可汗学院数学题目-33Datacollectionandconclusions

合集下载

(完整版)数据挖掘_概念和技术[第三版]部分习题答案解析

1.4 数据仓库和数据库有何不同？有哪些相似之处？答：区别：数据仓库是面向主题的，集成的,不易更改且随时间变化的数据集合，用来支持管理人员的决策，数据库由一组内部相关的数据和一组管理和存取数据的软件程序组成,是面向操作型的数据库,是组成数据仓库的源数据。

它用表组织数据，采用ER数据模型.相似:它们都为数据挖掘提供了源数据，都是数据的组合.1。

3 定义下列数据挖掘功能:特征化、区分、关联和相关分析、预测聚类和演变分析。

使用你熟悉的现实生活的数据库，给出每种数据挖掘功能的例子。

答:特征化是一个目标类数据的一般特性或特性的汇总。

例如，学生的特征可被提出，形成所有大学的计算机科学专业一年级学生的轮廓,这些特征包括作为一种高的年级平均成绩（GPA：Grade point aversge)的信息,还有所修的课程的最大数量.区分是将目标类数据对象的一般特性与一个或多个对比类对象的一般特性进行比较。

例如，具有高GPA 的学生的一般特性可被用来与具有低GPA 的一般特性比较.最终的描述可能是学生的一个一般可比较的轮廓，就像具有高GPA 的学生的75％是四年级计算机科学专业的学生,而具有低GPA 的学生的65％不是。

关联是指发现关联规则，这些规则表示一起频繁发生在给定数据集的特征值的条件.例如,一个数据挖掘系统可能发现的关联规则为：major(X，“computing science”) ⇒ owns(X, “personal computer”)［support=12％， confidence=98%] 其中，X 是一个表示学生的变量。

这个规则指出正在学习的学生，12％（支持度)主修计算机科学并且拥有一台个人计算机。

这个组一个学生拥有一台个人电脑的概率是98％(置信度，或确定度）.分类与预测不同，因为前者的作用是构造一系列能描述和区分数据类型或概念的模型（或功能），而后者是建立一个模型去预测缺失的或无效的、并且通常是数字的数据值。

可汗学院新SAT语法真题下载

可汗学院新SAT语法真题下载到目前为止，新版SAT可汗学院官方不断放出更多真题，已经放出了68篇阅读，且之前已经和大家分享过可汗学院新SAT阅读真题，想要下载的同学，请点击：新SAT阅读真题下载（共68篇，且已全）目前可汗学院一共放出41篇新SAT数学真题！想要下载的同学，请点击：新SAT数学真题下载（共41篇）分享了可汗学院的数学和阅读真题后，还有我们的可汗学院SAT语法真题。

截止到6月前，可汗学院一共放出了48套新SAT语法真题，想要吗下载请点击：新SAT语法真题下载（共48篇）（网址：）可汗学院新SAT语法真题（部分）Questions 1-5 are based on the following passage. 1Searching for GuinevereStories of kings and queens have captivated readers for centuries, and arguably, the tales of King Arthur and Guinevere are among the most enchanting. Arthur ruled the kingdom of Camelot, and Guinevere was his queen. But were they real people or fictional characters The debate has continued for centuries. Though many scholars have found evidence that the legendary Arthur was, at the very least, based on a real person who lived in Britain roughly between 450 and （1） 500 CE. They continue to search for the historical identity of Guinevere. Guinevere first appeared as King Arthur’s queen in one of the most widelystudied works of Arthurian literature, （2）The History of the Kings of Britain. This book was written by Geoffrey of Monmouth around 1135 CE. Geoffrey’s historical treatment of the legend is often（3）sited as evidence that the queen of Camelot existed, as the book chronicles the lives of a number of historical rulers.*God help those who help themselves. We help those who trust us. Contact Wechat:satxbs123, help is waiting.1A) NO CHANGEB) 500 CE. ContinuingC) 500 CE, continuingD) 500 CE, they continue2Which choice most effectively combines the sentences at the underlined portionA) The History of the Kings of Britain, and this bookB) The History of the Kings of Britain, whichC) a book called The History of the Kings of Britain,as thisD) a book called The History of the Kings of Britain,and this3A) NO CHANGEB) insightedC) citedD) incitedGuinevere is identified by Geoffrey as a noblewoman of Roman descent who met King Arthur in the court of Duke Cador of Cornwall, where she lived as a ward. (4)In Malory’s portrayal, Guinevere had no real power as a monarch but served as a kind of spiritual leader, providing guidance and moral support to the knights in their roles as defenders of the kingdom. Le Morte d’Arthur was also one of the first works to reference Guinevere’s romance with the knight, Sir many Arth urian scholars know, the distinction between history and literature was blurred in the Middle Ages. Consequently, the true identity of Guinevere may never be known with certainty. Yet regardless of whether Guinevere was real or fictional, her story (5) had endured centuries—and through each retelling, she continues to live on in the imaginations of people around the world.4At this point, the author wants to add a sentence which effectively sets up the portrayal of Guinevere discussed in the rest of the paragraph. Which choice best accomplishes this goalA) Three centuries later, however, Thomas Malory painted a very different portrait of Guinevere in Le Morte d’Arthur.B) Sir Thomas Malory was an English knight and Member of Parliament who also wrote extensively about the history of the British monarchy.C) Many historians believe that the portrayal of Arthur and Guinevere in Sir Thomas Malory’s Le Morte d’Arthur was actually a political commentary on the War of the Roses (1455-1487CE).D) In Le Morte d’Art hur, Sir Thomas Malory describes an idyllic England under King Arthur and Guinevere, which eventually collapses into chaos and political unrest. E. I would be guessing.5A) NO CHANGEB) was enduringC) would have enduredD) has enduredQuestions 1-5 are based on the following passage. 1Cometary Missions: Trajectory for SuccessScientists have been launching cometary missions since 1978. The first one, a joint mission by the European Space Agency, and the National Aeronautics and Space Administration (NAS A), was a “flyby” in which the spacecraft collected data while passing around Comet Giacobini-Zinner. (1)However, the landing of the Rosetta space probe on comet 67P/Churyumov-Gerasemenko in 2014 was different: it marked the first time that a probe landed on a( 2 )comet and giving scientists an unprecedented opportunity to study the surface of a comet. In order to continue this valuable research, additional missions are needed; thus, it is critical that more funding be allocated for this 2014 Rosetta mission provided a rare opportunity for scientists to test a number of hypotheses regarding the composition of (3) comets; the distribution of organic compounds in our solar system and the origins of life on Earth. Unlike other cometary missions, the Rosetta spacecraft contained a probe, Philae, that was able to land on the surface of a comet. *Rack your brain and you don't know Wechat: satxbs123, she can help you!1At this point, the writer wants to add accurate information from the graph. Which choice best accomplishes this goalA) From 1978 to 2014, the number of successful missions increased from 28 percent to 72 percent.B) Before 2014, the majority of attempted cometary missions were considered unsuccessful.C) Between then and 2014, 72 percent of the cometary missions were successful.D) Of the missions attempted since then, 44 percent have been successful.2A) NO CHANGEB) comet, but it gaveC) comet, yet givesD) comet, giving3A) NO CHANGEB) comets, the distribution of organic compounds in our solar system,C) comets, the distribution of organic compounds in our solar system;D) comets; the distribution of organic compounds in our solar system,。

数学之旅测试地的题目

1爱因斯坦创立广义相对论时用到了下列什么重要的数学工具？黎曼几何2下面这个方程有没有整数解？方程有没有整数解？有3下列哪个是孪生素数对？(17,19)4圆与椭圆在下列哪个数学分支中可看作一样？拓扑5具有同样周长的下列图形哪个面积更大？圆6以下汉字哪一个可以一笔不重复地写出？日7偶数与正整数哪个多？一样多8数列极限趋于0的直观定义的弱点是下面哪一点？缺乏可操作性9课程中费曼的故事告诉我们懂得一件事情最重要的是下面列出的哪一条？找到感觉10超弦理论中蜷缩的空间可以用下面那个空间来描述？Calabi-Yau空间11下面哪一位人物用穷竭法证明了圆的面积与其直径平方成正比？欧多克索斯12以下什么成果是阿基米德首先得到的?抛物线弓形的面积13阿基米德求几何级数的和用的是什么方法？几何的方法14欧多克索斯、阿基米德和刘徽等人对微积分的贡献主要体现在什么方面？定积分15《一种发展连续不可分量的新几何学的方法》是下列哪位数学家的著作？卡瓦列里16现在我们一直在用的“函数（function）”这个词是谁引进的？莱布尼兹17本课程提到的最美的风景点是指？牛顿-莱布尼兹公式18一直沿用至今的ε-δ语言是哪位数学家引入的？魏尔斯特拉斯19康托尔所创立的什么理论是实数以至整个微积分理论体系的基础？集合论20下面关于黎曼可积和勒贝格可积的论述那一项是正确的？黎曼可积函数类是不完备的，勒贝格可积函数类是完备的21试用阿基米德的方法求下面几何级数的和。

22计算加百列号的表面积与体积，并解释为何在这个号角里面灌满油漆，油漆的体积是有限的，但它却能够涂满无限的表面积？23举例说明黎曼积分中积分号和极限号有时不可交换，并给出可交换时需要的条件。

24下列四个定义中，哪个不能作为Rn中的度量(距离)？25度量的三个基本属性中不包括下列哪一个？连续性（三角不等式，正定型，对称性）26下列关于度量和范数的说法中正确的是？由范数可以定义距离，但由距离不可以定义范数27下列说法中不正确的是？对，若为的范数，则下列说法中不正确的是？若为实数，则有28以下现象可以用什么原理来解释？在三维空间中，波的传播有清晰的前后阵面，但是在二维空间中却没有？惠更斯原理29下列选项中正确的是？以下向量组中哪个不能构成的基向量？（0，1，1），（2，1，1），（1，0，0）30下列哪个选项是正确的？若向量a=（1，0，5，2），b=（3，-2，3，-4），c=（-1，1，t，3）线性相关，那么t的值为？1 31下列选项正确的是？向量和的夹角为？32下列说法哪一个是正确的？向量组线性无关的充分必要条件是？齐次线性方程组只有零解33下列哪个属性不是内积所具有的？三角不等式（对称性，对第一个变元的线性性，正定性）34给定一个集合，试验证下面两个集族是否构成集合M上的拓扑？1).2).35随着网络的迅速发展，人们越来越多的使用e-mail联系和交流。

可汗学院数学题目-41Complexnumbers[1]

