(完整word版)soa复习题
(完整word版)软件缺陷跟踪复习题
(完整word版)软件缺陷跟踪复习题一、选择：1．导致软件缺陷的最主要原因是（）。
A．软件系统越来越复杂，开发人员不可能精通所有的技术B．软件的需求说明书不规范C．硬件配置不对、缺乏，或处理器缺陷导致算术精度丢D．软件设置不对、缺乏，或操作系统错误导致无法释放资源、工具软件的错误，编译器的错误等2．软件的质量根本上由( ）决定。
A．编程技术B．测试技术C．过程质量D．开发工具3．下面关于软件缺陷的定义正确的是( ）：A．软件缺陷是计算机软件或程序中存在的某种破坏正常运行能力的问题、错误，或者隐藏的功能缺陷B．软件缺陷指软件产品（包括文档、数据、程序等）中存在的所有不希望或不可接受的偏差，这些偏差会导致软件的运行与预期不同，从而在某种程度上不能满足用户的需求C．从产品内部看，缺陷是软件产品开发或维护过程中存在的错误、毛病等各种问题；从产品外部看，缺陷是系统所需要实现的某种功能的失效或违背D．以上都对4．( )指软件缺陷对软件质量的破坏程度，即此缺陷的存在将对软件的功能和性能产生怎样的影响.A。
缺陷优先级 B. 缺陷严重程度C. 缺陷发生频率D. 缺陷类别5．下面关于软件缺陷管理的说法错误的是（）：A. 软件缺陷管理（Defect Management)是指对软件开发过程中的缺陷发现、确认、定位、修复、评审、关闭等一系列行为进行跟踪管理的过程，也就是在软件生命周期中获取、管理、沟通任何变更请求的过程，是软件研发过程中的一项过程管理B. 软件缺陷跟踪管理在现代软件开发中已经占据了很重要的位置，和软件开发的项目管理、需求、设计、开发、测试均严密相关C. 软件缺陷管理是在软件生命周期中为确保缺陷被跟踪和管理所进行的活动D。
软件开发过程中，只需要在测试阶段进行缺陷管理6．（ )是软件缺陷管理的核心,也是软件缺陷预防的核心任务。
A. 缺陷报告B。
缺陷分析 C. 缺陷库 D. 缺陷修复7．软件缺陷发现手段有多种。
(完整word版)软工复习题答案.doc
5位数字串（6）
位数字串（
18）
（2）设计测试用例见表A.2.
表A2测试用例表
测试数据
测试范围
期望结果
（1）（）5323456
等价类（1）（5）(6)
有效
（2）（01）8093216
等价类（2）（5）(6)
有效
（3）（0871）3356789
等价类（4）（5）(6)
有效
（4）（067）5065432
A．简化.压缩的
C．彻底的
A）需求分析
B．详细的
D．深入的
.设计过程。
8．下面说法正确的是（C
）。
A．经过测试没有发现错误说明程序正确
B．测试的目标是为了证明程序没有错误
C．成功的测试是发现了迄今尚未发现的错误的测试
D．成功的测试收是没有发现错误的测试
9．面向对象的主要特征除对象唯一性、封装、继承外，还有（
A
）。
A．多态性B．完整性
C．可移植性D．兼容性
四、简述题（第1、2、4答案不确定，需自己去找找）
1：简述需求分析的过程。
回答一：
（1）问题识别
（2）分析与综合
（3）编制需求分析阶段的文档
（4）需求验证
建立软件原型）3（分析需求可行性）4（确认需求优先级）5（建立需求分析模型）
6（创建数据字典）回答二：
（3）特殊需求
说明此用例的特殊要求
（4）前置条件
说明此用例开始执行的前提条件，如参与者登陆成功等
（5）后置条件
说明此用例执行结束后，结果应传给什么参与者
8：简述如何识别分析类及应当注意的问题？
分析类包括实体类、边界类和控制类
软件体系结构复习题
软件体系结构复习题15个中选择10个回答60分名词解释：SCA, SOA, OSOA, OASIS, CORBA, DCOM, URI, URL, WSDL, Web Service, OGSA, GloBus, x/OPEN, IIOP, GIOP问答题：1.网格与P2P体系结构的联系与区别2.OGSA的五层体系结构与TCP/IP五层体系结构的联系与区别3.三层C/S结构的主要特点4.SCA与SOA的联系5.解释流媒体信道模型中提高服务用户数的基本方法其它部分CORBA，流媒体:信道模型，方法，用体系结构图加以解释体系结构中超级节点形成的对等架构采用SCA模型，从XML找出体系结构图，及构件关系答案：名词解释：SCA:服务构件架构(Service Component Architecture) 致力于为使用广泛的编程语言来构造服务构件提供一种编程模型，并且也为把这些服务构件组装为一个业务上的解决方案提供了一种模型，这种组装的活动正是采用面向服务的架构(service-oriented architecture)来搭建应用系统的核心。
SOA:面向服务的体系结构（Service-oriented architecture）是构造分布式系统的应用程序的方法。
它将应用程序功能作为服务发送给最终用户或者其他服务。
它采用开放标准、与软件资源进行交互并采用表示的标准方式。
OSOA：OSOA（Open Service Oriented Architecture）协作组织目前正在起草一系列的规范，并以免版税的许可方式提供给业界使用。
这个站点集中包括了已经完成的规范和那些还处在早期的草案，我们希望能够得到来自于社区的反馈。
CORBA: CORBA（Common Object Request Broker Architecture,公共对象请求代理体系结构，通用对象请求代理体系结构）是由OMG组织制订的一种标准的面向对象应用程序体系规范。
OA复习题
知识点题型题目内容word判断题页码文字的大小根本不可以更改。
word单选题两节之间的分节符被删除后，以下哪个说法正确？word单选题如果需要对插入的图片在调整时最方便，那么图片与文字的环绕方式应该选择___word多选题以下关于模板的理解正确的有word单选题选定文本中一词的技巧方法是word单选题在一篇100页的文档中，如何快速的将光标移到第80页word单选题[格式]工具栏上的段落对齐方式按钮分别为word单选题如何快速的在文档的每一页都加上大小、位置均相同的图片word单选题以下有关WORD 2003中"项目符号"的说法错误的是word单选题要知道某中文文字的英文含义应word单选题若要改变打印时的纸张大小, 正确的是word单选题设置标题与正文之间距离的正规方法为word单选题如何在一篇内容很多的文档中，将重复过多次的“中国”快速的加上突出显示格式word单选题关于格式刷说法正确的是word单选题在[打印]对话框中"页面范围"选项卡下的"当前页"是专指word单选题要想将打开的多个文件同时关闭，可以执行下列哪项操作word单选题有关WORD 2002的[工具]菜单的[字数统计]命令的说法错误的是word单选题关于页面边框的说法正确的是word单选题删除整个表格中的某一个单元格，并使其右侧单元格左移，应先将插入点置于该格，然后word单选题要完全清除文本中的底纹效果，应单击[格式]菜单中[边框与底纹]，再单击[底纹]选项页，然后选择word单选题"页眉与页脚"处在那个菜单下word单选题一篇100页的文档，下列打印页码范围正确的是word单选题关于用[插入表格]命令，下面说法错误的是word单选题如何选定整个表格word单选题对于[拆分表格]，正确的说法是word单选题使用什么命令可使本来放在下面的图，移置于上面word单选题当插入时间域后出现的是99-10-17，此时要想看到域结果应该采用哪一个命令word单选题Word中文本下面的红色和绿色下划波浪线分别代表什么含义word单选题以下关于表格自动套用格式的说法中，正确的是word多选题有一篇文档，编完之后想检查错误，可发现自己看不清，有一点小，以下说法中哪几个可以改善这个效果？word多选题以下关于WORD中模板的描述正确的是word多选题下列说法中正确的有：word多选题下述设置项目符号的操作正确的有：word多选题以下关于表格中文本格式的说法，正确的是：word多选题在项目编号中，下面哪些说法正确？word多选题以下关于自选图形的说法正确的是？word多选题以下选项中，在Word2003中可以新建的文件类型是：word多选题下面关于对页眉页脚操作的说法中正确的有：word多选题下列关于大纲视图的说法中正确的有：word多选题哪几种对齐方式是Word格式工具栏中所列的对齐方式？word多选题关于Word2003的文本框，哪些说法是正确的？word多选题修改页眉页脚可以通过哪些途径？word多选题打印时如果指定的纸型小于实际的纸型，将会：word多选题交叉引用所能够引用的类型包括：word判断题在Word 2003中能够打开Excel文件，并且能够以Excel形式保存它word判断题在设置图片格式对话框中，选中“锁定纵横比”复选框，然后拖动图片四个角上选择柄可使用图片的纵横比不变。
(完整word版)期末复习题
