其它01045,22x y y x y x f求：随机变量Y 的边缘密度函数()y f Y ．解：当10<<y 时， ()()()()⎰⎰⎰----+∞∞-+=+==yyyY dx y xdx y x dx y x f y f 1021122545,………….3分()()()6211511312531252123103y y y y y xy x yx +-=⎪⎭⎫ ⎝⎛-+-⋅=⎪⎭⎫⎝⎛+⨯=-=．…….3分所以，随机变量Y 的边缘密度函数为()()⎪⎩⎪⎨⎧<<+-=其它01062115y y y y f Y ．………….2分八．（本题满分10分）设n X X X ,,,21Λ是n 个独立同分布的随机变量，1X 服从参数为λ的指数分布．令{}n X X X T ,,,m in 21Λ=，求随机变量T 的密度函数．解：对于任意的实数x ，随机变量T 的分布函数为 ()(){}()x X X X P x T P x F n T ≤=≤=,,,m in 21Λ{}()x X X X P n >-=,,,m in 121Λ()x X x X x X P n >>>-=,,,121Λ …………………….2分()()()x X P x X P x X P n >>>-=Λ211()()()()()()()()nX n x F x X P x X P x X P --=≤-≤-≤--=11111121Λ．………….3分所以，随机变量T 的密度函数为()()()()()x f x F n x F x f X n X T T 11--='=． ………….2分如果1X 服从参数为λ的指数分布，则1X 的密度函数为()⎩⎨⎧≤>=-0x x e x f xX λλ ．分布函数为()()⎩⎨⎧≤>-==-∞-⎰0001x x e dt t f x F xxX X λ ．………….1分因此此时{}n X X X T ,,,m in 21Λ=的密度函数为()()()()()x n x n xX n X T e n e e n x f x F n x f λλλλλ-----=⋅⋅=-=111，()0>x ．………….2分九．（本题满分8分）设随机向量()321,,X X X 间的相关系数分别为312312,,ρρρ，且，()()()0321===X E X E X E ，()()()02321>===σX D X D X D ．令：211X X Y +=，322X X Y +=,133X X Y +=．证明：321,,Y Y Y 两两不相关的充要条件为1312312-=++ρρρ．证明：充分性：如果1312312-=++ρρρ，则有01312312=+++ρρρ．而 ()()322121,cov ,cov X X X X Y Y ++= ()()()()32223121,cov ,cov ,cov ,cov X X X X X X X X +++=()()()()()()()3223231132112var X D X D X X D X D X D X D ⋅++⋅+⋅=ρρρ ()0121323122232213212=+++=+++=σρρρσρσσρσρ………….3分这说明随机变量1Y 与2Y 不相关．同理可得 ()0,cov 32=Y Y ，()0,cov 13=Y Y ，这就证明了随机变量321,,Y Y Y 两两不相关． ………….1分必要性：如果随机变量321,,Y Y Y 两两不相关，则有()0,cov 21=Y Y ，()0,cov 32=Y Y ，()0,cov 13=Y Y而由上面的计算，得()()01,cov 213231221=+++=σρρρY Y ， ………….3分由于02>σ，所以1132312+++ρρρ，即1132312-=++ρρρ． ………….1分十．（本题满分8分）设总体X 的密度函数为()⎩⎨⎧<<-=其它若011x xx f()5021,,,X X X Λ是从X 中抽取的一个样本，X 与2S 分别表示样本均值与样本方差．求()X E ，()X D ，()2S E ．解：因为()()011=⋅==⎰⎰-+∞∞-dx x x dx x xf X E ，()()2121311222==⋅==⎰⎰⎰-+∞∞-dx x dx x xdx x f x XE ，所以，()()()()2122=-=X E X E X D ．所以，()()0==X E X E ，………….2分()()10015021===n X D X D ，………….3分 ()()212==X D S E ．………….3分十一．（本题满分8分）设总体()4,0~N X ，()921,,,X X X Λ是取自该总体中的一个样本．求系数a 、b 、c ，使得统计量()()()298762543221X X X X c X X X b X X a T ++++++++=服从2χ分布，并求出自由度．解：因为()921,,,X X X Λ是取自总体()4,0N 中的简单随机样本，所以()4,0~N X i ，()9,,2,1Λ=i而且921,,,X X X Λ相互独立．所以()8,0~21N X X +，()12,0~543N X X X ++，()16,0~9876N X X X X +++．…….2分所以，()1,0~821N X X +，()1,0~12543N X X X ++，()1,0~169876N X X X X +++．…….2分因此，()()()()3~161282298762543221χX X X X X X X X X ++++++++．…….2分因此，当161,121,81===c b a 时，统计量()()()()3~161282298762543221χX X X X X X X X X T ++++++++=，自由度为3．………….2分十二．（本题满分8分）一家有500间客房的旅馆的每间客房装有一台kW 2（千瓦）的空调机，该旅馆的开房率为%80．求需要多少电力，才能有%99的可能性保证有足够的电力使用空调机．解：设X ：该旅馆开房数目，则()8.0,500~B X ．………….2分a ：向该旅馆供应的电力．则若电力足够使用空调机，当且仅当a X ≤2．因此()⎪⎪⎪⎪⎭⎫⎝⎛⨯⨯⨯-Φ≈⎪⎪⎪⎪⎭⎫⎝⎛⨯⨯⨯-≤⨯⨯⨯-=⎪⎭⎫ ⎝⎛≤=≤2.08.05008.050022.08.05008.050022.08.05008.050022a a X P a X P a X P ．由题设，99.02.08.05008.05002≥⎪⎪⎪⎪⎭⎫⎝⎛⨯⨯⨯-Φa ，………….3分查表，得33.22.08.05008.05002≥⨯⨯⨯-a，………….1分所以有 ()68.8412.08.050033.28.05002=⨯⨯⨯+⨯⨯≥a ．即至少向该旅馆供电842千瓦，才能保证该旅馆的空调机正常使用．………….2分十三．（本题满分8分）设总体X 的密度函数为()()⎩⎨⎧≤>=+-cx cx x c x f 01θθθ．其中0>c 是已知常数，而1>θ是未知参数．()n X X X ,,,21Λ是从该总体中抽取的一个样本，试求参数θ的最大似然估计量．解：似然函数为()()()()()121111+-=+-====∏∏θθθθθθθn n n ni i n i i x x x c x c x f L Λ………….2分所以，()()∑=+-+=ni i x c n n L 1ln 1ln ln ln θθθθ．所以，()∑=-+=ni i x c n nL d d 1ln ln ln θθθ．………….2分令：()0ln =θθL d d，即0ln ln 1=-+∑=ni i x c n n θ，………….2分得到似然函数的唯一驻点cn x nni iln ln 1-=∑=θ．所以参数θ的最大似然估计量为cn Xnni iln ln ˆ1-=∑=θ．………….2分。