Complex numbers1.(−8+4i)(1−i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−12+4iB.−12+12iC.−4+12iD.−4+4iCorrect answer: C Difficulty level: 22.(4+i)2Which of the following is equivalent to **plex number shown above?Note: i=√−1A.15+8iB.15−8iC.17+8iD.17−8iCorrect answer: A Difficulty level: 23.(8−2i)(4−2i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.28−24iB.28+8iC.36−24iD.36+8iCorrect answer: A Difficulty level: 24.(5+i)(7−3i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.32+8iB.32−8iC.38+8iD.38−8iCorrect answer: D Difficulty level: 25.i4+4i2+4Which of the following is equivalent to **plex number shown above?Note: i=√−1A.1B.−1C.i+4D.i−4Correct answer: A Difficulty level: 26.(−3−i)(4−2i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−14−2iB.−14+2iC.−10−2iD.−10+2iCorrect answer: B Difficulty level: 27.(6+2i)2Which of the following is equivalent to **plex number shown above?Note: i=√−1A.40+4i2B.40+24iC.32+24iD.32+4i2Correct answer: C Difficulty level: 28.(1+i)(1−i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.2-2iB.2iC.0D.2Correct answer: D Difficulty level: 29.i(7−3i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.4iB.10iC.7i−3D.7i+3Correct answer: D Difficulty level: 210.i2−16i+4Which of the following is equivalent to **plex number shown above?Note: i=√−1A.i−4B.i+4C.−i−4D.−i+4Correct answer: A Difficulty level: 211.(3+i)(2−4i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.2−10iB.2−14iC.10−10iD.10−14iCorrect answer: C Difficulty level: 212.i101Which of the following is equivalent to **plex number shown above?Note: i=√−1A.1B.−1C.iD.−iCorrect answer: C Difficulty level: 213.(5−i)2Which of the following is equivalent to **plex number shown above?Note: i=√−1A.24−10iB.24+10iC.26−10iD.26+10iCorrect answer: A Difficulty level: 214.−8(7i−3i2)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−80iB.−56i−24C.−56+24iD.−32iCorrect answer: B Difficulty level: 315.32+iWhich of the following is equivalent to **plex number shown above?Note: i=√−1A.2−iB.2+iC.6+3i5D.6+3i5Correct answer: D Difficulty level: 316.(3−i)3Which of the following is equivalent to **plex number shown above?Note: i=√−1A.8−26iB.18−26iC.27−26iD.30−26iCorrect answer: B Difficulty level: 317.(5−7i+i2)+(8i3+12)**plex expression above is equivalent to the expression a+bi for the integer constants a and b.What is the value of a?Note: i=√−1A.16B.17C.18D.19Correct answer: A Difficulty level: 318.(−3+2i)(1−i3)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−5−iB.−5+5iC.−1−iD.−1+5iCorrect answer: A Difficulty level: 319.i11+i13Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−2iB.2iC.0D.2Correct answer: C Difficulty level: 320.51+3iWhich of the following is equivalent to **plex number shown above?Note: i=√−1A.1+3i2B.1−3i2C.−5(1+3i)8D.−5(1−3i)8Correct answer: B Difficulty level: 321.(10−8i3)−(6+i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.4−7iB.4+7iC.4+9iD.4−9iCorrect answer: B Difficulty level: 322.11−iWhich of the following is equivalent to **plex number shown above?Note: i=√−1A.2−2iB.2+2iC.1−i2D.1+i2Correct answer: D Difficulty level: 323.21−iWhich of the following is equivalent to **plex number shown above?Note: i=√−1A.1−iB.1+iC.2−iD.2+iCorrect answer: B Difficulty level: 324.8ix=−5What is the value of x in the equation above?Note:i=√−1A.−8i5B.8i5C.−5i8D.5i8Correct answer: D Difficulty level: 325.(2−3i)3Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−46−9iB.−26−9iC.26−9iD.46−9iCorrect answer: A Difficulty level: 326.(12+i)(8−6i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−2+5iB.2+2iC.10+5iD.14+2iCorrect answer: C Difficulty level: 327.11−6i −11+6iWhich of the following is equivalent to **plex number shown above? Note: i=√−1A.1237iB.−1237iC.1237D.−1237Correct answer: A Difficulty level: 328.(23+12i)(12−13i)**plex expression above is equivalent to the expression a+bi for the rational constants a and b. What is the value of b?Note: i=√−1A.b=16B.b=−16C.b=496D. b=529Correct answer: D Difficulty level: 329.P(x)=2x2+3x−17If x=8−2i, what is the value of the polynomial P above?Note:i=√−1A.15−2iB.23−6iC.127−70iD.135−62iCorrect answer: C Difficulty level: 330.(6+i2)(2−2i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.6−8iB.8−8iC.10−8iD.12−8iCorrect answer: B Difficulty level: 331.3i +2i2Which of the following is equivalent to **plex number shown above?Note: i=√−1A.3i+2B.3i−2C.−3i+2D.−3i−2Correct answer: D Difficulty level: 332.21+iWhich of the following is equivalent to **plex number shown above?Note: i=√−1A.−1+iB.−1−iC.1+iD.1−iCorrect answer: D Difficulty level: 433.3i10+i11Which of the following is equivalent to **plex number shown above?Note: i=√−1A.3+iB.−3+iC.3−iD.−3−iCorrect answer: D Difficulty level: 434.P(n)=n2−5n−7What is the value of P(−3i)?Note:i=√−1A.−4+15iB.−7+12iC.−7+24iD.−16+15iCorrect answer: D Difficulty level: 435.√3t2+5t+√27=0Which of the following is a solution to the equation above?Note:i=√−1A.t=−4√11i6B.t=−3√33i6C.t=−5+√11i6D.t=−5√3+√33i6Correct answer: D Difficulty level: 436.29=3(x+7)2+41Which of the following is a solution to the equation above?Note:i=√−1A.x=−7+2iB.x=−42−12iC.x=−76+√4436iD.x=−7−√1233iCorrect answer: A Difficulty level: 437.(8−2i)2(8+2i)Which of the following is equivalent to **plex number shown above?Note: i=√−1A.60B.68C.480−120iD.544−136iCorrect answer: D Difficulty level: 438.√54i41√27i101Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−√2iB.−√2C.√2D.√2iCorrect answer: C Difficulty level: 439.22−i −22+iWhich of the following is equivalent to **plex number shown above? Note: i=√−1A.4i5B.−4i5C.2i3D.−2i3Correct answer: A Difficulty level: 440.1+2i1−2i ÷1−2i1+2iWhich of the following is equivalent to **plex number shown above? Note: i=√−1A.1B.−1C.−725+2425iD.−725−2425iCorrect answer: D Difficulty level: 441.i3+i2Which of the following is equivalent to **plex number shown above?Note: i=√−1A.−1B.−2C.−1+iD.−1−iCorrect answer: D Difficulty level: 442.m2+6m+10=0Which of the following are solutions to the equation above?I. -3+iII. -3-iIII. 3+iNote:i=√−1A.I onlyB.I and II onlyC.I and III onlyD.I, II, and IIICorrect answer: B Difficulty level: 443.704i1776Which of the following is equivalent to **plex number shown above?Note: i=√−1A.704B.−704C.704iD.−704iCorrect answer: A Difficulty level: 444.12+5i −4+3i3−iWhich of the following is equivalent to **plex number shown above?Note: i=√−1A.10+26i(2+5i)(3−i)B.10−26i(2+5i)(3−i)C.10+27i(2+5i)(3−i)D.10−27i(2+5i)(3−i)Correct answer: D Difficulty level: 445.5+7i6−3iWhich of the following is equivalent to **plex number shown above?Note: i=√−1A.9+57i45B.9+57i3C.51+57i45D.51+57i3Correct answer: A Difficulty level: 446.9+7i184−i**plex expression above is equivalent to the expression a+bi for the rational constants a and b.What is the value of b?Note: i=√−1A.b=215B.b=217C.b=−1D.b=−7Correct answer: B Difficulty level: 447.5−i+(11−i)z=40+18iWhat is the value of z in the equation above?Note:i=√−1A.z=−19+24iB.z=24+20iC.z=3+2iD.z=3.2+145iCorrect answer: C Difficulty level: 448.2i+4ℎ−14=2iℎWhat is the value of h in the equation above?Note:i=√−1A.ℎ=3+iB.ℎ=72C.ℎ=7−iD.ℎ=83Correct answer: A Difficulty level: 4。

数据挖掘_国防科技大学中国大学mooc课后章节答案期末考试题库2023年

数据挖掘_国防科技大学中国大学mooc课后章节答案期末考试题库2023年1.某超市研究销售纪录数据后发现，买啤酒的人很大概率也会购买尿布，这种属于数据挖掘的哪类问题？()答案:关联规则发现2.下列有关SVM说法不正确的是（）答案:SVM因为使用了核函数，因此它没有过拟合的风险3.影响聚类算法效果的主要原因有：（）答案:特征选取_聚类准则_模式相似性测度4.7、朴素贝叶斯分类器不存在数据平滑问题。

( )答案:错误5.决策树中包含一下哪些结点答案:内部结点（internal node）_叶结点（leaf node）_根结点（root node) 6.标称类型数据的可以利用的数学计算为：众数7.一般，k-NN最近邻方法在( )的情况下效果较好答案:样本较少但典型性好8.考虑两队之间的足球比赛：队0和队1。

假设65%的比赛队0胜出、P(Y=0)=0.65。

剩余的比赛队1胜出、P(Y=1)=0.35。

队0获胜的比赛中只有30%在队1的主场、P(X=1|Y=0)=0.3，而队1获胜的比赛中75%是主场获胜、P(X=1|Y=1)=0.75。

则队1在主场获胜的概率即P(Y=1|X=1)为：（）答案:0.579.一组数据的最小值为12,000，最大值为98,000，利用最小最大规范化将数据规范到[0,1]，则73,000规范化的值为：（）答案:0.71610.以下哪个分类方法可以较好地避免样本的不平衡问题：（）答案:KNN11.简单地将数据对象集划分成不重叠的子集，使得每个数据对象恰在一个子集中，下列哪些不属于这种聚类类型层次聚类_模糊聚类_非互斥聚类12.数据点密度分布不均会影响K-means聚类的效果。

答案:正确13.数据集成需要解决模式集成、实体识别、数据冲突检测等问题答案:正确14.决策树模型中应处理连续型属性数据的方法之一为：根据信息增益选择阈值进行离散化。

答案:正确15.数据库中某属性缺失值比较多时，数据清理可以采用忽略元组的方法。

Summa Scalable universal matrix multiplication algorithm

will consider the formation of the matrix products C C C C = = = = AB + C AB T + C AT B + C AT B T + C (1) (2) (3) (4)
These are the special cases implemented as part of the widely used sequential Basic Linear Algebra Subprograms 11]. We will assume that each matrix X is of dimension mX nX , X 2 fA; B; C g. Naturally, there are constraints on these dimensions for the multiplications to be well de ned: We will assume that the dimensions of C are m n, while the \other" dimension is k.
This work is partially supported by the NASA High Performance Computing and Communications Program's Earth and Space Sciences Project under NRA Grant NAG5-2497. Additional support came from the Intel Research Council. Jerrell Watts is being supported by an NSF Graduate Research Fellowship.

新北师大版高中数学高中数学选修2-3第三章《统计案例》测试卷(包含答案解析)(5)