1 processes P1, P2, and P3 ，Define semaphores, and synchronize the execution of P1, P2, and P3 by wait() and signal() on these semaphores.2 Draw three Gantt charts，What is the turnaround time of each process for SJF, and RRWhat are the average waiting time and the average turnaround time3 Consider the following page-reference string:(1) LRU page replacement(2) Optimal page replacement(3) FIFO replacement algorithmpage fault times，Page faults rate4 (1) Is the system in a safe or unsafe state? Why?(2) If P i request resource of (0, 1, 0, 0), can resources be allocated to it? Why?In a demand paging system, the page size is 1k bytes, and the page table is as follows(assuming use decimal values), translate a logical address 10 into its corresponding physical address, and why?A．8202 B．4106 C．2058D．1034In a demand paging system, the page size is 1024 bytes, and the page table is as follows, would the following virtual addresses (assuming use decimal values) result in a page fault? And why?(1)2500(2)5100Page table(1) 2500=2*1024+452 page 2 is valid, so no page fault(2) 5100=4*1024+1004 page 4 is invalid, a page fault occursConsider a simple paging system with a page table containing 1024 entries of 14 bits (including one valid/invalid bit) each, and a page size of 1024 bytes (5 points)(a) how many bits are in the logical address?(b) how many bits are in the physical address?(c) what is the size of the logical address space?(d) how many bits in the logical address specify the page number?(e) how many bits in the physical address specify the offset within the frame?1) 202) 233) 220B4) 105) 10(a)There are 10+10=20 bits in the logical address(b)There are 13+10=23 bits in the physical address(c)The size of the logical address space is 220 bytes(d)There are 10 bits in the logical address specifying the page number(e)There are 10 bits in the physical address specifying the offset within theframeConsider a system using segmentation with paging management scheme, whose physical memory is of 235 bytes. The logical address space consists of up to 8 segments. Each segment can be up to 213 pages, and page size of 512 bytes,(1) How many bits in the logical address specify the page number?(2) How many bits are there in the entire logical address?(3) What is the size of a frame?(4) How many bits in the physical address specify the frame number?(5) How many bits in the physical address specify the frame offset?(6) How many entries are there in the page table for each segmentation, i.e. how longis the page table for each segmentation?(1)Each segment consists of up to 213 pages, so log2(213) = 13 bits in the logicaladdress specify the page number(2)The entire logical address consists of segmentation and offset in segmentation,and offset in segmentation is divided into page number and page offset, so entire logical address haslog2 (8) + log2(213) + log2 (512) = 3 + 13 +9 =25 bits(3)the size of a frame is equal to that of a page, i.e., 512 bytes(4)the total size of physical memory is 235 bytes, and each frame is 29 = 512 bytes,so there are 26 = (35 –9) bits in the physical address specify the frame number(5)the number of bits in the physical address specifying the frame offset dependson the size of the frame, so log2 (29) = 9 bits in the physical address specify the frame offset(6)each entry in the page table corresponds to a page in the segment, eachsegment can be up to 213 pages, so there are 213 entries in the page table for each segment.A file system uses 256-byte physical blocks. Each file has a directory entry givingthe file name, location of the first block, length of file, and last block position.Assuming the last physical block read is 100, block 100 and the directory entry are already in main memory.For the following two file allocation algorithms, how many physical blocks must be read to access the specific block 600 (including the reading of block 600 itself), and why?