北京交通大学901软件工程2010-2015年(12-13年含答案)考研专业历年真题汇编

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20. （本小题共 13 分）若数列 An : a1 , a2 ,..., an (n 2) 满足 ak 1 ak 1(k 1, 2,..., n 1) ，数列 An 为 E 数列，记 S ( An )

2011年普通高等学校招生全国统一考试数学文试题(北京卷,含答案)

2011年普通高等学校招生全国统一考试数学文试题（北京卷，含答案）本试卷共5页，150分。

考试时长120分钟。

考生务必将答案答在答题卡上，在试卷上作答无效。

考试结束后，将本试卷和答题卡一并交回。

第一部分（选择题共40分）一、选择题共8小题，每小题5分，共40分。

在每小题列出四个选项中，选出符合题目要求的一项。

（1）已知全集U=R ，集合{}21P x x =∣≤，那么U P =ð(A)(,1-∞-) (B)(1,+∞) (C)（-1，1) (D)()()11-∞,-,+∞ （2）复数212i i-=+ (A)i (B )i - (C)4355i -- (D)4355i -+ （3）如果1122log log 0x y <<，那么（A ）1y x << (B)1x y << (C)1x y << (D)1y x <<（4）若p 是真命题，q 是假命题，则（A ）p q ∧是真命题 (B)p q ∨是假命题题 (C)p ⌝是真命题 (D)q ⌝是真命（5）某四棱锥的三视图如图所示，该四棱锥的表面积是(A)32(B)16+(C)48(D)16+（6）执行如图所示的程序框图，若输入A 的值为2，则输出的P 值为(A)2(B)3(C)4(D)5（7）某车间分批生产某种产品，每批的生产准备费用为800元。

若每批生产x 件，则平均仓储时间为8x 天，且每件产品每天的仓储费用为1元。

为使平均到每件产品的生产准备费用与仓储费用之和最小，每批应生产产品（A ）60件 (B)80件（C ）100件（D ）120件（8）已知点()()0,2,2,0A B 。

若点C 在函数2y x =的图象上，则使得ABC 的面积为2的点C 的个数为（A ）4 (B)3 (C)2 (D)1第二部分（非选择题共110分）二、填空题共6小题，每小题5分，共30分。