一、选择题1．在一次对性别与是否说谎有关的调查中，得到如下数据，根据表中数据判断如下结论中正确的是()A．在此次调查中有95%的把握认为是否说谎与性别有关B．在此次调查中有99%的把握认为是否说谎与性别有关C．在此次调查中有99.5%的把握认为是否说谎与性别有关D．在此次调查中没有充分证据显示说谎与性别有关2．某高校为调查学生喜欢“应用统计”课程是否与性别有关，随机抽取了选修课程的55名学生，得到数据如下表：临界值参考：（参考公式：22()()()()()n ad bcKa b c d a c b d-=++++，其中n a b c d=+++）参照附表，得到的正确结论是（）A．在犯错误的概率不超过0.1%的前提下，认为“喜欢“应用统计”课程与性别有关”B ．在犯错误的概率不超过0.1%的前提下，认为“喜欢“应用统计”课程与性别无关”C ．有99.99%以上的把握认为“喜欢“应用统计”课程与性别有关”D ．有99.99%以上的把握认为“喜欢“应用统计”课程与性别无关”3．为了调查某校高二学生的身高是否与性别有关，随机调查该校64名高二学生，得到2×2列联表如表：附：K 2()()()()2()n ad bc a b c d a c b d -=++++由此得出的正确结论是（）A ．在犯错误的概率不超过0.01的前提下，认为“身高与性别无关”B ．在犯错误的概率不超过0.01的前提下，认为“身高与性别有关”C ．有99.9%的把握认为“身高与性别无关”D ．有99.9%的把握认为“身高与性别有关” 4．已知两个统计案例如下：①为了探究患肺炎与吸烟的关系，调查了339名50岁以上的人，调查结果如下表：②为了解某地母亲与女儿身高的关系，随机测得10对母女的身高如下表：母亲身高(cm) 159 160160 163 159 154 159 158 159 157 女儿身高(cm) 158 159 160 161 161 155 162 157 162 156则对这些数据的处理所应用的统计方法是（） A ．①回归分析，②取平均值 B ．①独立性检验，②回归分析 C ．①回归分析，②独立性检验D ．①独立性检验，②取平均值5．某科研机构为了研究中年人秃发与患心脏病是否有关,随机调查了一些中年人的情况,具体数据如表,根据表中数据则可判定秃发与患心脏病有关,那么这种判定出错的可能性为( ) 患心脏病情况秃发情况患心脏病无心脏病秃发 20 300 不秃发5450A ．0.1B ．0.05C ．0.01D ．0.996．设导弹发射的事故率为0.01，若发射10次，其出事故的次数为ξ，则下列结论正确的是（） A ．0.1E ξ=B ．•01D ξ=C ．10()0.01?0.99k k P k ξ-==D ．1010()0.99?0.01kkkP k C ξ-==7．某研究型学习小组调查研究学生使用智能手机对学习的影响．部分统计数据如下表：附表：经计算2K 的观测值10k =，则下列选项正确的是（） A ．有99.5％的把握认为使用智能手机对学习有影响 B ．有99.5％的把握认为使用智能手机对学习无影响C ．有99.9％的把握认为使用智能手机对学习有影响D ．有99.9％的把握认为使用智能手机对学习无影响8．为了检验设备M 与设备N 的生产效率，研究人员作出统计，得到如下表所示的结果，则（）附：参考公式：22()()()()()n ad bc K a b c d a c b d -=++++，其中n a b c d =+++.A ．有90%的把握认为生产的产品质量与设备的选择具有相关性B ．没有90%的把握认为生产的产品质量与设备的选择具有相关性C ．可以在犯错误的概率不超过0.01的前提下认为生产的产品质量与设备的选择具有相关性D ．不能在犯错误的概率不超过0.1的前提下认为生产的产品质量与设备的选择具有相关性 9．为考察数学成绩与物理成绩的关系,在高二随机抽取了300名学生,得到下面的列联表:现判断数学成绩与物理成绩有关系,则犯错误的概率不超过 ( ) A ．0.005B ．0.01C ．0.02D ．0.0510．以下四个命题中：①某地市高三理科学生有15000名，在一次调研测试中，数学成绩ξ服从正态分布()2100,N σ，已知()801000.40P ξ<≤=，若按成绩分层抽样的方式抽取100分试卷进行分析，则应从120分以上（包括120分）的试卷中抽取15分；②已知命题:p x ∀∈R ，sin 1x ≤，则:p x ⌝∃∈R ，sin 1x >；③在[]4,3-上随机取一个数m ，能使函数()222f x x mx =++在R 上有零点的概率为37； ④在某次飞行航程中遭遇恶劣气候，用分层抽样的20名男乘客中有5名晕机，12名女乘客中有8名晕机，在检验这些乘客晕机是否与性别有关时，采用独立性检验，有97%以上的把握认为与性别有关．()2P k k ≥0．15 0．1 0．05 0．025 0k 2．0722．7063．8415．024其中真命题的序号为（） A ．①②③ B ．②③④C ．①②④D ．①③④11．有下列数据： x123y35.9912.01下列四个函数中，模拟效果最好的为（） A ．B ．C ．D ．12．某家具厂的原材料费支出x 与销售量y （单位：万元）之间有如下数据，根据表中提供的全部数据，用最小二乘法得出y 与x 的线性回归方程为ˆ8ˆy x b =+，则^b为（） x 2 4 5 6 8 y2535605575A ．5B ．15C ．10D ．20二、填空题13．回归方程ˆˆ 2.50.2x y=+在样本(4,1.2)处的残差为________. 14．如果根据性别与是否爱好运动的列联表得到K 2≈3．852＞3．841，则判断性别与是否爱好运动有关，那么这种判断犯错的可能性不超过________． 15．若两个分类变量X 与Y 的列联表为：y 1 y 2 x 1 10 15 x 24016则“X 与Y 之间有关系”这个结论出错的可能性为________． 16．给出下列命题：①线性相关系数越大，两个变量的线性相关越强；反之，线性相关性越弱； ②由变量和的数据得到其回归直线方程：，则一定经过；③从越苏传递的产品生产流水线上，质检员每10分钟从中抽取一件产品进行某项指标检测，这样的抽样是分层抽样；④在回归分析模型中，残差平方和越小，说明模型的拟合效果越好； ⑤在回归直线方程中，当解释变量每增加一个单位时，预报变量增加0．1个单位，其中真命题的序号是___________．17．某单位为了了解用电量y 度与气温x ℃之间的关系，随机统计了某4天的用电量与当天气温. 气温（℃）14 12 86用电量（度） 22 26 34 38由表中数据得线性方程x b a yˆˆ+=中2ˆ-=b ，据此预测当气温为5℃时，用电量的度数约为 .18．以下4个命题中，正确命题的序号为_________．①“两个分类变量的独立性检验”是指利用随机变量2K 来确定是否能以给定的把握认为“两个分类变量有关系”的统计方法； ②将参数方程cos sin x y θθ=⎧⎨=⎩（θ是参数，[]0,θπ∈）化为普通方程，即为221x y +=；③极坐标系中，22,3A π⎛⎫⎪⎝⎭与()3,0B 19 ④推理：“因为所有边长相等的凸多边形都是正多边形，而菱形是所有边长都相等的凸多边形，所以菱形是正多边形”，推理错误在于“大前提”错误.19．在吸烟与患肺病这两个分类变量的计算中，下列说法正确的是_____________． ①若K 2的观测值满足K 2≥6.635，我们有99%的把握认为吸烟与患肺病有关系，那么在100个吸烟的人中必有99人患有肺病；②从独立性检验可知有99%的把握认为吸烟与患病有关系时，我们说某人吸烟，那么他有99%的可能患有肺病；③从统计量中得知有95%的把握认为吸烟与患肺病有关系，是指有5%的可能性使得推断出现错误．20．某班主任对全班50名学生的积极性和对待班级工作的态度进行了调查，统计数据如下表所示：则至少有________的把握认为学生的学习积极性与对待班级工作的态度有关.（请用百分数表示）.注：独立性检验界值表三、解答题21．为了解某企业生产的某产品的年利润与年广告投入的关系，该企业对最近一些相关数据进行了调查统计，得出相关数据见下表：根据以上数据，研究人员分别借助甲、乙两种不同的回归模型，得到两个回归方程：方程甲，2(1)(1) 2.75yb x =-+＾＾；方程乙，(2)1.6yc x =-＾＾.（1）求b ＾(结果精确到0.01)与c ＾的值.（2）为了评价两种模型的拟合效果，完成以下任务.①完成下表(备注：i i ie y y =-＾＾，i e ＾称为相应于点(x i ，y i )的残差)；②分别计算模型甲与模型乙的残差平方和Q1及Q2，并通过比较Q1，Q2的大小，判断哪个模型拟合效果更好.22．为考察某种药物预防禽流感的效果，进行动物家禽试验，调查了100个样本，统计结果为：服用药的共有60个样本，服用药但患病的仍有20个样本，没有服用药且未患病的有20个样本．（1）根据所给样本数据画出22⨯列联表；（2）请问能有多大把握认为药物有效？附公式：()()()()()22=n ad bcKa b c d a c b d-++++．23．为提高全民身体素质，加强体育运动意识，某校体育部从全校随机抽取了男生､女生各100人进行问卷调查，以了解学生参加体育运动的积极性是否与性别有关，得到如下列联表(单位：人)：男生 70 30 100 女生 60 40 100 合计13070200（1）根据以上数据，判断能否在犯错误的概率不超过10%的情况下认为该校参加体育运动的积极性与性别有关；（2）用频率估计概率，现从该校所有女生中随机抽取3人.记被抽取的3人中“偶尔运动或不运动”的人数为X ，求X 的分布列､期望()E X 和方差()D X .附：22()()()()()n ad bc K a b c d a c b d -=++++，其中n a b c d =+++.()20P K k 0.150.10 0.05 0.025 0k 2.0722.7063.8415.02424．在中国，不仅是购物，而且从共享单车到医院挂号再到公共缴费，日常生活中几乎全部领域都支持手机支付，出门不带现金的人数正在迅速增加．某机构随机抽取了一组市民，并统计他们各自出门随身携带现金（单位：元）的情况，制作出如图所示的茎叶图．规定：随身携带的现金在100元以下（不含100元）的为“手机支付族”，其他为“非手机支付族”．（1）根据茎叶图的数据，完成答题卡上的22⨯列联表；男生女生合计手机支付族非手机支付族合计45（2）根据（1）中的列联表，判断是否有99%的把握认为“手机支付族”与“性别”有关．附：()20P K k ≥0.050 0.010 0.001 0k 3.8416.63510.82822()()()()()()n ad bc K n a b c d a b c d a c b d -==+++++++25．某地为响应国家“脱贫攻坚战”的号召，帮助贫困户脱贫，安排贫困人员参与工厂生产．现用A ，B 两条生产线生产某产品．为了检测该产品的某项质量指标值（记为Z ），现随机抽取这两种这两条生产线的产品各100件，由检测结果得到如下频率分布直方图．（Ⅰ）分别估计A ，B 两条生产线的产品质量指标值的平均数（同一组数据中的数据用该组区间的中点值作代表），从平均数结果看，哪条生产线的质量指标值更好？（Ⅱ）计算A 生产线的产品质量指标值的众数和中位数（中位数计算结果精确到小数点后两位）．（Ⅲ）该公司规定当92Z ≥时，产品为超优品．根据所检测的结果填写22⨯列联表，并判断是否有95%的把握认为“生产超优品是否与生产线有关”．附：()()()()()22n ad bc K a b c d a c b d -=++++，n a b c d =+++()20P K k ≥0.050 0.010 0.005 0.001 0k3.8416.6357.87910.82822⨯列联表26．某学生兴趣小组随机调查了某市100天中每天的空气质量等级和当天到某公园锻炼的人次，整理数据得到下表（单位：天）：（2）求一天中到该公园锻炼的平均人次的估计值（同一组中的数据用该组区间的中点值为代表）；（3）若某天的空气质量等级为1或2，则称这天“空气质量好”；若某天的空气质量等级为3或4，则称这天“空气质量不好”．根据所给数据，完成下面的2×2列联表，并根据列联表，判断是否有95%的把握认为一天中到该公园锻炼的人次与该市当天的空气质量有关？附：2()()()()()n ad bc K a b c d a c b d -=++++，【参考答案】***试卷处理标记，请不要删除一、选择题 1．D 解析：D 【解析】根据上表数据可求得20.027 1.323k ≈<，再结合课本上的概率附表可知在此次调查中没有充分证据显示说谎与性别有关，故选D2．A解析：A 【分析】计算212.010.828K ≈>，对比临界值表得到答案. 【详解】()222552020105()53912.010.828()()()()3025302545n ad bc K a b c d a c b d ⨯-⨯-===≈>++++⨯⨯⨯，故在犯错误的概率不超过0.1%的前提下，认为“喜欢“应用统计”课程与性别有关”. 故选：A. 【点睛】本题考查了独立性检验，意在考查学生的计算能力和应用能力.3．D解析：D 【分析】根据22⨯列联表，计算2k ，与临界值表比较即可得出结论. 【详解】K 的观测值：K 2264(862426)34303232⨯⨯-⨯=≈⨯⨯⨯20.330；由于20.330＞10.828，∴有99.9%的把握认为“身高与性别有关”，即在犯错误的概率不超过0.001的前提下，认为“身高与性别有关” 故选：D ．【点睛】本题主要考查了独立性检验的应用问题，K 2的计算，22⨯列联表，考查了运算能力，属于中档题．4．B解析：B 【分析】根据独立性检验和回归分析的概念，即可作出判定，得到答案．【详解】由题意，独立性检验通常是研究两个分类变量之间是否有关系，所以①采用独立性检验，回归分析通常是研究两个具有相关关系的变量的相关程度，②采用回归分析，综上可知①是独立性检验，②是回归分析，故选B ．【点睛】本题主要考查了独立性检验和回归分析的概念及其判定，其中解答中熟记独立性检验和回归分析的概念是解答的关键，着重考查了分析问题和解答问题的能力，属于基础题．5．C解析：C 【分析】首先列出22⨯联表，通过计算出2K 的值，然后作统计推断，得出正确的结论. 【详解】列出22⨯联表如下图所示：()277520450530015.96825750455320K ⨯⨯-⨯=≈⨯⨯⨯ 6.635>，故判断错误的概率不超过0.01，故选C .【点睛】本小题主要考查补全22⨯联表，考查2K 的计算以及独立性检验的概念，属于基础题. 独立性检验的步骤：(1)根据样本数据制成22⨯列联表；(2)根据公式22n ad bc K a b c d a c b d -=++++（）（）（）（）（），计算2K 的观测值；(3)比较2K 与临界值的大小关系作统计推断． 6．A解析：A 【解析】【分析】由题意知本题是在相同的条件下发生的试验，发射的事故率都为0.01，实验的结果只有发生和不发生两种结果，故本题符合独立重复试验，由独立重复试验的期望公式得到结果. 【详解】由题意知本题是在相同的条件下发生的试验，发射的事故率都为0.01，故本题符合独立重复试验，即ξ～(10,0.01)B . ∴100.010.1E ξ=⨯= 故选A ．【点睛】解决离散型随机变量分布列和期望问题时，主要依据概率的有关概念和运算，同时还要注意题目中离散型随机变量服从什么分布，若服从特殊的分布则运算要简单的多．7．A解析：A 【解析】【分析】由题意结合2K 的观测值k 由独立性检验的数学思想给出正确的结论即可. 【详解】由于2K 的观测值10k =7.879>,其对应的值0.0050.5%=，据此结合独立性检验的思想可知：有99.5％的把握认为使用智能手机对学习有影响. 本题选择A 选项. 【点睛】独立性检验得出的结论是带有概率性质的，只能说结论成立的概率有多大，而不能完全肯定一个结论，因此才出现了临界值表，在分析问题时一定要注意这点，不可对某个问题下确定性结论，否则就可能对统计计算的结果作出错误的解释．8．A解析：A 【解析】将表中的数据代入公式，计算得22100(487243) 3.0535050919K ⨯⨯-⨯=≈⨯⨯⨯，∵3.053 2.706>，∴有90%的把握认为生产的产品质量与设备的选择具有相关性，故选A ．9．D解析：D 【解析】因为K 2的观测值k=2300(371433585)12217872228⨯-⨯⨯⨯⨯≈4.514>3.841, 所以在犯错误的概率不超过0.05的前提下认为数学成绩与物理成绩有关系. 选D.10．B解析：B 【解析】对于①，在一次调研测试中，数学成绩ξ服从正态分布N （100，σ2），∴数学成绩ξ关于ξ=100对称，∵P （80＜ξ≤100）=0.40，∴P （ξ＞120）=P （ξ＜80）=0.5-0.40=0.1，则该班数学成绩在120分以上的人数为0.1×100=10，故①错误；对于②，已知命题p ：∀x ∈R ，sinx≤1，则￢p ：∃x ∈R ，sinx ＞1，故②正确；对于③，由)2−8≥0，解得m≤-2或m≥2，∴在[-4，3]上随机取一个数m ，能使函数()22f x x =+在R 上有零点的概率为37，故③正确；对于④，填写2×2列联表如下：则k 2的观测值k ＝()23215854 5.398 5.024********⨯⨯-⨯≈>⨯⨯⨯有97%以上的把握认为晕机与性别有关.故④对故选B11．A解析：A 【解析】当x ＝1，2，3时，分别代入求y 值，离y 最近的值模拟效果最好，可知A 模拟效果最好．故选A.考点：非线性回归方程的选择.12．C解析：C 【详解】由题意可得：2456855x ++++==，2535605575505y ++++==，回归方程过样本中心点，则：5285,1ˆˆ0bb =⨯+∴=. 本题选择C 选项.二、填空题13．【分析】根据残差的定义直接计算即可【详解】由题当x=4时故所以回归方程在样本处的残差为故答案为：【点睛】本题主要考查了残差的概念考查了运算能力属于容易题解析：9-【分析】根据残差的定义直接计算即可.【详解】由题当x =4时，4ˆ 2.50.210.2y=+=⨯，故1.210.29-=-所以回归方程ˆˆ 2.50.2x y=+在样本(4,1.2)处的残差为9-. 故答案为：9- 【点睛】本题主要考查了残差的概念，考查了运算能力，属于容易题.14．【解析】∵P(K2≥3841)≈005∴判断性别与是否爱好运动有关出错的可能性不超过5点睛：根据卡方公式计算再与参考数据比较就可确定可能性解析：5%【解析】∵P (K 2≥3．841)≈0．05．∴判断性别与是否爱好运动有关，出错的可能性不超过5%．点睛：根据卡方公式计算2K ，再与参考数据比较，就可确定可能性.15．1【解析】由题意可得K2的观测值k ＝≈7227∵P(K2≥6635)≈1所以x 与y 之间有关系出错的可能性为1解析：1% 【解析】由题意可得K 2的观测值k ＝210154016)(10164015)(1015)(4016)(1040)(1516)+++⨯-⨯++++（≈7．227，∵P (K 2≥6．635)≈1%, 所以“x 与y 之间有关系”出错的可能性为1%16．②④⑤【解析】试题分析：线性相关系数为当越接近1时两个变量的线性相关越强当越接近0时两个变量的线性相关越弱①错；由变量和的数据得到其回归直线方程：则一定经过②正确；每10分钟从中抽取一件产品进行某项解析：②④⑤ 【解析】试题分析：线性相关系数为，当越接近1时，两个变量的线性相关越强，当越接近0时，两个变量的线性相关越弱，①错；由变量和的数据得到其回归直线方程：，则一定经过，②正确；每10分钟从中抽取一件产品进行某项指标检测，这样的抽样是系统抽样，③错；相关指数用来刻画回归的效果，其计算公式是，在含有一个解释变量的线性模型中，恰好等于相关系数的平方．显然，取值越大，意味着残差平方和越小，也就是模型的拟合效果越好，④正确；在回归直线方程表示解释变量每增加一个单位时，预报变量增加0．1个单位，⑤正确，故填②④⑤．考点：线性相关，线性回归直线方程，抽样方法，残差．17．【解析】试题分析：由回归方程过样本平均数点则：由代入可得：由当气温为5℃时用电量的度数约为：40考点：回归方程的性质及应用解析：【解析】试题分析：由回归方程过样本平均数点(,)x y ，则：10,30x y ==，由2ˆ-=b代入x b a yˆˆˆ+=可得： ˆ50a=，由ˆ502y x =-当气温为5℃时，用电量的度数约为：40 考点：回归方程的性质及应用.18．①③④【解析】①是独立性检验的应用①对②中由于所以显然是半个圆②错③中由极坐标中两点距离公式=③对④中所有边长相等的凸多边形都是正多边形为大前提是错误的因为只需要正多边形挤压变形使之仍为凸多边形即可解析：①③④ 【解析】①是独立性检验的应用，①对．②中由于[]0,θπ∈，所以01y ≤≤，显然是半个圆，②错．③中，由极坐标中两点距离公式2221212212cos()AB ρρρρθθ=+--=14912()19,2+-⨯-=19AB ③对．④中“所有边长相等的凸多边形都是正多边形”为大前提，是错误的，因为只需要正多边形挤压变形，使之仍为凸多边形即可．④对．所以填①③④．19．③【解析】推断在100个吸烟的人中必有99人患有肺病说法错误排除①有99的把握认为吸烟与患病有关系时与99的可能患有肺病是两个不同概念排除②故填③解析：③ 【解析】推断在100个吸烟的人中必有99人患有肺病，说法错误，排除①，有99%的把握认为吸烟与患病有关系时，与99%的可能患有肺病是两个不同概念,排除②，故填③．20．【分析】根据列联表计算可得由可得结果【详解】由题意得：至少有的把握认为学生的学习积极性与对待班级工作的态度有关故答案为：【点睛】本题考查独立性检验问题的求解考查基础公式的应用解析：99.9%【分析】根据22⨯列联表计算可得2K ，由210.828K >可得结果.【详解】由题意得：()225018197611.53810.82825252426K ⨯⨯-⨯=≈>⨯⨯⨯， ∴至少有10.1%99.9%-=的把握认为学生的学习积极性与对待班级工作的态度有关.故答案为：99.9%. 【点睛】本题考查独立性检验问题的求解，考查基础公式的应用.三、解答题21．（1）0.33b ≈＾，2c =＾；（2）①表格见解析，②Q 1<Q 2，所以模型甲的拟合效果更好. 【分析】（1）对于方程甲：设t=(x-1)2，则(1)2.75ybt =+＾＾，代入数据，求出,t y ，代入方程即可求出b ＾，对于方程乙，求出x 的值，代入方程，即可求出c ＾；（2）①将数据分别代入两方程，计算求解，即可完成表格，②分别计算模型甲与模型乙的残差平方和Q 1及Q 2，进行比较，即可得答案. 【详解】（1）对于方程甲：设t=(x-1)2，则(1)2.75y bt =+＾＾，所以t =15(1+4+9+16+25)=11，1(346811) 6.45y =++++=，所以6.411 2.75b =⨯+＾，解得0.33b ≈＾. 对于方程乙：1(23456)45x =++++=，所以6.44 1.6c =⨯-＾，解得2c =＾. （2）①经计算，可得下表：12因为Q 1<Q 2，所以模型甲的拟合效果更好. 【点睛】本题考查回归直线的求法与应用，残差平方的计算与分析，计算难度偏大，考查分析理解，计算求值的能力，属中档题.22．（1）列联表见解析；（2）大概有90%把握认为药物有效．【分析】（1）根据服用药的共有60个样本，服用药但患病的仍有20个样本，没有服用药且未患病的有20个样本，根据各种数据，列好表格，填好数据，得到列联表.（2）根据列联表数据，代入临界值公式，做出观测值，进行比较，即可得出结果. 【详解】（1）根据服用药的共有60个样本，服用药但患病的仍有20个样本，没有服用药且未患病的有20个样本，得到列联表()()()()()22n ad bc K a b c d a c b d -=++++()210040202020 2.77860406040⨯⨯-⨯=≈⨯⨯⨯由()22.7060.10P K ≥=，所以大概有90%把握认为药物有效．【点睛】本题考查了完善列联表和独立性检验，考查了数据分析能力和计算能力，属于基础题目. 23．（1）不能在犯错误的概率不超过10%的情况下认为该校参加体育运动的积极性与性别有关；（2）分布列答案见解析，6()5E X =，18()25D X =. 【分析】（1）代入2K 即可得出结论；（2）X 服从二项分布，分别求出概率，即可得出X 的分布列，然后代入数据求出期望和方差即可. 【详解】（1）由列联表可知2200(70406030)2002.1981307010010091k ⨯⨯-⨯==≈⨯⨯⨯，因为2.198 2.706<，所以不能在犯错误的概率不超过10%的情况下认为该校参加体育运动的积极性与性别有关. （2）由题意可知2(3,)5XB ，X 的所有可能取值为0,1,2,3，033327(0)()5125P X C ===，1232354(1)()()55125P X C ==⨯=，2232336(2)()55125P X C ==⨯=，33328(3)()5125P X C ===. 所以X 的分布列为()355E X =⨯=，()3(1)5525D X =⨯⨯-=.【点睛】本题主要考查独立性检验原理以及利用二项分布求期望和方差.属于中档题. 24．（1）列联表见解析；（2）有99%的把握认为”“手机支付族”与“性别”有关．【分析】（1）根据茎叶图提供的数据可计数可得出列联表；（2）计算出2K 可得结论．【详解】解：（1）（2）由于245(1516410)7.287 6.63519262520K ⨯⨯-⨯=≈>⨯⨯⨯，因此有99%的把握认为”“手机支付族”与“性别”有关．【点睛】本题考查列联表，考查独立性检验，正确认识茎叶图是解题关键．25．（Ⅰ）81.68；80.4；A 生产线的质量指标值更好；（Ⅱ）众数为80；中位数约为81.58；（Ⅲ）列联表见解析，有.【分析】（Ⅰ）同一组数据中的数据用该组区间的中点值作估值结合频率可计算出均值；（Ⅱ）频率最大的那组数据中间值为众数，中位数要计算频率不0.5的那一点，它在区间[]76,84上．（Ⅲ）根据频率分布直方图可得各数据，得列联表，计算2K 后可得结论．【详解】解：（Ⅰ）设A ，B 两条生产线的产品质量指标值的平均数分别为x ，y ，由直方图可得(640.00625720.01825800.05375880.035960.01125)881.68x =⨯+⨯+⨯+⨯+⨯⨯=，同理80.4y =，x y >，因此A 生产线的质量指标值更好．（Ⅱ）A 生产线的产品质量指标值的众数为80由A 生产线的产品质量指标值频率分布直方图，前两组频率为0.0062580.0187580.20.5⨯+⨯=<前三组频率为0.0062580.0187580.0537580.630.5⨯+⨯+⨯=> 故中位数在区间[]76,84，设为x ，则()0.0062580.0187580.05375760.5x ⨯+⨯+⨯-=，解得 5.587681.58x ≈+=，故A 生产线的产品质量指标值的中位数约为81.58．（Ⅲ）()229982912004.714 3.84110010011189K ⨯-⨯⨯=≈>⨯⨯⨯故有95%的把握认为“生产超优品是否与生产线有关”．【点睛】本题考查频率分布直方图，考查用频率分布直方图估计众数，中位数，均值等，考查独立性检验．考查了学生的数据处理能力和运算求解能力，属于中档题．26．（1）该市一天的空气质量等级分别为1、2、3、4的概率分别为0.43、0.27、0.21、0.09；（2）350；（3）有，理由见解析.【分析】（1）根据频数分布表可计算出该市一天的空气质量等级分别为1、2、3、4的概率；（2）利用每组的中点值乘以频数，相加后除以100可得结果；（3）根据表格中的数据完善22⨯列联表，计算出2K 的观测值，再结合临界值表可得结论. 【详解】（1）由频数分布表可知，该市一天的空气质量等级为1的概率为216250.43100++=，等级为2的概率为510120.27100++=，等级为3的概率为6780.21100++=，等级为4的概率为7200.09100++=；（2）由频数分布表可知，一天中到该公园锻炼的人次的平均数为100203003550045350100⨯+⨯+⨯=（3）22⨯列联表如下：()221003383722 5.820 3.84155457030K ⨯⨯-⨯=≈>⨯⨯⨯，因此，有95%的把握认为一天中到该公园锻炼的人次与该市当天的空气质量有关.【点睛】本题考查利用频数分布表计算频率和平均数，同时也考查了独立性检验的应用，考查数据处理能力，属于基础题.。