(1)contiguous allocation(2) linked allocation(1) 1Contiguous allocation supports direct access on arbitrtary physical blocks in a file. To access physical block 600, the bolck 600 can be directory read into memory.(2) 500Linked allocation is used only for sequential-access files, file system can only read blocks one by one, directed by file pointers.To find the phyical block 600 in the file, file system locates on the 101th block following the 100th block just read, and reads block 101, block 102, …, until block 600.Consider a file system in which a directory entry can store up to 32 disk block addresses. For the files that is not larger than 32 blocks, the 32 addresses in the entry serves as the file’s index table. For the files that is larger than 32 blocks, each of the 32 addresses points to an indirect block that in turn points to 512 file blocks on the disk, and the size of a block is 512-bytes. What is the largest size of a file?32 x 512 =16384 file blocks16384 x 512 = 8388608 bytesor25 x 29 x 29 = 223 bytesIn the file system on a disk with physical block sizes of 512 bytes, a file is made up of 128-byte logical records, and each logical record cannot be separately stored in two different blocks. The disk space of the file is organized on the basis of indexed allocation, and a block address is stored in 4 bytes. Suppose that 2-level index blocks be used to manage the data blocks of the file, answer the following questions:1) What is the largest size of the file?2) Given 2000, the number of a logical record in the file, how to find out the physical address of the record 2000 in accordance with the 2-level index blocks.1)512/4=128128*128*512=128*64 KB=8192KBor =223 bytes = 8 MBor = 83886802)512/128=42000/4=500in 3 block in the first –level index blockin 116 block in the second-level index blockA file is made up of 128-byte fix-sized logical records and stored on the disk in the unit of the block that is of 1024 bytes. The size of the file is 10240 bytes. Physical I/O operations transfer data on the disk into an OS buffer in main memory, in terms of 1024-byte block. If a process issues read requests to read the file’s recor ds in the sequential access manner, what is the percentage of the read requests that will result in I/O operations?1024/128=8 record10240/8=80 record10240/1024=10 block10/80=1/8=0.125or 12.5%Consider a paging system with the page table stored in memory.a. If a memory reference takes 300 time unit, how long does it take to access an instruction or data in a page that has been paged into memory?b. If we add TLB (translation look-aside buffers), and 80 percent of all page-table entries can be found in the TLB, what is the effective memory access time? (Assume that finding a page-table entry in the associative registers takes 20 time unit, if the entry is there.)a.300 x 2=600b.0.8 x (300+20) + 0.2 x (600+20)=256+124=380。
SOA 2010 Exam P 官方样题
SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETYEXAM P PROBABILITYP SAMPLE EXAM QUESTIONSCopyright 2009 by the Society of Actuaries and the Casualty Actuarial Society Some of the questions in this study note are taken from past SOA/CAS examinations.U.S.A.PRINTEDIN1. A survey of a group’s viewing habits over the last year revealed the followinginformation:(i) 28% watched gymnastics(ii) 29% watched baseball(iii) 19% watched soccer(iv) 14% watched gymnastics and baseball(v) 12% watched baseball and soccer(vi) 10% watched gymnastics and soccer(vii) 8% watched all three sports.Calculate the percentage of the group that watched none of the three sportsduring the last year.