口语考试题

北京交通大学2011级2级起点期末口语考试题Part oneRead the following paragraphHow do A students like these do it? Brains aren’t the only answer. The most gifted students do not necessarily perform best in exams. Knowing how to make the most of one’s abilities counts for much more. Hard work isn’t the whole story either. Some of these high-achieving students actually put in fewer hours than their lower-scoring classmates. The students at the top of the class get there by mastering a few basic techniques that others can easily learn. Here, according to education experts and students themselves, are the secrets of A students.This explains why it can be so difficult to get a western-style discussion going with Japanese students of English. Whenever I serve a volleyball, everyone just stands back and watches it fall. No one hits it back. Everyone waits until I call on someone to take a turn. And when that person speaks, he doesn't hit my ball back. He serves a new ball. Again, everyone just watches it fall. So I call on someone else. This person does not refer to what the previous speaker has said. He also serves a new ball. Everyone begins again from the same starting line, and all the balls run parallel. There is never any back and forth.Just after this success, however, tragedy struck. In August of 1973, Stevie was involved in a serious car accident. For nearly a week he lay in a coma, unable to speak or walk. “We don’t know when he’ll be out of danger,” the doctor said. Everyone waited and prayed. Suddenly, it didn’t matter that Stevie was a musical genius or that he had conquered blindness and poverty. All he had left was his faith and strong will. She was a small woman, old and wrinkled. When she started washing for us, she was already past seventy. Most Jewish women of her age were sickly and weak. All the old women in our street had bent backs and leaned on sticks when they walked. But this washwoman, small and thin as she was, possessed a strength that came from generations of peasant forebears. Mother would count out to her a bundle of laundry that had accumulated over several weeks. She would then lift the bundle, put it on her narrow shoulders, and carry it the long way home.I remember when they took their first holiday together. Ted wanted to do something energetic, because he didn’t usually get much exercise during the year. Mary’s job meant that she was on her feet most of the time. All she wanted to do was lie in the sun. Ted hated the idea of lying on a beach; Mary hated the idea of being too active. They compromised, and took their holiday in mid-summer, high in the Alps. Mary was able to lie in the sun by the hotel swimming pool, while Ted went off for long walks in the mountains with a group of hikers. In the evening they met at the hotel, both content with their day, happy to eat a leisurely meal together and dance a little afterwards.How can a country where typical parents are ashamed of their daughter studying mathematics instead of going dancing, or of their son reading Weber while his friends play baseball be expected to compete in the technology race with Japan or remain a leading political and cultural force in Europe? How long can America remain aworld-class power if we constantly put social skills and physical strength over academic achievement and intellectual ability?My father, Winston Churchill, began his love affair with painting in his 40s, amid disastrous circumstances. As First Lord of the Admiralty in 1915, he had been deeply involved in a campaign in the Dardanelles that could have shortened the course of a bloody world war. But when the mission failed, with great loss of life, Churchill paid the price, both publicly and privately: He was removed from the Admiralty and lost his position of political influence.This belief in hard work is the first of three main factors contributing to Asian students’ outstanding performance. It springs from Asians’ common heritage of Confucianism, the philosophy of the 5th-century-BC Chinese sage whose teachings have had a profound influence on Chinese soci ety. One of Confucius’s primary teachings is that through effort, people can perfect themselves.Joint ventures involving Western and Japanese companies often run into conflicts — a multitude of little things that escalate into big emotional battles in which all the parties keep exclaiming: “What’s wrong with them!? Can they understand that …?!” But because the conflicts are mainly due to cultural differences, neither side can understand —unless they have a “cultural translator”.Part twoDescribe the following pictureKey words:Children’s education, standard, pressure, abilityMy View on High Expectations from ParentsThere's such an interesting picture. The parents are holding two ends of a rope high above their head. They urge their kid “Honey, jump over it. Be quick!"This picture shows something quite meaningful and serious. In today's China, thereare too many parents who have high expectations on their kids and are willing to do everything possible as long as their kids have bright future.Facing fierce competition in the future, no wonder our parents are very much concerned about us. They lay great hope on us and want us to be strong enough to face challenges. We understand and respect their feelings.However, if the expectations are high beyond our ability, it is a kind of heavy pressure on us. We will feel hard to breathe and are easy to be sad when we feel that we are unable to meet their requirements. As a result, it will have bad effect on our study and life. It is harmful to our healthy development.I would like to say to all parents that we are sure to do our best and not to let you down through hard working. Please don't give us too much pressure and make us relaxed. We need your encouragement so that we will have more confidence to do things better.Key words:filial piety, housework, family, careDear students, our way of life