动画演示数据结构与算法

动画演示数据结构与算法## English Response.### Introduction.Data structures and algorithms are fundamental concepts in computer science. They provide the building blocks for organizing and manipulating data efficiently. Visualizing these concepts through animations can greatly enhance understanding and retention. This article presents a comprehensive guide to animated demonstrations of data structures and algorithms.### Types of Data Structures.Arrays: A linear collection of elements that can be accessed randomly.Linked Lists: A linear collection of elements that are connected by pointers.Stacks: A last-in, first-out (LIFO) data structure that follows the principle of a stack of plates.Queues: A first-in, first-out (FIFO) data structurethat follows the principle of a queue of people waiting in line.Trees: A hierarchical data structure that organizes elements in a parent-child relationship.Hash Tables: A data structure that maps keys to values, enabling efficient retrieval based on key lookup.Graphs: A data structure that represents a network of nodes connected by edges, often used for representing relationships.### Types of Algorithms.Sorting Algorithms: Algorithms that arrange elements in a specific order, such as ascending or descending.Searching Algorithms: Algorithms that find an element within a data structure.Tree Traversal Algorithms: Algorithms that visit nodes in a tree in a specific order.Graph Traversal Algorithms: Algorithms that visit nodes in a graph in a specific order.Hashing Algorithms: Algorithms that map keys to values in a hash table.Dynamic Programming Algorithms: Algorithms that solve complex problems by breaking them down into smaller subproblems and storing the solutions.Greedy Algorithms: Algorithms that make locally optimal choices at each step, aiming for a globally optimal solution.### Animation Tools and Resources.Visualgo: An online platform that provides interactive visualizations of data structures and algorithms.Khan Academy: A non-profit educational organizationthat offers animated videos explaining data structures and algorithms.Coursera: An online learning platform that offers courses on data structures and algorithms, many of which include animated demonstrations.edX: Another online learning platform that offers courses on data structures and algorithms with animated content.YouTube: A vast repository of videos, including many animated demonstrations of data structures and algorithms.### Benefits of Animation.Enhanced Understanding: Visualizing data structures andalgorithms in motion allows students to grasp their behavior more intuitively.Increased Retention: Animated demonstrations can create a lasting impression, making it easier for students to remember and apply the concepts.Improved Problem-Solving Skills: By witnessing thestep-by-step execution of algorithms, students can develop stronger problem-solving abilities.Engaging Learning Experience: Animations add an element of interactivity and engagement, making the learning process more enjoyable.Foundation for Real-World Applications: Understanding data structures and algorithms is essential for building efficient and effective software applications.### Conclusion.Animated demonstrations of data structures andalgorithms are a powerful tool for enhancing understanding, retention, and problem-solving skills. By utilizing the resources available online, educators and students can leverage these visualizations to make learning more effective and engaging.## 中文回答：### 介绍。

An imaging K-band survey - I The catalogue, star and galaxy counts

ible stars (Black, 1986). They may have been detected, via infrared imaging, as companions to white dwarfs by Zuckerman & Becklin (1987) and Becklin & Zuckerman (1988) and a systematic survey, based on I plates, by Hawkins & Bessel (1988) shows that low-luminosity stars are found as an extension of the stellar luminosity function. Because the blackbody ﬂux from such objects peaks at K , and the objects are additionally reddened by the presence of strong absorption bands in the optical, a K -band survey opens up the possibility of revealing and quantifying new populations of such objects. Secondly the survey provides a catalogue of galaxies selected by near-infrared ﬂux, which allows a new approach to the study of galaxy evolution. It is now well established that the faint galaxy number-magnitude counts in the optical show a blue excess population over that predicted without evolution (e.g. Peterson et al. 1979, Kron 1980, Hall & Mackay 1984, Tyson 1988). Deep redshift surveys by Broadhurst et al. (1988), Colless et al. (1990) and Lilly et al. (1991) are beginning to show that the excess is not due to high redshift objects, as would be expected in simple models of luminosity evolution in the stellar populations, but appears to be due to lower-luminosity objects with a higher space density. At redder wavelengths the light becomes dominated by stars with progressively longer lifetimes. The near-infrared light from a galaxy is dominated by the “old-stellar population”. This consists of old, evolved, stars on the giant branch with lifetimes of gigayears. Thus the relative contribution

application to privacy

2
University of Waterloo rjain@cs.uwaterloo.ca Tata Institute of Fundamental Researchjaikumar@tifr.res.in, pgdsen@tcs.tifr.res.in
1
Abstract. We prove the following theorem about relative entropy of quantum states. Substate theorem: Let ρ and σ be quantum states in the same Hilbert space with relative entropy S (ρ σ ) := Tr ρ(log ρ − log σ ) = c. Then for all > 0, there is a state ρ such that the 2 trace distance ρ − ρ tr := Tr (ρ − ρ)2 ≤ , and ρ /2O(c/ ) ≤ σ . It states that if the relative entropy of ρ and σ is small, then there is a state ρ close to ρ, i.e. with small trace distance ρ − ρ tr , that when scaled down by a factor 2O(c) ‘sits inside’, or becomes a ‘substate’ of, σ . This result has several applications in quantum communication complexity and cryptography. Using the substate theorem, we derive a privacy trade-oﬀ for the set membership problem in the twoparty quantum communication model. Here Alice is given a subset A ⊆ [n], Bob an input i ∈ [n], and they need to determine if i ∈ A. Privacy trade-oﬀ for set membership: In any two-party quantum communication protocol for the set membership problem, if Bob reveals only k bits of information about his input, then Alice must reveal at least n/2O(k) bits of information about her input. We also discuss relationships between various information theoretic quantities that arise naturally in the context of the substate theorem. Most of the results presented here have previously appeared in [JRS02].