(A) 24(B)36(C)41(D)52(E)602.The probability that a visit to a primary care physician’s (PCP) office results in neitherlab work nor referral to a specialist is 35% . Of those coming to a PCP’s office, 30% are referred to specialists and 40% require lab work.Determine the probability that a visit to a PCP’s office results in both lab work andreferral to a specialist.0.05(A)(B)0.120.18(C)0.25(D)(E) 0.353.You are given P[A∪B] = 0.7 and P[A∪B′] = 0.9 .Determine P[A] .(A) 0.20.3(B)0.4(C)0.6(D)(E) 0.84.An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16 red balls and anunknown number of blue balls. A single ball is drawn from each urn. The probabilitythat both balls are the same color is 0.44 .Calculate the number of blue balls in the second urn.(A) 4(B)2024(C)44(D)(E) 645.An auto insurance company has 10,000 policyholders. Each policyholder is classified as(i) young or old;(ii) male or female; and(iii) married or single.Of these policyholders, 3000 are young, 4600 are male, and 7000 are married. Thepolicyholders can also be classified as 1320 young males, 3010 married males, and 1400 young married persons. Finally, 600 of the policyholders are young married males.How many of the company’s policyholders are young, female, and single?(A) 280(B) 423(C) 486880(D)(E) 8966. A public health researcher examines the medical records of a group of 937 men who diedin 1999 and discovers that 210 of the men died from causes related to heart disease.Moreover, 312 of the 937 men had at least one parent who suffered from heart disease,and, of these 312 men, 102 died from causes related to heart disease.Determine the probability that a man randomly selected from this group died of causesrelated to heart disease, given that neither of his parents suffered from heart disease.(A) 0.1150.173(B)0.224(C)0.327(D)(E) 0.5147.An insurance company estimates that 40% of policyholders who have only an auto policywill renew next year and 60% of policyholders who have only a homeowners policy will renew next year. The company estimates that 80% of policyholders who have both anauto and a homeowners policy will renew at least one of those policies next year.Company records show that 65% of policyholders have an auto policy, 50% ofpolicyholders have a homeowners policy, and 15% of policyholders have both anauto and a homeowners policy.Using the company’s estimates, calculate the percentage of policyholders that willrenew at least one policy next year.(A)2029(B)41(C)(D)53(E) 708.Among a large group of patients recovering from shoulder injuries, it is found that 22%visit both a physical therapist and a chiropractor, whereas 12% visit neither of these.The probability that a patient visits a chiropractor exceeds by 0.14 the probability thata patient visits a physical therapist.Determine the probability that a randomly chosen member of this group visits aphysical therapist.0.26(A)(B)0.380.40(C)0.48(D)(E) 0.629.An insurance company examines its pool of auto insurance customers and gathers thefollowing information:(i) All customers insure at least one car.(ii) 70% of the customers insure more than one car.(iii) 20% of the customers insure a sports car.(iv) Of those customers who insure more than one car, 15% insure acar.sportsCalculate the probability that a randomly selected customer insures exactly one car and that car is not a sports car.0.13(A)0.21(B)0.24(C)0.25(D)(E) 0.3010.An insurance company examines its pool of auto insurance customers and gathers thefollowing information:(i) All customers insure at least one car.(ii) 64% of the customers insure more than one car.(iii) 20% of the customers insure a sports car.(iv) Of those customers who insure more than one car, 15% insure a sports car.What is the probability that a randomly selected customer insures exactly one car, andthat car is not a sports car?0.16(A)0.19(B)0.26(C)0.29(D)0.31(E)11.An actuary studying the insurance preferences of automobile owners makes the followingconclusions:(i) An automobile owner is twice as likely to purchase collision coverage asdisability coverage.