is always sunny, blue skies, the brightest ray of sunshine which in the end? Some people say that good academic performance, was said to give to help others ... and I think that our way of life in the most brilliant sunshine should belong famous, thanks to everyone helping us grow. Y es, be grateful, is a feeling, be grateful, but also a sentiment.I will never forget the parents of my love, my love and care. What can I do for them? I often asked myself. Even if it is for them to hang Chuijian, wash dishes, give them aria song, to accompany them to shop, walk, my heart will feel comfort. Grateful, learn repay, I seemed to have grown up: I was careful to learn, let them worry about me; I picked rushing to wash dishes, so that they can get more rest for a while; I often hum songs, so full of family laughter ... I do everything I can give parents a good timeto leave the most memorable, so they are happy, so proud of them,I love my parents and the whole world children love their parents. Let us say to parents: "We love you!" Let us act together, Grateful, grateful. Winter is no longer cold, dark night is no longer a long, happy to accompany you on time around me. Students, we come from? Heard this, I am sure you will say, the parents took us to the world up to. Y es, ah, one day a decade ago, our parents smile with tears and happiness to greet our arrival. But when we come into the world the moment, parents have more of a heavy work - to take care of us. Although this is a heavy burden, but they relied on to support the parents I grew up. In order to give us a comfortable living environment, they are always so hard, so hard. Small, I always saw it as a matter of course, because I do not know, do not know their parents hard. Now, I grew up, I know that with a grateful heart to sympathize with parents, should take up, care, responsibility of caring for their parents.because my parents have, it gives me the opportunity to appreciate this colorful world of well-being of life, enjoy life joy and happiness, they gave me life, gave me excellent care. Children have a happy, happy most of the parents, children with depression, is also worried about most of the parents. Wife licking the calf, pare ntal love, as deep as the sea. Therefore, no matter the social status of parents, how the level of knowledge and other qualities, they are our life's biggest benefactor, is a person worthy of our love forever.However, the students, do you ask yourself too: I miss their parents and how many? Do you pay attention to the birthday of their parents? A folk saying: child birthday, Mother bitter day. When you celebrate for his birthday when you thought with the death-like pain, so you birth mother? Ever conceived in good faith to bless your life mother heard it? We Chinese are an ancient civilization, ancient emphasis on filial piety, Confucius, saying: "Parents of the year, not know that also. The one hand, happy and fear." That is, say, the health of parents, children should always be worried about the mind. However, according to reports, a middle school in Beijing this year have shown a sample survey: nearly 50% of the students did not know the birthday of their parents, let alone parents of the birthday wishes. Students, may soon wish nothing for themselves, but for parents, this sound better blessing than anything else, are memorable, are sufficient to enable them to tears!Filial piety, the man of this also, a people who only know how grateful the parents can be a complete person. Students, parents, let us be grateful to him now! With a grateful heart to treat their parents with a sincere heart to go with their parents, not to be taken for granted that parents do anything to help us, they took us to this beautiful world, is great enough, and our parenting adults, not asking for anything, we have to pay for silence, we stop blindly to demand their pay, Thanksgiving it, thank parents for their bit by bit.Key words:Honesty, morality, success, college studentsHonesty Is GoldHonesty means speaking the truth and being fair and upright in act. He who lies and cheats is dishonest .Those who gain fortunes not by hard labour but by other means is dishonest.Honesty is a good virtue. If you are honest all the time,you’ll be trusted and respected by others. A liar is always looked down upon and regarded as a black sheep by the people around. Once you lie, people will never believe you even if you speak the truth.However, in the tide of commodity economy today, it seems that more and more people believe in money at the sacrifice of honesty. To them, among such things as health, beauty, m6ney, intelligence, honesty, reputation and talent, honesty is the only thing that can be east away. They don’t understand or pretend not to understand that honesty is the biggest fortune humans own, and that it is the prerequisite for doing everything well. I think these people areto be pitied.In short, honesty is gold. Honest, your reputation will become great;dishonest, your name will be spoiled and your personality degraded. Therefore, we should never make such an excuse as "A little dishonesty is only a trifle thing. We should eradicate immediately the seed of dishonesty once it is sowed in our minds.Key words:overindulgence, education, personality, independenceFirst, give children space to move on his ownBaby of course like to live in her mother's arms, but he can not always live this way. There is such a mother, the child has the second grade, and send him to school but also effortlessly carrying him away, until tens of meters away from the school, for fear of seeing the teacher, the child was reluctant to put down ... ... This is the mother of the child care grew, his autonomy talk about it? As parents, should be based on the child's own characteristics and capabilities, expanding free space for children, such as encouraging a friend to find his own play, let him in this space as