可汗学院数学题目-31Datainferences

Data inferences1.A random sample of 35 four-door passenger vehicles had a mean gas mileage, in miles per gallon (mpg), of 25.9 mpg. The estimate had a margin of error of 2.6 mpg at a 98% confidence level. Of the following, which is most plausible value for the true mean of the mileage of four-door passenger vehicles in general?A.24 mpgB.29 mpgC.32 mpgD.35 mpgCorrect answer: A Difficulty level: 22.Based on random samples of river heights, oceanographers estimate that 4,800 cubic kilometers (km3) of freshwater is discharged into the Arctic Ocean annually. The estimate has a margin of error of 240km3at the 90%confidence level. Which of the following is the most reasonable claim about the volume of freshwater discharged annually into the Arctic Ocean?A.It is between 4,800 and 5,040 km3.B.It is between 4,560 and 5,040 km3.C.It is between 240 and 4800 km3.D.It is between 240 and 4320 km3.Correct answer: B Difficulty level: 23.A survey of 9,000 randomly selected dentists practicing in the United States (US) found that 3,600 of them used text messaging to remind clients of upcoming appointments. If there were 147,000 practicing dentists in the US, approximately how many of them used text messaging for that purpose?A.12,600 dentistsB.58,800 dentistsC.159,600 dentistsD.367,500 dentistsCorrect answer: B Difficulty level: 24.A random sample of international flights in 2013 showed that 79% of the flights arrived within 15 minutes of their scheduled arrival time, and this estimate had a margin of error of 3% at a 95% confidence level. Which of the following conclusions is best?A.Exactly 79% arrived within 15 minutes of their scheduled times.B.Exactly 76% arrived within 15 minutes of their scheduled times.C.Between 76% and 82% arrived within 15 minutes of their scheduled times.D.Between 92% and 98% arrived within 15 minutes of their scheduled times.Correct answer: C Difficulty level: 25.In a poll of 1,578 randomly selected American adults, 44.8% of the respondents said that airlines should allow in-flight calls on airplanes. The results had a margin of error of 2.5%at a 90% confidence level. Which of the following is most likely to be equal to the percentage of all American adults who would say that airlines should allow in-flight calls?A.40%B.43%C.48%D.90%Correct answer: B Difficulty level: 26.In a study of 40randomly selected captive Octopus vulgaris, researchers found that the octopi could learn to integrate visual and tactile information to find food after a mean of 119 trials. The estimate had a margin of error of 18 trials at a 95% confidence level. Which of the following is the most reasonable conclusion about the number of trials it would take all captive Octopus vulgaris in general to learn to find the food?A.The evidence suggests that half of them would require 119 or fewer trials.B.The evidence suggests that the mean number of trials would be 119 trials.C.The evidence suggests that the mean number of trials would be between 119 and 137 trials.D.The evidence suggests that the mean number of trials would be between 101 and 137 trials. Correct answer: D Difficulty level: 37.Researchers measured the heart rates of several randomly selected astronauts exercising on stationary bicycles during long-term space missions. The researchers found the mean heart rate of the astronauts was 155 beats per minute with a margin of error of 12 beats per minute at a 90% confidence level. Which of the following is the most plausible value for the mean heart rate of all astronauts exercising on stationary bicycles during long-term space missions?A. 130 beats per minuteB. 140 beats per minuteC. 160 beats per minuteD. 175 beats per minuteCorrect answer: C Difficulty level: 38.In a poll of 14,697 randomly selected adults in the United States, those polled spent an average of $95 per day in November of last year, as compared with $91 per day in November two years ago. The estimates had a margin of error of $4 at the 95% confidence level. Which of the following is a reasonable claim to make based on this sample?A. All adults in the United States spent between $91 and $99 daily last November.B. 95% of adults in the United States spent between $91 and $99 daily last November.C. It is plausible that average daily spending of adults in the United States remained the same inNovember of last year as it was in November two years ago.D. Between 91% and 99% of adults in the United States spent $4 more daily last November thanin November two years ago.Correct answer: C Difficulty level: 39.In a particular county, a sample of the population showed that 84% of the households lived in the same residence as they had the previous year. The estimate had a margin of error of 1.5% at the 90% confidence level. If the county has 50,000 households, which of the following best estimates the number of households that lived in the same residence as they had the previous year, at the 90% confidence level?A. 34,500 to 49,500 residentsB. 37,500 to 50,000 residentsC. 41,250 to 42,750 householdsD. 44,250 to 45,750 householdsCorrect answer: C Difficulty level: 310.Prior to the 2014 elections, 1,000 randomly selected Louisiana voters were surveyed about what single issue would most likely influence their vote. Of those surveyed, 560voters answered that the state of the economy would most influence their vote. Based on this information, which statement about all voters in Louisiana is most appropriate?A.Exactly 56% of all Louisiana voters thought the state of the economy would most influencetheir voting.B.Approximately 56%of all Louisiana voters thought the state of the economy would mostinfluence their voting.C.Exactly 56% of Louisiana voters would vote for the candidate with the best plan to improvethe economy.D.Approximately 56% of Louisiana voters would vote for the candidate with the best plan toimprove the economy.Correct answer: B Difficulty level: 311.A **pany sells bottles of water. A random sample of 50 water bottles found that the bottles contained, on average, 0.96 liters of water, and the estimate had a margin of error of 0.08 liters at the 95% confidence level. Which of the following is a reasonable claim to make based on this sample?A.**pany always fills its water bottles with less than 1 liter of water.B.**pany usually fills its water bottles with between 0.96 liters and 1.04 liters of water.C.**pany always fills its water bottles with between 0.88 liters and 1.04 liters of water.It is plausible that, on average, **pany fills its water bottles with 1 or more liters of water.Correct answer: D Difficulty level: 412.An archaeologist uses an accelerator mass spectrometer to find the age of a buried branch. At the 68% confidence level, the spectrometer estimates that the branch was 10,000 years old with a margin of error of 200 years. Which of the following could the spectrometer estimate as the age of the branch at the 95% confidence level?A.9,500 years old, with a margin of error of 500 yearsB.10,000 years old, with a margin of error of 400 yearsC.9,500 years old, with a margin of error of 50 yearsD.10,000 years old, with a margin of error of 40 yearsCorrect answer: B Difficulty level: 413.A random sample of500residents of a town included 173residents who primarily spoke a language other than English at home, with a margin of error of 25 residents and a confidence level of 98%. If the town has 25,000 residents, how many residents primarily speak a language other than English at home, with the 98% confidence level?A.7,400 to 9,900 residentsB.8,625 to 8,675 residentsC.24,475 to 24,525 residentsD.23,250 to 25,750 residentsCorrect answer: A Difficulty level: 414.In the National Health and Nutrition Examination Survey, 80.5% of the participants had healthy levels of high-density lipoprotein (HDL) cholesterol, the estimate has a margin of error of 1.5% at a confidence level of 95%. The participants were a random sample of United States (US) residents. If the US population at the time of the study was 316 million people, what is the estimate at a 95% confidence level of the number of people in the US who had healthy levels of HDL cholesterol?A.237 million to 246 million peopleB.245million to 264 million peopleC.250 million to 259 million peopleD.254 million to 300 million peopleCorrect answer: C Difficulty level: 415.A university had 150 students registered for an introductory sociology class. The students were randomly assigned to either a section taught with case studies or with lectures. Both sections took the same final exam. Both margins of error are at a 90% confidence level. Which of the following conclusions is the most reasonable regarding the sociology class?A.The university should only offer the lecture teaching method, because students in that sectionhad a higher final exam average.B.The university should not offer either teaching method, case studies or lectures, becauseneither section achieved the target 90% average.C.The university should only offer the lecture teaching method, because the final exam averageplus the margin of error achieved the target 90%.D.The university does not have strong evidence to indicate which teaching method is moreeffective.Correct answer: D Difficulty level: 4。

可汗学院新SAT语法真题下载

可汗学院新S A T语法真题下载-CAL-FENGHAI-(2020YEAR-YICAI)_JINGBIAN可汗学院新SAT语法真题下载到目前为止，新版SAT可汗学院官方不断放出更多真题，已经放出了68篇阅读，且之前已经和大家分享过可汗学院新SAT阅读真题，想要下载的同学，请点击：新SAT阅读真题下载（共68篇，且已全）目前可汗学院一共放出41篇新SAT数学真题！想要下载的同学，请点击：新SAT数学真题下载（共41篇）分享了可汗学院的数学和阅读真题后，还有我们的可汗学院SAT语法真题。

截止到6月前，可汗学院一共放出了48套新SAT语法真题，想要吗？下载请点击：新SAT语法真题下载（共48篇）（网址：/sat/news/606541.html）可汗学院新SAT语法真题（部分）Questions 1-5 are based on the following passage. 1Searching for GuinevereStories of kings and queens have captivated readers for centuries, and arguably, the tales of King Arthur and Guinevere are among the most enchanting. Arthur ruled the kingdom of Camelot, and Guinevere was his queen. But were they real people or fictional charactersThe debate has continued for centuries. Though many scholars have found evidence that the legendary Arthur was, at the very least, based on a real person who lived in Britain roughly between 450 and （1） 500 CE. They continue to search for the historical identity of Guinevere. Guinevere first appeared as King Arthur’s queen in one of the most wide lystudied works of Arthurian literature, （2）The History of the Kings of Britain. This book was written by Geoffrey of Monmouth around 1135 CE. Geoffrey’s historical treatment of the legend is often（3）sited as evidence that the queen of Camelot existed, as the book chronicles the lives of a number of historical rulers.*God help those who help themselves. We help those who trust us. Contact Wechat:satxbs123, help is waiting.1A) NO CHANGEB) 500 CE. ContinuingC) 500 CE, continuingD) 500 CE, they continueWhich choice most effectively combines the sentences at the underlined portion?A) The History of the Kings of Britain, and this bookB) The History of the Kings of Britain, whichC) a book called The History of the Kings of Britain,as thisD) a book called The History of the Kings of Britain,and this3A) NO CHANGEB) insightedC) citedD) incitedGuinevere is identified by Geoffrey as a noblewoman of Roman descent who met King Arthur in the court of Duke Cador of Cornwall, where she lived as a ward. (4)In Malory’s portrayal, Guinevere had no real power as a monarch but served as a kind of spiritual leader, providing guidance and moral support to the knights in their roles as defenders of the kingdom. Le Morte d’Arthur was also one of the first works to reference Guinevere’s romance with the knight, Sir Lancelot.As many Arthurian scholars know, the distinction between history and literature was blurred in the Middle Ages. Consequently, the true identity of Guinevere may never be known with certainty. Yet regardless of whether Guinevere was real or fictional, her story (5) had endured centuries—and through each retelling, she continues to live on in the imaginations of people around the world.4At this point, the author wants to add a sentence which effectively sets up the portrayal of Guinevere discussed in the rest of the paragraph. Which choice best accomplishes this goal?A) Three centuries later, however, Thomas Malory painted a very different portrait of Guinevere in Le Morte d’Arthur.B) Sir Thomas Malory was an English knight and Member of Parliament who also wrote extensively about the history of the British monarchy.C) Many historians believe that the portrayal of Arthur and Guinevere in Sir Thomas Malory’s Le Morte d’Arthur was actually a political commentary on the War of the Roses (1455-1487CE).D) In Le Morte d’Arthur, Sir Thomas Malory describes an idyllic England under King Arthur and Guinevere, which eventually collapses into chaos and political unrest. E. I would be guessing.A) NO CHANGEB) was enduringC) would have enduredD) has enduredQuestions 1-5 are based on the following passage. 1Cometary Missions: Trajectory for SuccessScientists have been launching cometary missions since 1978. The first one, a joint mission by the European Space Agency, and the National Aeronautics and Space Administration (NASA), was a “flyby” in which the spacecraft collected data while passing around Comet Giacobini-Zinner. (1)However, the landing of the Rosetta space probe on comet 67P/Churyumov-Gerasemenko in 2014 was different: it marked the first time that a probe landed on a( 2 )comet and giving scientists an unprecedented opportunity to study the surface of a comet. In order to continue this valuable research, additional missions are needed; thus, it is critical that more funding be allocated for this purpose.The 2014 Rosetta mission provided a rare opportunity for scientists to test a number of hypotheses regarding the composition of (3) comets; the distribution of organic compounds in our solar system and the origins of life on Earth. Unlike other cometary missions, the Rosetta spacecraft contained a probe, Philae, that was able to land on the surface of a comet. *Rack your brain and you don't know why.Follow Wechat: satxbs123, she can help you!1At this point, the writer wants to add accurate information from the graph. Which choice best accomplishes this goal?A) From 1978 to 2014, the number of successful missions increased from 28 percent to 72 percent.B) Before 2014, the majority of attempted cometary missions were considered unsuccessful.C) Between then and 2014, 72 percent of the cometary missions were successful.D) Of the missions attempted since then, 44 percent have been successful.2A) NO CHANGEB) comet, but it gaveC) comet, yet givesD) comet, givingA) NO CHANGEB) comets, the distribution of organic compounds in our solar system,C) comets, the distribution of organic compounds in our solar system;D) comets; the distribution of organic compounds in our solar system,。

Purification and Three-Dimensional Electron Microscopy Structure of the Neisseria meningiti