(ii) The event that an automobile owner purchases collision coverage isindependent of the event that he or she purchases disability coverage.thatautomobile owner purchases both collisionan(iii)Theprobabilityand disability coverages is 0.15 .What is the probability that an automobile owner purchases neither collision nordisability coverage?0.18(A)(B) 0.33(C) 0.480.67(D)(E) 0.8212. A doctor is studying the relationship between blood pressure and heartbeat abnormalitiesin her patients. She tests a random sample of her patients and notes their blood pressures (high, low, or normal) and their heartbeats (regular or irregular). She finds that:(i) 14% have high blood pressure.(ii) 22% have low blood pressure.(iii) 15% have an irregular heartbeat.(iv) Of those with an irregular heartbeat,one-third have high blood pressure.(v) Of those with normal blood pressure,one-eighth have an irregular heartbeat.What portion of the patients selected have a regular heartbeat and low blood pressure?(A) 2%(B) 5%(C) 8%(D) 9%(E) 20%13.An actuary is studying the prevalence of three health risk factors, denoted by A, B, and C,within a population of women. For each of the three factors, the probability is 0.1 thata woman in the population has only this risk factor (and no others). For any two of thethree factors, the probability is 0.12 that she has exactly these two risk factors (but notthe other). The probability that a woman has all three risk factors, given that she has Aand B, is 1.3What is the probability that a woman has none of the three risk factors, given that she does not have risk factor A?(A) 0.280(B) 0.311(C) 0.467(D) 0.484(E) 0.70014. In modeling the number of claims filed by an individual under an automobile policyduring a three-year period, an actuary makes the simplifying assumption that for all integers n ≥ 0, p p n n +=11, where p n represents the probability that the policyholder files n claims during the period.Under this assumption, what is the probability that a policyholder files more than oneclaim during the period?(A) 0.04(B) 0.16(C) 0.20(D) 0.80(E) 0.9615.An insurer offers a health plan to the employees of a large company. As part of thisplan, the individual employees may choose exactly two of the supplementary coverages A, B, and C, or they may choose no supplementary coverage. The proportions of thecompany’s employees that choose coverages A, B, and C are 115,,and ,4312 respectively.Determine the probability that a randomly chosen employee will choose nosupplementary coverage.(A) 0(B) 47144(C) 12(D) 97144(E) 7916. An insurance company determines that N , the number of claims received in a week, is arandom variable with P[N = n ] = 11,2n + where 0n ≥. The company also determines that the number of claims received in a given week is independent of the number of claims received in any other week.Determine the probability that exactly seven claims will be received during a giventwo-week period. (A) 1256(B) 1128(C) 7512(D) 164(E) 13217.An insurance company pays hospital claims. The number of claims that includeemergency room or operating room charges is 85% of the total number of claims.The number of claims that do not include emergency room charges is 25% of the total number of claims. The occurrence of emergency room charges is independent of theoccurrence of operating room charges on hospital claims.Calculate the probability that a claim submitted to the insurance company includesoperating room charges.0.10(A)0.20(B)0.25(C)0.40(D)(E) 0.8018.Two instruments are used to measure the height, h, of a tower. The error made by theless accurate instrument is normally distributed with mean 0 and standard deviation0.0056h . The error made by the more accurate instrument is normally distributed withmean 0 and standard deviation 0.0044h .Assuming the two measurements are independent random variables, what is theprobability that their average value is within 0.005h of the height of the tower?