their own masters. Second, give the child time to make his own arrangementsMany parents think that children are small, do not know how to organize their activities. However, if the adult child completely arranged the timing of the child only to perform, then the child will never develop the autonomy does not come out.A father, his 3 year old child when the child on a daily free time he can, as long as no risk, children can make their own arrangements to do that he is willing to do: playing, watching TV, painting, puzzles, or doing nothing ... ... bored, he will eventually take the initiative to come to their parents, their parents give them some guidance in the recommendations. Over time, the child will be coming to understand the value of time, learned to arrange a time.Third, children conditions to allow him to exercise hisDestructive Enthusiasm among children with the objective of this violation of law practice, is certainly to fail, but to passively completely "go with the flow" attitude is not conducive to the child's growth. In accordance with objective laws, and actively create conditions to allow children to exercise, is what we should take the right approach.A 5-year-old mother to see interest in children's dishes, prepared for the children on a small stool, the child said: "I know that you especially love to work, like their dishes,but the faucet is too high, you is not enough that my mother prepared a small stool for you ... "the children shouted excitedly:" Thank you Mom! "immediately boarded a small stool pleased to learn the way to wash the adults.Fourth, children, and let him find the answers themselvesChildren ask questions, once the usual practice is for adults to tell him the answer. This seems simple and easy way, but such children grow up, they will not think the problem, hoping someone can provide ready-made answers. This directly hindered the children's intellectual autonomy.A successful experience of the parents is: kids asked me the word, although I know, but I do not tell him, but let him go to the dictionary. Later, and then did not even know the words, he did not ask me, but to their own dictionary.Key words:Help, raincoat, umbrellaWe are all living under the same sky, dancing on the same land. Everything we use, we share. Every problem we face, we solve it together. No matter how tall is the mountain, when we climb it together, it will be under our feet. No matter how deep is the ocean, when we swim together, he can reach the bottom.Y ou and me, all of us are connected, everything we do, think and feel has an effect upon the whole. For example: today, over 26,500 children died around the world. The silent killers are poverty, hunger, easily preventable diseases and illnesses, and other related causes. If we work together, we can prevent this from happening.One can't solve the problem, neither two can. For tens might be hard, for hundreds might be easy. Since we were born, there is a thread which connects all of us together. One goes down, everyone will be going down. If we give the one who goes down a little bit support, we will be all heading forward.If the world is held by piece, then it is help by piece of love, piece of hope, piece of help, piece of smile from everyone of us. We are all part of the piece puzzle, onemissing won't make the whole perfect.As the song sang,"if we hold on together, our dream will never die", yes, if we hold on together, we can help out others promblems, we can dream the dream we never dreamed.Let's go hand in hand to create a better world. Remember we all need your help.Key words:Parents’ expectation, cage, paly, studyKey words:Donation, love, Project Hope, childrenKey words:Independence, housework, self-training, abilitiesKey words:Water, waste, save, responsibilityWater Is the Source of LifeAs the source of life, water is indispensable (不可缺少的) to human beings. People consume much water in production and daily life. Agriculture and industry account for most of mankind＇s water usage. Without adequate water supplies, human future is threatened.Nowadays water shortages touch a large percentage of world population. According to one UN study in 1999, nearly one in five people lacks safemclean waterg WaterWelated diseases in some areas are killing people. Population explosion and water pollution make water crisis increasingly urgent. Water related problems are likely to worsen in the foreseeable future.The situation is really alarming (令人震惊的). Therefore, immediate and effectivemeasures must be taken to protect and exploit water resources. Remember, where there is water, there is life. If we could not make proper use of water resources on earth, the last drop of water would be our teardrop.Key words:Environment, protection, energy, pollutionMake Our Cities GreenerMany things have been done and great progress has been made in the greenization of our cities in recent years. Many trees have been planted and much care has been taken of the planted trees. But, in spite of all this, greenization in the cities in China is far from satisfactory compared with cities in other countries in the world. For example, the green space for every people in Shanghai is only 2 square meters while that of the people in London exceeds 20 square meters.We can benefit much from the greenization of our cities. First, trees absorb carbon dioxide which is harmful, but produce oxygen which is essential to human beings. Secondly, with trees all around, our cities will look more beautiful. Thirdly, trees can improve the climate of the cities, making it neither too hot in summer nor too cold in winter.Since we can get so much benefit from making our cities greener, we should spare no effort to do so. We should plant even more trees and take even better care of them while they are growing. Besides, we should not cut down trees any more. In a word, we should do everything to add to the greenization of our cities.。