J OURNAL OF B ACTERIOLOGY,Sept.2007,p.6389–6396Vol.189,No.17 0021-9193/07/$08.00ϩ0doi:10.1128/JB.00648-07Copyright©2007,American Society for Microbiology.All Rights Reserved.Puriﬁcation and Three-Dimensional Electron Microscopy Structure of the Neisseria meningitidis Type IV Pilus Biogenesis Protein PilGᰔRichard F.Collins,‡Muhammad Saleem,‡and Jeremy P.Derrick* Faculties of Life Sciences and Engineering/Physical Sciences,Manchester Interdisciplinary Biocentre,University ofManchester,131Princess Street,Manchester,United KingdomReceived25April2007/Accepted26June2007Type IV pili are surface-exposed retractableﬁbers which play a key role in the pathogenesis of Neisseriameningitidis and other gram-negative pathogens.PilG is an integral inner membrane protein and a componentof the type IV pilus biogenesis system.It is related by sequence to the extensive GspF family of secretoryproteins,which are involved in type II secretion processes.PilG was overexpressed and puriﬁed from Esche-richia coli membranes by detergent extraction and metal ion afﬁnity chromatography.Analysis of the puriﬁedprotein by perﬂuoro-octanoic acid polyacrylamide gel electrophoresis showed that PilG formed dimers andtetramers.A three-dimensional(3-D)electron microscopy structure of the PilG multimer was determinedusing single-particle averaging applied to samples visualized by negative staining.Symmetry analysis of theunsymmetrized3-D volume provided further evidence that the PilG multimer is a tetramer.The reconstructionalso revealed an asymmetric bilobed structure approximately125Åin length and80Åin width.The larger lobewithin the structure was identiﬁed as the N terminus by location of Ni-nitrilotriacetic acid nanogold particlesto the N-terminal polyhistidine tag.We propose that the smaller lobe corresponds to the periplasmic domainof the protein,with the narrower“waist”region being the transmembrane section.This constitutes theﬁrstreport of a3-D structure of a member of the GspF family and suggests a physical basis for the role of theprotein in linking cytoplasmic and periplasmic protein components of the type II secretion and type IV pilusbiogenesis systems.The gram-negative pathogen Neisseria meningitidis is the causative agent of meningococcal meningitis and septicemia and a serious public health problem.Type IV pili are long, thin,and mechanically robustﬁbers which extend from the surface of the bacterium and bind to host cell surface receptors (37).They have been shown to play a pivotal role in coloniza-tion and infection by N.meningitidis(27).Type IV pilus tip adhesins mediate binding to epithelial cell receptors(37)and so play a role in colonization(27).Type IV pili are also known to be involved in autoagglutination(38)and natural compe-tence for DNA uptake(18).The ability of type IV pili to retract has been linked to twitching motility,a form of bacterial propulsion along solid surfaces(3).Despite the importance of type IV pilus biogenesis,the present understanding of the process at a molecular level is rudimentary.A survey has iden-tiﬁed15different proteins involved in the biogenesis of type IV pili in N.meningitidis(5).Of these,only six(PilD,-F,-M,-N, -O,and-P)are essential for assembly of the major pilus pro-tein,PilE,into pili(4).Other proteins appear to be involved in the functional maturation of the pilus(PilC,-I,-J,-K,and-W), its retraction(PilT),or its transit across the outer membrane (PilQ).Interestingly,although pilG mutants in pathogenic Neisseria are devoid of pili(40),Carbonnelle et al.found that a pilG pilT double mutant was piliated,with adhesive and aggregative properties which were close to those of the wild-type strain(4).These authors concluded that PilG plays a role in the counterretraction of the pilusﬁber rather than a primary role in pilus assembly.Several lines of evidence have pointed to similarities be-tween type IV pilus biogenesis and the type II secretion system (23,28).For example,the GspE family(GSP is for g eneral type II s ecretory p athway)of hexameric ATPases,which are in-volved in type II secretion(23,43),have orthologs which ap-pear to carry out similar functions in type IV pilus biogenesis (4,14,20).The PulD and PilQ secretins,which mediate the transport of pullulanase and type IV pili,respectively,across the outer membrane,show some homology in sequence within their C-terminal regions and have broad structural similarities (6,12).PilG is a member of the large GspF family of integral inner membrane proteins,which includes proteins involved in type IV pilus biogenesis and type II secretion.Evidence from studies on both systems suggests that GspF orthologs function as a complex with other proteins involved in the secretion process.For example,Py et al.showed that in the type II secretion system in Erwinia chrysanthemi,GspF formed part of a complex with GspE,-L,and-M(30).Crowther et al.,working on the mechanism of bundle-forming pilus formation in entero-pathogenic Escherichia coli,showed that the GspF ortholog BfpE is involved in recruiting the ATPase BfpD to the cyto-plasmic membrane(13).They also provided evidence that BfpE interacted with a separate ATPase,BfpF,to promote pilus retraction.Further work has shown that peptides derived from another pilus biogenesis component protein,BfpC,and BfpE stimulated the ATPase activity of BfpD in an allosteric manner(14).These observations support the view that the*Corresponding author.Mailing address:Faculties of Life Sciences and Engineering/Physical Sciences,Manchester Interdisciplinary Bio-centre,University of Manchester,131Princess Street,Manchester, United Kingdom.Phone:001-44-0612004207.Fax:001-44-0612360409.E-mail:Jeremy.Derrick@.‡Both authors contributed equally to the work.ᰔPublished ahead of print on6July2007.6389 at Penn State Univ on April 16, 2008 Downloaded fromBfpE/PilG pilus biogenesis components play a crucial role in the formation of a multiprotein complex which could traverse the inner membrane.To date,there have been no reports of three-dimensional (3-D)structures of members of the GspF family.Blank and Donnenberg used a fusion protein approach to study the to-pology of BfpE and identiﬁed four transmembrane segments, three within the center of the gene and one at the C terminus (1).These observations led to a proposed topology for BfpE, with a large cytoplasmic N-terminal domain,three transmem-brane␣-helices,a smaller periplasmic domain,and a fourth C-terminal transmembrane␣-helix.Single-particle averaging (SPA)is a method which can be applied to determine the3-D structure of a large protein(Ͼ200kDa)by a process of aver-aging projections of many thousands of individual protein par-ticles(17).We have previously described the application of SPA methods to determine the3-D structure of another mem-brane-associated component from the Neisseria type IV pilus biogenesis system,the outer membrane secretin PilQ(10,12). The results established that PilQ is a dodecamer and adopts a cage-like structure with a large internal cavity which isﬁlled when PilQ associates with type IV pili(9,11).These observa-tions illustrate the potential of SPA methods to provide struc-tural information which can suggest clues to the function of the protein.Here we describe the3-D structure of PilG from Neisseria meningitidis by SPA.We show that PilG forms a tetrameric bilobed structure,which agrees well with the Blank and Donnenberg topology model for the ortholog BfpE.The signiﬁcance of the structure,in terms of its implications for the function of this class of pilus biogenesis proteins,is discussed.MATERIALS AND METHODSMaterials.BD Co-TALON resin was obtained from BD Biosciences.n-Do-decyl-D-maltoside(DDM)was from Anatrace Ltd.,United Kingdom.Oligonu-cleotide primers were purchased from Sigma-Aldrich,and Pfu polymerase was purchased from Epicenter Biotechnologies.The E.coli strain DH5␣was ob-tained from Invitrogen Ltd.,United Kingdom.E.coli strains BL21(DE3)*and BL21(DE3)pLysS were from Novagen(Merck Biosciences Ltd.,United King-dom).Perﬂuoro-octanoic acid(PFO)was supplied by Fluorochem(Old Glossop, Derbyshire,United Kingdom).Ni-nitrilotriacetic acid(NTA)nanogold was from Nanoprobes(Yaphank,NY).Cloning and overexpression of recombinant PilG.DNA sequence coding for the integral membrane protein PilG was ampliﬁed from genomic DNA isolated from Neisseria meningitidis MC58(39)by use of high-ﬁdelity Pfu polymerase and the following two primers:5Ј-CGGGGCTAGCATATGGCTAAAAACG and 5Ј-GGATCTGTGCCTCGAGTCAGGCGACCACG.An NdeI site was intro-duced into the5Јend of the PCR product before the start codon,and an XhoI site was positioned after the stop codon at the3Јend.The PCR product was ligated directionally into the corresponding sites of pET-28a(Novagen)to yield the expression vector pNm-pilG,which introduces a hexahistidine tag at the N terminus of the protein.For expression,pNm-pilG was transformed into E.coli BL21(DE3)*(Invitrogen),and transformants were grown in2YT medium(34) to an optical density at600nm of0.8to1.0at37°C.At this point,expression was induced by the addition of isopropyl-␤-D-thiogalactoside(IPTG)to0.1mM,and the cells were transferred to a temperature of25°C.Growth was allowed to continue for a further16h after induction at25°C,before the cells were har-vested by centrifugation at6,000ϫg av for20min.Puriﬁcation of recombinant His-tagged PilG.The bacterial cell pellet was resuspended in lysis buffer(50mM Tris[pH8.0],200mM NaCl,1mM EDTA, and0.5mM dithiothreitol);10ml of lysis buffer was used per liter of culture. Lysozyme and DNase were also added to aid cell lysis.To minimize proteolysis, a“cocktail”of protease inhibitors(complete,EDTA free,catalog no. 11873580001;Roche)was added at the standard concentration recommended by the manufacturer.Cells were broken using a probe sonicator,and unbroken cells were removed by low-speed centrifugation at10,000ϫg av for20min.The supernatant wasﬁltered through a0.45-␮m syringeﬁlter and retained.A mem-brane-rich preparation was subsequently isolated by centrifugation at100,000ϫg av for1h at4°C.The supernatant was discarded and the pellet washed with50 mM Tris(pH8.0)and200mM NaCl,to which protease inhibitors had been added.PilG was solubilized from this membrane-rich preparation by resuspen-sion in50mM Tris(pH8.0),200mM NaCl,1%(wt/vol)DDM,also including proteinase inhibitors,at20°C for1h.Insoluble material was subsequently re-moved by centrifugation at100,000ϫg av for30min at22°C.Co-TALON resin, which had been preequilibrated in50mM Tris(pH8.0),200mM NaCl,0.08% (wt/vol)DDM,was added to the supernatant,and the suspension was mixed gently for16h at4°C.The Co-TALON resin was packed into a benchtop column and washed with10column volumes of Tris-NaCl-DDM buffer.PilG was eluted by application of50mM Tris-HCl,200mM NaCl,0.08%(wt/vol)DDM,and200 mM imidazole(pH8.0).The imidazole was removed by dialysis against50mM Tris-HCl,200mM NaCl,0.08%(wt/vol)DDM(pH8.0),and the puriﬁed PilG was concentrated by ultraﬁltration.The identity of the eluted protein was con-ﬁrmed by matrix-assisted laser desorption ionization–time ofﬂight mass spec-trometry.PFO-PAGE of oligomeric PilG.PFO polyacrylamide gel electrophoresis (PFO-PAGE)was performed using a variation on previously reported methods (31).Ten microliters of the sample,at a protein concentration of600␮g/ml,was mixed1:1with PFO sample buffer(100mM Tris[pH7.5],8%[wt/vol]Na-PFO, and20%[vol/vol]glycerol)and incubated at4°C for10min.Samples were then loaded onto7.5%slab gels containing375mM Tris-HCl(pH7.5)and run in a buffer precooled to4°C and containing25mM Tris-HCl(pH7.5),200mM glycerol,and8%(wt/vol)Na-PFO for2h at4°C.CD.Circular dichroism(CD)spectra of PilG were obtained using a JASCO J-10spectrometer and a quartz cell with a1-mm path length.PilG was at a concentration of0.130mg/ml in50mM Tris-HCl(pH7.5),200mM NaCl,and 0.08%(wt/vol)DDM.The spectrum was recorded at20°C,and data from a blank,containing the same buffer but with no protein,were subtracted.Mea-surements were made only down to the wavelength where the instrument dynode voltage indicated the detector was still in its linear range.Electron microscopy of PilG.Samples of PilG were diluted to25␮g/ml in50 mM Tris-HCl,200mM NaCl,and0.08%(wt/vol)DDM and negatively stained with uranyl acetate as previously described(11).Transmission electron micros-copy(TEM)images were recorded with a Tecnai10operating at100KeV onto Kodak SO-163ﬁlm.Data were scanned using a UMAX2000-Powerlook scanner at1,600dots per inch,providing a specimen level increment of3.66Å/pixel. Determination of PilG3-D structure.A total of9,440particles were interac-tively selected using BOXER(25)and contrast normalized.The contrast transfer function(CTF)for each particle in the data set was determined using the program CTFIT(25),and corrections for amplitudes and phases were applied using a GroEL structure factor data set(provided by S.Ludtke).The data set was initially analyzed by calculating55reference-free projection averages that re-vealed the complex orientated in multiple particle positions.By following meth-ods previously described(12),a preliminary3-D model was determined from SPA classes that represented different views of the oligomer.In outline,this process involved the generation of reference-free class averages followed by the imposition of a low-pass resolutionﬁlter to20Å.Poorly sampled or noisy images were deleted manually.A variety of classes that were representative of different sample orientations,i.e.,exhibiting rotational and bilateral symmetry as well as a variety of intermediate orientations,were subsequently selected.Classes which were apparently duplicated were deleted.From these classes,an unsymmetrized start model was generated and then reﬁned,with the appropriate symmetry imposed.As part of the symmetry analysis process,a range of symmetries was applied in separate structural calculations:C1(i.e.,no rotational symmetry imposed),D2,C2,and C4symmetries were imposed during independent com-putations.The relative orientations of the characteristic views were determined using a Fourier common-lines routine,and the resulting averages were combined to generate the preliminary3-D model.The3-D structure was subsequently reﬁned using eight rounds of iterative projection matching.Each reﬁnement was assessed by examining convergence with the Fourier shell correlation(FSC)of the3-D models generated from each iteration.The resolution was determined to be22Åby FSC analysis as described previously(12).FSC effectively splits the particles into two independent datasets:calculation of a3-D volume from each data set generated two structures which had essentially the same features as the ﬁnal structure.Furthermore,the principal features of theﬁnal model were not dependent on the starting model:a repeat of the procedure using an amorphous “blob”as a start model yielded a similar reconstruction.The labeling of PilG with Ni-NTA nanogold particles was carried out as described for the secretin protein PilQ(19).A total of1,100gold-labeled particles were selected interactively,and6390COLLINS ET AL.J.B ACTERIOL.at Penn State Univ on April 16, 2008 Downloaded froma 3-D structure was calculated using the reﬁned PilG structure as a start model for the gold-labeled data set.Symmetry analysis.Density map correlation coefﬁcients were calculated by rotation of the unsymmetrized (C1)3-D map about the long axis of the PilG complex by use of MAPROT (36),and the density map correlation coefﬁcient was calculated using OVERLAPMAP (2);both programs are from the CCP4suite (8).RESULTSExpression and puriﬁcation of PilG.As part of a program to investigate the structures and functions of the proteins in-volved in the biogenesis of type IV pili in Neisseria ,we sought to express and purify the integral membrane protein PilG in quantities suitable for structural analysis.An expression vector for PilG was constructed by insertion of an NdeI-XhoI frag-ment incorporating the entire pilG open reading frame into pET-28a to create pNm-pilG.Following the transformation of pNm-pilG into E.coli BL21(DE3)*,expression was monitored by Western blotting under a variety of different growth condi-tions.Optimal expression of PilG was obtained by transferring the cells to grow at 25°C after induction with IPTG (data not shown).Following cell breakage by sonication,total mem-branes were isolated by ultracentrifugation before solubiliza-tion with the detergent DDM.PilG was then puriﬁed by metal afﬁnity chromatography using a Co-TALON afﬁnity resin.Ex-amination of the puriﬁed protein by sodium dodecyl sulfate (SDS)-PAGE gave a band with an estimated subunit molecular mass of about 43kDa,close to the predicted value of 48.2kDa expected from the sequence (Fig.1A).The procedure pro-duced a yield of about 1mg puriﬁed protein from 6liters of bacterial culture.Biophysical properties of puriﬁed PilG.Previous work using a yeast two-hybrid system had suggested that the PilG ortholog BfpE is multimeric,but the precise quaternary structure of any member of the GspF family has yet to be established (13).PFO-PAGE is a technique that allows the quaternary structure of an oligomeric complex to be quantitatively assessed because the detergent PFO maintains subunit interactions in many protein oligomers (31).Separation of puriﬁed PilG by PFO-PAGE identiﬁed two main bands which migrated at estimated molecular masses of ϳ40kDa and ϳ80kDa (Fig.1B)and were identiﬁed as monomeric and dimeric species,respectively.A higher-molecular-mass band,estimated at ϳ180kDa,was also detectable and could correspond to a trimer or tetramer.The quaternary structure of PilG was also examined by ultracen-trifugation:sedimentation equilibrium analysis produced an overall molecular mass estimate of 124Ϯ4kDa.Sedimenta-tion velocity analysis showed the presence of three species:around two-thirds of the sample was dimeric,with the remain-ing fraction consisting mostly of tetramers and,to a lesser extent,higher-order species (data not shown).The ultracen-trifugation was carried out using PilG solubilized in DDM:this detergent appeared to preserve the quaternary structure of the protein better than PFO.Nevertheless,the results provided clear evidence for the formation of dimers and tetramers by PilG.Secondary structure bioinformatic predictions (33)indicated a high content of ␣-helix in PilG,at 73%.In order to establish that the puriﬁed PilG sample was correctly folded,a CD spec-trum was obtained (Fig.2).The results show a double mini-mum in the CD signal at 208and 220nm,characteristic of a protein with high ␣-helical content.Noise in the signal in the 190-to 200-nm region complicated an estimate of the propor-tion of ␣-helix in the protein.TEM of puriﬁed PilG.Figure 3A shows a sample area of a TEM of puriﬁed PilG visualized by negative staining.The population of PilG oligomers was homogeneous,generally nonaggregated,and uniform in size (80to 130Å).Determina-tion of a 3-D structure by SPA using the EMAN software (25)requires that a preliminary 3-D model be generated and then subjected to an iterative reﬁnement procedure to generate the ﬁnal structure.To generate this preliminary model,a set of reference-free class averages needs to be generated from the particle data set:a representative selection of those for PilG is shown in Fig.3B.Each class average represents an averaged projection of a particular orientation of the PilG particle.The results show a variety of different projection types,demonstrat-ing that the PilG oligomer was positioned on the carbon sup-port ﬁlm in different orientations.Although many projection classes were apparent,two types of high-contrastprojectionsFIG.1.SDS-PAGE and PFO-PAGE of puriﬁed PilG.(A)SDS-PAGE of puriﬁed PilG on a 4to 20%continuous-gradient gel.Mo-lecular mass markers are indicated on the right of the gel.(B)PFO-PAGE of PilG on a 4to 12%gradient nes:1,ferritin (440kDa);2,immunoglobulin G (150kDa);3,bovine serum albumin (66kDa);4,PilG.Both gels were stained with Coomassieblue.FIG.2.CD spectrum of puriﬁed PilG.degM,●●●.V OL .189,20073-D STRUCTURE OF THE PilG PILUS BIOGENESIS PROTEIN 6391at Penn State Univ on April 16, 2008 Downloaded fromcould be broadly distinguished from the majority of the data.One type of projection (Fig.3B,top left)was ϳ80to 90Åin width,had a squared four-lobed appearance,and was well stained in the center.The other type of high-contrast projec-tion (Fig.3B,bottom right)was more elongated.The other class averages represented various intermediate orientations between these two projections (Fig.3B,middle rows)and so were suitable for calculating a 3-D structure using Euler angle determination and cross-common lines projection matching (25).The readily apparent visual correlation between each class average (Fig.3B,right column)and its equivalent back-projection from the reﬁned 3-D structure (Fig.3B,left col-umn)demonstrates an excellent correspondence between the data and the ﬁnal structure.3-D structure of solubilized PilG.A total of 9,440unique PilG particles were processed to produce a 3-D reconstruction at an estimated resolution of 22Å.Structure calculation car-ried out with no symmetry imposed (i.e.,C1),generated a volume which was analyzed by map density correlation analysis (Fig.4).The results showed a strong signal for twofold sym-metry,with slightly weaker peaks at 90°and 270°.A top view of the unsymmetrized volume showed an apparent fourfold sym-metrical arrangement (Fig.4,inset).Given the results of the biochemical analysis of the quaternary structure detailed above,the most plausible symmetry arrangement would be D2,C2,or C4.Repeating the structure calculation,but with D2symmetry imposed,produced a rectangle-shaped volume,which was inconsistent with the unsymmetrized class averages shown in Fig.3(data not shown).The possibility of D2sym-metry was therefore discounted.3-D structural processing was repeated,imposing C2or C4rotational symmetry:theresult-FIG.3.Analysis of PilG by TEM.(A)Sample TEM data of PilG negatively stained with 2%(wt/vol)uranyl acetate.PilG particles are highlighted in square boxes.The image has been CTF corrected and contrast enhanced for presentation.Scale bar ϭ500Å.(B)Selected class averages generated from SPA as applied to the PilG data set.The 14average pairs shown are representative samples from the total classes and show a back-projection from the C4-symmetrized 3-D volume paired with the corresponding unsymmetrized class average used in the calculation.Excellent visual correlation between the pairs is apparent.Box size is 234Åby 234Å.FIG.4.Rotational symmetry analysis of SPA PilG volume.Real-space self-correlation analysis of the unsymmetrized PilG volume is shown as a function of the angle of rotation about the main axis of the complex.Correlation coefﬁcients were calculated from a 3-D volume produced with no symmetry averaging applied (C1).The volume was calculated using procedures described previously (12).The inset shows a top-view back-projection of the unsymmetrized volume.6392COLLINS ET AL.J.B ACTERIOL .at Penn State Univ on April 16, 2008 Downloaded froming reconstructions (Fig.5A)were essentially similar in all major features.Volumes from both structures,when con-toured at 2.6␴above the mean density,would accommodate an estimated molecular mass of about 240kDa.This is consis-tent with a PilG tetramer,assuming a contribution of 192kDa from the protein and 48kDa from the bound DDM micelle.It is not possible,however,to distinguish between a dimer-of-dimers quaternary arrangement (i.e.,C2)and a tetramer re-lated by fourfold rotational symmetry (C4)on the basis of these data.As the C2and C4models show essentially the same features,further analysis will use the C4structure,although the conclusions drawn also pertain to the C2reconstruction.It should be noted that the sample would undoubtedly have con-tained PilG dimers,but their size (ϳ120kDa)is too small to be visualized as a particle well stained for SPA.Consequently,the particles selected and analyzed by SPA corresponded only to the tetrameric species,and the dimeric species were effectively omitted.The PilG tetramer is 125Åin length and has a bilobed,asymmetric,“missile-like”appearance when viewed from the side.A narrowed waist divides the molecule into two unequal halves,which can be characterized as “ﬁn-like”and “cone-like”features (Fig.5A).Viewed from the top,the particle has a distinctive square appearance,maximally 80Åwide (Fig.5).In order to identify which end of the molecule corresponded to the N-terminal domain,we employed selective labeling with Ni-NTA nanogold,which binds to the N-terminal polyhistidine tag.In a separate reconstruction,a data set of 1,100particles of the Ni-NTA nanogold–PilG complex was collected:some ex-amples of the images used are shown in ing theunliganded PilG structure as a starting model,a volume for the Ni-NTA nanogold–PilG complex was determined (Fig.6B).A major strength of the nanogold labeling methodology is that the nanogold particles have scattering characteristics different from those of the PilG volume in negative staining:the positive density in Fig.6B,in green,is attributable to PilG,and the negative density,in yellow,corresponds to the nanogold par-ticles.The location of the nanogold label can therefore be readily distinguished from background noise.The PilG portion of the structure has the characteristics of the PilG reconstruc-tion alone but is inevitably noisier,due to the lower number of particles used in the reconstruction.The main features of the PilG structure—the “ﬁns”at the top,the narrower “waist”region,and the “cone”at the bottom of the structure—are readily apparent.It is clear that the majority of the nanogold labels associate with the bottom half of the structure.It should be noted that the yellow nanogold density in Fig.6B and C is somewhat diffuse,possibly because the N-terminal polyhisti-dine tag is surface exposed and consequently mobile.Never-theless,the results provide clear evidence that the N terminus is associated with the “cone”portion of the PilG structure.A composite image of the (unliganded)PilG structure in green with the nanogold density peaks superimposed in yellow is shown on the left side of Fig.6C.DISCUSSIONThe members of the GspF family are not well understood at the structural level,despite their established importance in type II secretion and type IV pilus biogenesis.The datapre-FIG.5.3-D reconstruction of PilG.(A)Surface-rendered symmetrized volumes of the PilG complex:the structure in the left column has C2symmetry imposed,and the structure on the right has C4symmetry.Both volumes are displayed at a threshold of 2.6␴above the mean density.The volume was low-pass ﬁltered to a 22-Åresolution.The main features and dimensions of the complex are indicated.(B)Slabs taken through the C4-symmetrized 3-D volume from panel A,highlighting structural features at 3␴(red),3.5␴(orange),4␴(yellow),4.5␴(pale blue),and 5␴(dark blue)above the mean density.V OL .189,20073-D STRUCTURE OF THE PilG PILUS BIOGENESIS PROTEIN 6393at Penn State Univ on April 16, 2008 Downloaded fromsented here have provided the ﬁrst indication of the tertiary structure of a member of the GspF family.In order to correlate the 3-D reconstruction of PilG with its predicted topology in the membrane,we analyzed the PilG sequence for transmem-brane helices with a variety of prediction methods.Most meth-ods—HMMTOP (41),DAS (15),PHDhtm (32),and TMPred (22)—predicted four helices.However,TMHMM (35)and SOSUI (21)found only three,missing the third transmem-brane helix,which was detected only weakly by most methods,probably due to the relatively high number of charged residues within this portion of the sequence.The prediction of the number and location of transmembrane helices in membrane proteins is not completely reliable,although methods based purely on hydrophobicity scales are generally less accurate (7).In view of the preponderance of predictions in favor of four helices in PilG,rather than three,and considering the similar-ities between the roles of PilG and BfpE in pilus biogenesis,we propose that PilG is most likely to adopt a four-helix topology.This topological model would effectively divide PilG into two unequal halves:an N-terminal cytoplasmic domain of around 170residues and a C-terminal periplasmic domain of 100res-idues,separated by three transmembrane helices (Fig.6C,right).Such an arrangement would have an obvious correlation with the 3-D reconstruction,with the N-terminal domain align-ing with the bottom part of the structure,the ﬁn-shaped top of the structure corresponding to the C-terminal domain,and the narrow “waist”region comprising the central three transmem-brane helices (Fig.6C).This conclusion would suggest that PilG not only spans the bacterial inner membrane but has substantial segments of polypeptide chain exposed at both the cytoplasmic and periplasmic sides.We would expect this con-clusion to extend to other orthologs of PilG including,for example,PulF from Klebsiella pneumoniae (29)and OutF from Erwinia chrysanthemi (24).The overall size of the PilG particles is at the limits of the potential of single-particle 3-D reconstructions (17).Neverthe-less,the resulting 3-D reconstruction provided clear evidence for a tetrameric quaternary arrangement.ThisobservationFIG.6.Nanogold labeling of PilG and proposed orientation in the inner membrane.(A)Montage of raw multiple gold-labeled PilG complexes.Data were CTF corrected for presentation.Box size is 234Åby 234Å.(B)3-D reconstruction of the PilG-nanogold complex,ﬁltered to a 30-Åresolution.The PilG volume is shown by green netting contoured at 3␴above the mean density.Peaks associated with nanogold particles are shown in yellow at Ϫ4.1␴below the mean density.C4symmetry has been imposed.The volume was calculated using the PilG reconstruction determined previously (Fig.5)as a start model for the nanogold-labeled data set (n ϭ1,100)and processed as described previously (12).The volume is displayed following low-pass ﬁltering to 25Å.Scale bar ϭ50Å.(C)The left panel shows a composite ﬁgure,with the volume from the reconstruction of PilG alone displayed at 2.6␴(light green)and 5␴(dark green)above the mean density in green surface rendering with 50%opacity.Superimposed on this density are the peaks associated with the bound nanogold particles,shown at Ϫ5␴below the mean density in yellow surface rendering.Both volumes had C4symmetry imposed.The cartoon on the right shows the PilG topology,with transmembrane helices predicted using PHDhtm (32),matched approximately to the distribution of mass in the PilG 3-D structure on the left.6394COLLINS ET AL.J.B ACTERIOL .at Penn State Univ on April 16, 2008 Downloaded from。