(A)0.38(B)0.47(C)0.68(D)0.84(E) 0.9019.An auto insurance company insures drivers of all ages. An actuary compiled thefollowing statistics on the company’s insured drivers:Age of Driver Probabilityof AccidentPortion of Company’sInsured Drivers16-20 21-30 31-65 66-99 0.060.030.020.040.080.150.490.28A randomly selected driver that the company insures has an accident. Calculate the probability that the driver was age 16-20.(A)0.13(B)0.16(C)0.19(D)0.23(E) 0.4020.An insurance company issues life insurance policies in three separate categories:standard, preferred, and ultra-preferred. Of the company’s policyholders, 50% arestandard, 40% are preferred, and 10% are ultra-preferred. Each standard policyholder has probability 0.010 of dying in the next year, each preferred policyholder hasprobability 0.005 of dying in the next year, and each ultra-preferred policyholderhas probability 0.001 of dying in the next year.A policyholder dies in the next year.What is the probability that the deceased policyholder was ultra-preferred?0.0001(A)0.0010(B)0.0071(C)0.0141(D)(E) 0.281721.Upon arrival at a hospital’s emergency room, patients are categorized according to theircondition as critical, serious, or stable. In the past year:(i) 10% of the emergency room patients were critical;(ii) 30% of the emergency room patients were serious;(iii) the rest of the emergency room patients were stable;critical patients died;the(iv)of40%(vi) 10% of the serious patients died; and(vii) 1% of the stable patients died.Given that a patient survived, what is the probability that the patient was categorized as serious upon arrival?(A) 0.06(B) 0.29(C) 0.30(D) 0.39(E) 0.6422. A health study tracked a group of persons for five years. At the beginning of the study,20% were classified as heavy smokers, 30% as light smokers, and 50% as nonsmokers.Results of the study showed that light smokers were twice as likely as nonsmokersto die during the five-year study, but only half as likely as heavy smokers.A randomly selected participant from the study died over the five-year period.Calculate the probability that the participant was a heavy smoker.(A)0.20(B)0.25(C)0.35(D)0.42(E) 0.5723.An actuary studied the likelihood that different types of drivers would be involved in atleast one collision during any one-year period. The results of the study are presentedbelow.Type of driver Percentage ofall driversProbabilityof at least onecollisionTeen Young adult Midlife Senior8%16%45%31%0.150.080.040.05Total 100%Given that a driver has been involved in at least one collision in the past year, what is the probability that the driver is a young adult driver?(A)0.06(B)0.16(C)0.19(D)0.22(E) 0.2524. The number of injury claims per month is modeled by a random variable N withP[N = n ] =1( + 1)( + 2)n n , where 0n ≥. Determine the probability of at least one claim during a particular month, giventhat there have been at most four claims during that month.(A) 13(B) 25(C) 12(D) 35(E) 5625. A blood test indicates the presence of a particular disease 95% of the time when thedisease is actually present. The same test indicates the presence of the disease 0.5% of the time when the disease is not present. One percent of the population actually has the disease.Calculate the probability that a person has the disease given that the test indicates the presence of the disease.