2011级2级起点第一学期期末口语考试题

北京交通大学2011级2级起点第一学期期末口语考试题Part oneRead the following paragraph (21世纪)How do A students like these do it? Brains aren’t the only answer. The most gifted students do not necessarily perform best in exams. Knowing how to make the most of one’s abilities counts for much more. Hard work isn’t the whole story either. Some of these high-achieving students actually put in fewer hours than their lower-scoring classmates. The students at the top of the class get there by mastering a few basic techniques that others can easily learn. Here, according to education experts and students themselves, are the secrets of A students.This explains why it can be so difficult to get a western-style discussion going with Japanese students of English. Whenever I serve a volleyball, everyone just stands back and watches it fall. No one hits it back. Everyone waits until I call on someone to take a turn. And when that person speaks, he doesn't hit my ball back. He serves a new ball. Again, everyone just watches it fall. So I call on someone else. This person does not refer to what the previous speaker has said. He also serves a new ball. Everyone begins again from the same starting line, and all the balls run parallel. There is never any back and forth.Just after this success, however, tragedy struck. In August of 1973, Stevie was involved in a serious car accident. For nearly a week he lay in a coma, unable to speak or walk. “We don’t know when he’ll be out of danger,” the doctor said. Everyone waited and prayed. Suddenly, it didn’t matter that Stevie was a musical genius or that he had conquered blindness and poverty. All he had left was his faith and strong will. She was a small woman, old and wrinkled. When she started washing for us, she was already past seventy. Most Jewish women of her age were sickly and weak. All the old women in our street had bent backs and leaned on sticks when they walked. But this washwoman, small and thin as she was, possessed a strength that came from generations of peasant forebears. Mother would count out to her a bundle of laundry that had accumulated over several weeks. She would then lift the bundle, put it on her narrow shoulders, and carry it the long way home.I remember when they took their first holiday together. Ted wanted to do something energetic, because he didn’t usually get much exercise during the year. Mary’s jobmeant that she was on her feet most of the time. All she wanted to do was lie in the sun. Ted hated the idea of lying on a beach; Mary hated the idea of being too active. They compromised, and took their holiday in mid-summer, high in the Alps. Mary was able to lie in the sun by the hotel swimming pool, while Ted went off for long walks in the mountains with a group of hikers. In the evening they met at the hotel, both content with their day, happy to eat a leisurely meal together and dance a little afterwards.How can a country where typical parents are ashamed of their daughter studying mathematics instead of going dancing, or of their son reading Weber while his friends play baseball be expected to compete in the technology race with Japan or remain a leading political and cultural force in Europe? How long can America remain a world-class power if we constantly put social skills and physical strength over academic achievement and intellectual ability?My father, Winston Churchill, began his love affair with painting in his 40s, amid disastrous circumstances. As First Lord of the Admiralty in 1915, he had been deeply involved in a campaign in the Dardanelles that could have shortened the course of a bloody world war. But when the mission failed, with great loss of life, Churchill paid the price, both publicly and privately: He was removed from the Admiralty and lost his position of political influence.This belief in hard work is the first of three main factors contributing to Asian students’ outstanding performance. It springs from Asians’ common heritage of Confucianism, the philosophy of the 5th-century-BC Chinese sage whose teachings have had a pro found influence on Chinese society. One of Confucius’s primary teachings is that through effort, people can perfect themselves.Joint ventures involving Western and Japanese companies often run into conflicts — a multitude of little things that escalate into big emotional battles in which all the parties keep exclaiming: “What’s wrong with them!? Can they understand that …?!” But because the conflicts are mainly due to cultural differences, neither side can understand —unless they have a “cultural translator”.Read the following paragraph (交互)Book 1, Unit 1, College LifeReading 1Doing homework is important, but it’s not always enough. In order to really understand the ideas, it’s better to meet with a small group of two or three students after you study. Complete the assignments with them. Those students who did this understood the information better. They felt more connected to the class, and participated more in their classes.Reading 2Most colleges have attendance rules. At some colleges, you may be dropped for not attending classes, so ask your teachers about these policies. Missing classes is serious. You might miss an assignment, a quiz, a review for a test, or special information about an assignment. If you miss a class, you must find out what you missed. Ask some in the class to help you.Book 1, Unit 2, A World of Fast FoodReading 1The result is that Jamaicans are not eating healthy food. Fast food is cheaper than healthy food. Healthy food takes more time to shop for and prepare. Of course, Jamaicans would like to eat food that is both healthy and fast. To help with this problem, the yogurt shop TCBY has been creative. They have four stores in Jamaica that sell yogurt as well as low-fat ice cream.Reading 2Some teenagers came here on their first dates. Others came here after high school football games. Some got their first jobs here. Many