可汗学院数学系列课程单课时教学实践研究

可汗学院数学系列课程单课时教学实践研究作者：赵安琪崔佳丽来源：《求知导刊》2017年第19期本文根据部分现有文献及对可汗学院课程的亲身体验，对其数学系列课程单课时内的具体教学实践进行初步探讨，并对比传统课堂的教学实践。

希望能够从中找到帮助我国教师改善目前传统课堂教学模式的方法。

一、研究方法本研究浏览了可汗学院中幼儿园数学、一年级数学、高中几何、早期数学专题的全部248个课时的视频，高中代数、算数专题的部分课时的视频；完整学习幼儿园数学、一年级数学及高中代数专题中44个课时的视频，并完成对应课后作业；尝试学习任务测试功能，完成一年级数学中10个测试。

以可汗学院具体一课时的教学视频为案例，重点分析其教学实践流程，并进行总结归纳，为我国传统课堂信息化改革提供实践依据。

此外，通过广泛研读研究可汗学院及微课有关文献，掌握与本研究相关的成果，为本文提供理论基础。

二、研究结果1.可汗学院数学系列课程的设置可汗学院网站右上方的菜单分为“年级”与“科目”两大类，是两种不同的课程划分方式。

“年级”中的数学课程按年级从幼儿园到高中进行划分，根据年级的增长，难度循序渐进。

学生掌握学科基本知识结构，在头脑中形成相应的知识体系或编码系统。

“科目”中的课程按专题被分为早期数学、算数、代数等。

2.单课时案例：“斜率在平面直角坐标系中的应用”本课程节选自可汗学院中代数专题，时长7分钟。

本课内容是在学习过基本代数、平面直接坐标系、线性方程等相关概念之后，引入斜率这一新概念。

本课从平面直接坐标中的方程式入手，涉及三角函数值域、线性规划问题、不等式问题、导数的几何意义等方面，应用范围广泛。

学生熟练掌握利用直线斜率来处理数学问题，将定式总结成规律，并能迁移到类似问题的解决中，对开拓思维、提高解题能力具有积极意义。

3.可汗学院教学实践的特点（1）课程以视频方式呈现，时间都比较短，控制在10分钟之内。

课程内容从基础知识逐步螺旋上升，由易到难，互相衔接，符合学生的数学学习特点。

(压轴题)高中数学选修1-2第三章《推理与证明》检测(包含答案解析)