people have good memories of this Old McDonald’s. They’re angry that the building is in danger and that their memories are as well.Book 1, Unit 3, Healthy HabitsReading 1It’s obvious that eating right and exercising regularly are very beneficial. Eating good foods makes you feel great and helps prevent many types of illnesses. Exercise makes your body release chemicals called endorphins, which reduce stress and cause you to feel better emotionally. Exercise tones muscles, helps the body to support the spine, and produces healthier bones, lungs, and heart. Now doesn’t that sound like a magical potion?Reading 2At the gym, Zoe may try out an aerobics class and another day enroll in a spinning class. Zoe keeps on top of all the new offerings at her gym because she doesn’t want to miss out on anything new. She feels great after a workout; it’s good for her health, she has made new friends and there is always something new to try.Book 1, Unit 4, Personal heroesReading 1I’m excited by the possibilities I see for medicine, for education and, of course, for technology. And I believe that through our natural inventiveness, creativity and willingness to solve tough problems, we’re going to make some amazing achievements in all these areas in my lifetime. Reading 2The word people used most often in describing Lee was intense; he was notoriously impatient except when talk turned to martial arts. He laughed at his own jokes, and he had an annoying habit of singing to himself on car trips. Many people also way him as a kind and compassionate man who did favors for people he didn’t even know.Book 1, Unit 5, Bilingual EducationReading 1Bilingual education has always been and continues to be a controversial subject. It is controversial for a variety of reasons. Some critics argue that bilingual education places an unfair burden on schools, and taxpayer’s money should not be spent teaching immigrants in their native language. They reason that all people in the United States should have to read, write, and speak English. Reading 2She took me to the front of the room and spoke to the other boys and girls. She pointed at me but I did not understand her. Then the other boys and girls laughed and pointed at me. I did not feel so good. Thereafter, I kept away from the groups as much as I could and worked alone. I worked hard. I listened to the strange sounds. I learned new names, new words.Book 1, Unit 6, Love and MarriageReading 1Many people, including many women, support payment of lobola. They explain that in Tswana culture, a marriage means joining a family, not just marring an individual. The idea of marring into a family is very African. A woman does not get married to just the man. Through marriage, the young woman’s family loses their daughter as she then becomes a part of her husband’s family. Lobola is a way to show thanks. It is a form of generosity in African culture.Reading 2Men who aren’t rich and people who love romance stories will be happy with recent research findings. It turns out the Beatles were right: money can’t buy you love. When college women chose among hypothetical men to date or marry, the attractiveness of bucks ranked behind honesty, good looks, and having time for family life.Book 1, Unit 7, Money MattersReading 1The Internet is another important part of the youth market. The percentage of kids under eighteen who are online is large and growing. It is estimated that kids aging five to eighteen will spend over $ 3 billion via the Internet. The amount of time that teens spend online is, on average, over seven hours a week.Reading 2We’re paying for more and more things electronically. Debit cards can be used to pay for our gas, buy our movie tickets, and even purchase our groceries. In some places you can even use a card to pay for a taxi. You need a credit card to reserve a hotel room or rent a car. You can pay your bills form your home computer or at your banking machine without ever using cash or writing a check.Book 2, Unit 1, Love and CareReading 1Mother Teresa’s mission was specifically to “the poorest of the poor” because she believed that the worst poverty of all was to reject a human soul in need. Those who visited the homes she established in the worst slums of Calcutta emerged overwhelmed by the love and cheerfulness that permeated the wards, bring hope to the abandoned and dignity to the dying.Reading 2She was scheduled to start classes within days when she went on her last vacation with childhood friends. Early on August 18, their car skidded off a curving road in the Smokies, and Patti, not wearing a seat belt, was thrown from the vehicle.Book 2, Unit 2, Arts and MusicReading 1Furthermore, because of my heavy Asian art background, it has been an immense obstacle for me to transform most of my concepts of Eastern art into an American graphic point of view in such a short period of time. Therefore, I become frustrated when my instructor tells me not only to participate more in the class but also to think more conceptually.Reading 2All that can really be said is that Mozart does have a positive short-term effect on spatial perception. Whether listening to Mozart makes us smarter is uncertain. That music can affect our moods, exciting us to perform harder or calming use to concentrate better, this would seem to be a matter of the individual student.Part twoDescribe the following pictureKey words:Children’s education, standard, pressure, abilityKey words:filial piety, housework, family, careKey words:Honesty, morality, success, college studentsKey words:Spoil the child with overindulgence, education, personality, independenceKey words:Care for others, help, raincoat, umbrellaKey words:Parents’ expectation, cage, play, studyKey words:Donation, love, Project Hope, childrenKey words:Independence, housework, self-training, abilitiesKey words:Water, waste, save, responsibilityKey words:Environment, protection, energy, pollution。

(NEW)北京交通大学语言与传播学院211翻译硕士英语[专业硕士]历年考研真题及详解

目　录2011年北京交通大学语言与传播学院211翻译硕士英语考研真题及详解2012年北京交通大学语言与传播学院211翻译硕士英语考研真题及详解2013年北京交通大学语言与传播学院211翻译硕士英语考研真题及详解2011年北京交通大学语言与传播学院211翻译硕士英语考研真题及详解Part I. Vocabulary and, Grammar (30 points)Section A Multiple Choice (20 points).Directions: Beneath each sentence there are four words or phrases marked A, B, C, and D. Choose the answer that best completes the sentence. Mark your answers on your answer sheet.1．The performance of this machine calls for much ________.A. technologyB. scienceC. techniqueD. technicality【答案】C【解析】句意：这台机器的性能操作需要很多技巧。

technique技巧；手法。

technology科技。

science科学；技术。

technicality学术性；专门性。

2．His strange behavior aroused the ________ of the police.A. suspicionB. doubtC. disbeliefD. incredibility【答案】A【解析】句意：他的奇怪的行为引起了警察的怀疑。

suspicion怀疑；疑心；指对某人做某事的目的、意图有怀疑。

doubt对事物的真、假有怀疑。

disbelief怀疑；不相信。

incredibility不能相信；不可信的事物。

北京交通大学-901-2018-真题

工程考研真题和答案》
一、单项选择题（本大题共 20 个小题，每小题 2 分，共 40 分） 1、软件工程的出现主要是由于（B） A.程序设计语言的快速发展 B.软件危机的出现 C.计算机硬件的飞速发展 D.其他工程学科的影响
2、软件工程除了重视软件开发研究外，另一项重要的组成内容是软件的（C）和过程改进。 A.成本核算 B.工具开发 C.项目管理 D.人员培训
2. 软件测试作为软件质量保证的一个重要手段，请回答如下问题： (13 分）
（1）软件测试包括哪些步骤？这些步骤的测试目的分别是什么？ (8 分）【答案】：（1）主要包括：单元测试、集成测试、系统测试、验收测试 4 个步骤。单元测试：保证每个模块作为一个单元能正确运行。集成测试：把通过单元测试的模块集成起来，主要发现与接口有关的问题。系统测试：是对整个系统的测试，将硬件、软件、操作人员看作一个
1、“数据耦合指的是两个模块之间有调用关系，传递简单的数据值，这种耦合程度较低，模块的独立性较高。（✔）
2、瀑布模型不要求项目所有的活动都严格按照顺序执行，该模型可以有反馈。（X） 3、面向对象方法是将数据和对数据的操作紧密结合起来的方法，这是与传统结构化方法的主要区别。（✔）
4、原型模型是在设计阶段快速构建一部分系统的软件开发模型。（X） 5、MVC 模式将模型、视图与控制器三个相对独立的部分分隔开来，这样可以实
A.封装 B.多态 C.继承 D.泛化 16、在采用 UML 进行软件建模过程中，类图是系统的一种静态视图，用（）可明确表示两类事物之间存在的整体/部分形式的关联关系。 A.依赖关系 B.泛化关系 C.聚集关系 D.实现关系
17、一个故障已经被发现且排除了。为了检查修改是否引起了其他故障，这时应该进行（）。 A.程序走查 B.回归测试 C.软件评审 D.验收测试

2011年高考数学试题(理科)

2011年北京高考数学（理）试题及答案word版2011年普通高等学校招生全国统一考试数学（理）（北京卷）本试卷共5页，150分。

考试时间长120分钟。

考生务必将答案答在答题卡上，在试卷上作答无效。

考试结束后，将本试卷和答题卡一并交回。

第一部分（选择题共40分）一、选择题共8小题，每小题5分，共40分。

在每小题列出的四个选项中，选出符合题目要求的一项。

1．已知集合P=｛x︱x2≤1｝,M=｛a｝.若P∪M=P,则a的取值范围是A．(-∞, -1] B．[1, +∞）C．[-1，1] D．（-∞，-1] ∪[1，+∞）2．复数A．i B．-i C． D．3．在极坐标系中，圆ρ=-2sinθ的圆心的极坐标系是A． B．C． (1,0) D．(1， )4．执行如图所示的程序框图，输出的s值为A．-3B．-C．D．25．如图，AD，AE，BC分别与圆O切于点D，E，F，延长AF与圆O交于另一点G。

给出下列三个结论：①AD+AE=AB+BC+CA；②AF•AG=AD•AE③△AFB ～△ADG其中正确结论的序号是A．①② B．②③C．①③ D．①②③6．根据统计，一名工作组装第x件某产品所用的时间（单位：分钟）为（A，C 为常数）。

已知工人组装第4件产品用时30分钟，组装第A件产品用时15分钟，那么C和A的值分别是A．75，25 B．75，16 C．60，25 D．60，167．某四面体的三视图如图所示，该四面体四个面的面积中，最大的是A．8 B． C．10 D．8．设 , ， , .记为平行四边形ABCD内部（不含边界）的整点的个数，其中整点是指横、纵坐标都是整数的点，则函数的值域为A． B．C． D．第二部分（非选择题共110分）二、填空题共6小题，每小题5分，共30分。

9．在中。

若b=5，，tanA=2，则sinA=____________；a=_______________。

10．已知向量a=（，1），b=（0，-1），c=（k，）。

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