一、选择题1．A 、B 两种品牌各三种车型2017年7月的销量环比(与2017年6月比较)增长率如下表： A 品牌车型 A 1A 2A 3环比增长率 -7.29% 10.47% 14.70%B 品牌车型 B 1 B 2 B 3环比增长率-8.49% -28.06% 13.25%根据此表中的数据，有如下关于7月份销量的四个结论:①A 1车型销量比B 1车型销量多； ②A 品牌三种车型总销量环比增长率可能大于14.70%； ③B 品牌三款车型总销量环比增长率可能为正；④A 品牌三种车型总销量环比增长率可能小于B 品牌三种车型总销量环比增长率．其中正确结论的个数是( ) A ．1B ．2C ．3D ．42．如图，第（1）个图案由1个点组成，第（2）个图案由3个点组成，第（3）个图案由7个点组成，第（4）个图案由13个点组成，第（5）个图案由21个点组成，⋅⋅⋅，依次类推，根据图案中点的排列规律，第50个图形由多少个点组成（）A ．2449B ．2451C ．2455D ．24583．观察下列各式：211=，22343++=，2345675++++=，2456789+107+++++=，，可以得出的一般结论是（）A ．()()()21232n n n n n ++++++-=B ．()()()21231n n n n n ++++++-=C ．()()()()2123221n n n n n ++++++-=- D ．()()()()2123121n n n n n ++++++-=-4．学生的语文、数学成绩均被评定为三个等级，依次为“优秀”“合格”“不合格”．若学生甲的语文、数学成绩都不低于学生乙，且其中至少有一门成绩高于乙，则称“学生甲比学生乙成绩好”．如果一组学生中没有哪位学生比另一位学生成绩好，并且不存在语文成绩相同、数学成绩也相同的两位学生，那么这组学生最多有( ) A ．2人B ．3人C ．4人D ．5人5．如图中的三角形图案称为谢宾斯基三角形.在四个三角形图案中，着色的小三角形的个数依次构成数列{}n a 的前4项，则{}n a 的通项公式可以为（）A ．21n a n =-B ．21nn a =- C ．3nn a =D ．13-=n n a6．三角形的面积为1()2S a b c r =++⋅，其中,,a b c 为三角形的边长，r 为三角形内切圆的半径，则利用类比推理，可得出四面体的体积为（）A ．13V abc = B ．13V Sh =C ．1()3V ab bc ca h =++，（h 为四面体的高） D ．()123413V S S S S r =+++，（1234,,,S S S S 分别为四面体的四个面的面积，r 为四面体内切球的半径）7．已知三个月球探测器α，β，γ共发回三张月球照片A ，B ，C ，每个探测器仅发回一张照片.甲说：照片A 是α发回的；乙说：β发回的照片不是A 就是B ；丙说：照片C 不是γ发回的，若甲、乙、丙三人中有且仅有一人说法正确，则发回照片B 的探测器是（） A ．αB ．βC ．γD ．以上都有可能8．我国的刺绣有着悠久的历史，如图，(1)(2)(3)(4)为刺绣最简单的四个图案，这些图案都是由小正方形构成，小正方形个数越多刺绣越漂亮．现按同样的规律刺绣(小正方形的摆放规律相同)，设第n 个图形包含()f n 个小正方形，则()f n 的表达式为（）A ．()21f n n =-B ．2()2f n n =C ．2()22f n n n=-D ．2()221f n n n =-+9．我国南宋数学家杨辉在所著的《详解九章算法》一书中用如图所示的三角形解释二项展开式的系数规律，去掉所有为1的项，依次构成2，3，3，4，6，4，5，10，10，5，6…，则此数列的前50项和为（）A ．2025B ．3052C ．3053D ．304910．某中学在高二下学期开设四门数学选修课，分别为《数学史选讲》.《球面上的几何》.《对称与群》.《矩阵与变换》．现有甲.乙.丙.丁四位同学从这四门选修课程中选修一门，且这四位同学选修的课程互不相同，下面关于他们选课的一些信息：①甲同学和丙同学均不选《球面上的几何》，也不选《对称与群》：②乙同学不选《对称与群》，也不选《数学史选讲》：③如果甲同学不选《数学史选讲》，那么丁同学就不选《对称与群》．若这些信息都是正确的，则丙同学选修的课程是（） A ．《数学史选讲》 B ．《球面上的几何》 C ．《对称与群》D ．《矩阵与变换》11．已知222233+=，333388+=，44441515+=，⋅⋅⋅，若66n nm m+=（m 、n 均为正实数），根据以上等式，可推测m 、n 的值，则m n +等于（）A ．40B ．41C ．42D ．4312．数列中，则，则A ．B ．C ．D ．二、填空题13．从22211,2343,345675,=++=++++=中，可猜想第n 个等式为__________.14．我国古代数学名著《九章算术》的论割圆术中有：“割之弥细，所失弥少，割之又割，以至于不可割，则与圆周盒体而无所失矣.”它体现了一种无限与有限的转化过程.比如在表达式11111+++中“…”既代表无限次重复，但原式却是个定值，它可以通过方程11x x+=求得152x +=，类似上述过程，则33++=__________．15．我国古代数学名著《九章算术》中割圆术有：“割之弥细，所失弥少，割之又割，以至于不可割，则与圆周合体而无所失矣．”其体现的是一种无限与有限的转化过程，比如在222+++222?··中“”即代表无限次重复，但原式却是个定值x . 这可以通过方程2x x +=确定x ＝2，则11111+=++_______.16．设等差数列{}n a 的前n 项和为n S ，则4S ，84S S -，128S S -，1612S S -成等差数列．类比以上结论有：设等比数列{}n b 的前n 项积为n T ，则4T ，____________，1612T T 成等比数列． 17．在平面上，若两个正三角形的边长的比为1：2，则它们的面积比为1：4，类似地，在空间内，若两个正四面体的棱长的比为1：2，则它们的体积比为____18．如图，在圆内画1条线段，将圆分成2部分；画2条相交线段，将圆分割成4部分；画3条线段，将圆最多分割成7部分；画4条线段，将圆最多分割成11部分．则在圆内画n 条线段，将圆最多分割成______部分．19．已知数列：21，12，31，22，13，41，32，23，14，…根据它的前9项的规律，这个数列的第30项为__________．20．甲、乙、丙三位同学被问到是参加了学校组织的A 、B 、C 三个活动兴趣小组时，甲说：我参加的兴趣小组比乙多，但没参加过A 兴趣小组；乙说：我没参加过B 兴趣小组；丙说：我们三人参加了同一兴趣小组；由此可判断乙参加的兴趣小组为__________．三、解答题21．观察下列各等式：tan10tan 20tan 20tan60tan60tan101︒︒+︒︒+︒︒=tan 20tan30tan30tan 40tan 40tan 201︒︒+︒︒+︒︒= tan33tan 44tan 44tan13tan33tan131︒︒+︒︒+︒︒= （1）尝试再写出一个相同规律的式子；（2）写出能反映以上式子一般规律的恒等式；（3）并对你写出的（2）恒等式进行证明．22．某同学在研究相邻三个整数的算术平方根之间的关系时，发现以下三个式子均是正确的:①1322+<;②2423+<;③3524+<.（1）已知2∈(1.41，1.42)， 3∈(1.73，1.74)， 5∈(2.23，2.24)，请从以上三个式子中任选一个，结合此范围验证其正确性(注意不能近似计算．．．．．．．．)；（2）请将此规律推广至一般情形，并加以证明.23．数学研究性学习是高中学生数学学习的一个有机组成部分，是在基础性、拓展性课程学习的基础上，进一步鼓励学生运用所学知识解决数学的和现实的问题的一种有意义的主动学习，是以学生动手动脑主动探索实践和相互交流为主要学习方式的学习研究活动．某同学就在一次数学研究性学习中发现，以下五个式子的值都等于同一个常数． ①22sin 13171317o o o o cos sin cos +-； ②22sin 15151515o o o o cos sin cos +-； ③22sin 18121812o o o o cos sin cos +-； ④22sin 300300o o o o cos sin cos +-； ⑤()()22sin10401040ooo o cos sin cos -+--．（1）试从上述五个式子中选择一个，求出这个常数；（2）根据（1）的计算结果，归纳出一个三角恒等式；（3）利用所学知识证明这个结论． 24．用综合法或分析法证明：（1）如果 ,0a b >，则 lg lg lg22a b a b++≥；（2）610232+>+．25．将正整数排成如图的三角形数阵，记第n 行的n 个数之和为n a .（1）设*13521()n n S a a a a n N -=+++⋅⋅⋅+∈，计算2S ，3S ，4S 的值，并猜想n S 的表达式；（2）用数学归纳法证明（1）的猜想. 26．已知数列 {}n a 满足：112a =，()()11312111n n n n a a a a ++++=--，()101n n a a n +<≥；数列{}n b 满足：()2211n n n b a a n +=-≥．（1）求数列 {}n a ，{}n b 的通项公式；（2）证明：数列 {}n b 中的任意三项不可能成等差数列．【参考答案】***试卷处理标记，请不要删除一、选择题 1．B 解析：B 【分析】根据表中数据，对关于7月份销量的四个结论，分析正误即可．【详解】解：根据表中数据，对关于7月份销量的四个结论：对于①，A 1车型销量增长率比B 1车型销量增长率高，但销量不一定多，①错误；对于②，A 品牌三种车型中增长率最高为14.70%，所以总销量环比增长率不可能大于14.70%，②错误；对于③，B 品牌三款车型中有销量增长率为13.25%，所以它的总销量环比增长率也可能为正，③正确；对于④，由题意知A 品牌三种车型总销量环比增长率，也可能小于B 品牌三种车型总销量环比增长率，④正确；综上所述，其中正确的结论序号是③④．故选B ．【点睛】本题考查了合情推理与命题真假的判断，也考查了销售量与增长率的应用问题，是基础题．2．B解析：B 【分析】设第()n 个图案的点的个数为n a ，推测得到12(1)n n a a n --=-，利用1n -个式子相加，由等差数列的求和公式，即可求解. 【详解】设第()n 个图案的点的个数为n a ，由题意可得123451,3,7,13,21,a a a a a =====，可得213243542,4,6,8,a a a a a a a a -=-=-=-=，由此推测12(1)n n a a n --=-，则()()()()21324312462(1)n n a a a a a a a a n --+-+-+-=++++-，化简可得(1)(222)1(1)2n n n a n n -+--==-，所以(1)1n a n n =-+所以5050(501)12451a =⨯-+=. 故选：B. 【点睛】本题主要考查了归纳推理的应用，其中解答中构造数列并得出的数列的递推关系式，结合等差数列的求和公式求解是解答的关键，着重考查推理与运算能力.3．C解析：C 【解析】 1=12， 2+3+4=32， 3+4+5+6+7=52， 4+5+6+7+8+9+10=72， …，由上述式子可以归纳：左边每一个式子均有2n-1项，且第一项为n ，则最后一项为3n-2 右边均为2n-1的平方故选C点睛：归纳推理的一般步骤是：（1）通过观察个别情况发现某些相同性质；（2）从已知的相同性质中推出一个明确表达的一般性命题（猜想）．4．B解析：B 【解析】试题分析：用,,A B C 分别表示优秀、及格和不及格，显然语文成绩得A 的学生最多只有1个，语文成绩得B 也最多只有1个，得C 的最多只有1个，因此人数最多只有3人，显然(),(),()AC BB CA 满足条件，故选B ．考点：合情推理的应用．5．D解析：D 【分析】着色的小三角形个数构成数列{}n a 的前4项，分别得出，即可得出{}n a 的通项公式．【详解】着色的小三角形个数构成数列{}n a 的前4项，分别为：11a =，23a =，23333a =⨯=，234333a =⨯=，因此{}n a 的通项公式可以是：13-=n n a ．故选：D ．【点睛】本题考查了等比数列的通项公式，考查了观察分析猜想归纳推理能力与计算能力，属于中6．D解析：D 【分析】设四面体的内切球的球心为O ，则球心O 到四个面的距离都是r ，根据体积公式得到答案. 【详解】设四面体的内切球的球心为O ，则球心O 到四个面的距离都是r ，将O 与四顶点连起来，可得四面体的体积等于以O 为顶点，分别以四个面为底面的4个三棱锥体积的和， ∴V 13=（S 1+S 2+S 3+S 4）r . 故选：D ．【点睛】本题考查了类比推理，意在考查学生的空间想象能力和推断能力.7．A解析：A 【分析】结合题中条件，分别讨论甲对、乙对或丙对的情况，即可得出结果. 【详解】如果甲对，则β发回的照片是C ，故丙也对，不符合条件，故甲错误；如果乙对，则丙错误，故照片C 是γ发回的.得到照片A 是由β发回，照片B 是由α发回.符合逻辑，故照片B 是由α发回；如果丙对，则照片C 是由β发出，甲错误，可以推出α发出照片B ，γ发出照片A ，故照片B 是由α发出. 故选A 【点睛】本题主要考查推理分析，根据合情推理的思想，进行分析即可，属于常考题型.8．D解析：D 【分析】先分别观察给出正方体的个数为：1，14+，148++，⋯⋯，总结一般性的规律，将一般性的数列转化为特殊的数列再求解．【详解】解：根据前面四个发现规律： ()()2141f f -=⨯， ()()3242f f -=⨯，()()4343f f -=⨯，⋯⋯，()(1)4(1)f n f n n --=-，累加得： ()()2(1)14[12(1)]42(1)222n n f n f n n n n n --=⨯++⋯⋯+-=⨯=-=-， ()11f =2()221f n n n ∴=-+，故选：D ．本题主要考查了归纳推理，属于中档题．9．D解析：D 【分析】去除所有为1的项后,根据图可知前n 行共有(1)2n n +个数,从而得到前10行共55个数,然后用前10行的和减去后五项,即可得到此数列的前50项和. 【详解】解:去除所有为1的项后,由图可知前n 行共有(1)2n n +个数, 当n =10时,10(101)552⨯+=,即前10行共有55个数. 因为第n -1行的和为12122n n n n n C C C -+++=-, 所以前10行的和为231112(22)(22)(22)2244072-+-++-=-=.因为第10行最后5个数为1011C ,911C ,811C ,711C ,611C ,所以此数列的前50项的和为4072-11-55-165-330-462=3049. 故选:D . 【点睛】本题考查了归纳推理和等比数列前n 项和的求法,考查了推理能力,属难题.10．D解析：D 【分析】列举出所有选择可能，然后根据三个信息，确定正确的选项. 【详解】4个同学，选4门课，各选一门且不重复的方法共24种，如下：【点睛】本小题主要考查分析与推理，考查列举法，属于基础题. 11．B解析：B 【分析】根据前面几个等式归纳出一个关于k 的等式，再令6k =可得出m 和n 的值，由此可计算出m n +的值. 【详解】2222222232121+=+=--，2233333383131+=+=--，22444444154141+=+=--，由上可归纳出()222,11k kk k k k N k k *+=≥∈--，当6k =时，则有2266666161+=--，26135m ∴=-=，6n =，因此，41m n +=，故选B. 【点睛】本题考查归纳推理，解题时要根据前几个等式或不等式的结构进行归纳，考查推理能力，属于中等题.12．C解析：C 【解析】【分析】分别计算、、归纳出的表达式，然后令可得出的值。

可汗学院数学题目-33Datacollectionandconclusions

(完整版)数据挖掘_概念和技术[第三版]部分习题答案解析

可汗学院新SAT语法真题下载

数学之旅测试地的题目

可汗学院数学题目-41Complexnumbers[1]

最新北师大版高中数学高中数学选修2-3第三章《统计案例》检测题(包含答案解析)(1)

数据挖掘_国防科技大学中国大学mooc课后章节答案期末考试题库2023年

Summa Scalable universal matrix multiplication algorithm

新北师大版高中数学高中数学选修2-3第三章《统计案例》测试卷(包含答案解析)(5)

动画演示数据结构与算法

An imaging K-band survey - I The catalogue, star and galaxy counts

application to privacy

可汗学院数学题目-31Datainferences

可汗学院新SAT语法真题下载

Purification and Three-Dimensional Electron Microscopy Structure of the Neisseria meningiti

可汗学院数学系列课程单课时教学实践研究

(压轴题)高中数学选修1-2第三章《推理与证明》检测(包含答案解析)