(A)0.3240.657(B)0.945(C)0.950(D)(E) 0.99526.The probability that a randomly chosen male has a circulation problem is 0.25 . Maleswho have a circulation problem are twice as likely to be smokers as those who do nothave a circulation problem.What is the conditional probability that a male has a circulation problem, given that he isa smoker?(A) 14(B) 13(C) 25(D) 12(E) 2327. A study of automobile accidents produced the following data:Model year Proportion ofall vehiclesProbability ofinvolvementin an accident1997 0.16 0.051998 0.18 0.021999 0.20 0.03Other 0.46 0.04An automobile from one of the model years 1997, 1998, and 1999 was involvedin an accident.Determine the probability that the model year of this automobile is 1997 .(A)0.22(B)0.30(C)0.33(D)0.45(E) 0.5028. A hospital receives 1/5 of its flu vaccine shipments from Company X and the remainderof its shipments from other companies. Each shipment contains a very large number of vaccine vials.For Company X’s shipments, 10% of the vials are ineffective. For every other company, 2% of the vials are ineffective. The hospital tests 30 randomly selected vials from ashipment and finds that one vial is ineffective.What is the probability that this shipment came from Company X?0.10(A)0.14(B)0.37(C)0.63(D)(E) 0.8629.The number of days that elapse between the beginning of a calendar year and the momenta high-risk driver is involved in an accident is exponentially distributed. An insurancecompany expects that 30% of high-risk drivers will be involved in an accident during the first 50 days of a calendar year.What portion of high-risk drivers are expected to be involved in an accident during thefirst 80 days of a calendar year?(A) 0.15(B) 0.34(C) 0.43(D)0.57(E) 0.6630.An actuary has discovered that policyholders are three times as likely to file two claimsas to file four claims.If the number of claims filed has a Poisson distribution, what is the variance of thenumber of claims filed?(A) 131(B)(C) 22(D)(E) 431. A company establishes a fund of 120 from which it wants to pay an amount, C, to any ofits 20 employees who achieve a high performance level during the coming year. Eachemployee has a 2% chance of achieving a high performance level during the comingyear, independent of any other employee.Determine the maximum value of C for which the probability is less than 1% that thefund will be inadequate to cover all payments for high performance.(A) 24(B) 30(C) 40(D) 60(E) 12032. A large pool of adults earning their first driver’s license includes 50% low-risk drivers,30% moderate-risk drivers, and 20% high-risk drivers. Because these drivers have noprior driving record, an insurance company considers each driver to be randomly selected from the pool.This month, the insurance company writes 4 new policies for adults earning their firstdriver’s license.What is the probability that these 4 will contain at least two more high-risk drivers than low-risk drivers?0.006(A)0.012(B)0.018(C)0.049(D)(E) 0.07333. The loss due to a fire in a commercial building is modeled by a random variable Xwith density function0.005(20) for 020()0 otherwise.x x f x −<<⎧=⎨⎩Given that a fire loss exceeds 8, what is the probability that it exceeds 16 ?(A) 125(B) 19(C) 18(D) 13(E) 3734. The lifetime of a machine part has a continuous distribution on the interval (0, 40)with probability density function f , where f (x ) is proportional to (10 + x )− 2.Calculate the probability that the lifetime of the machine part is less than 6.(A) 0.04(B) 0.15(C) 0.47(D) 0.53(E) 0.9435. The lifetime of a machine part has a continuous distribution on the interval (0, 40)with probability density function f , where f (x ) is proportional to (10 + x )− 2.What is the probability that the lifetime of the machine part is less than 5?(A) 0.03(B) 0.13(C) 0.42(D) 0.58(E) 0.9736. A group insurance policy covers the medical claims of the employees of a smallcompany. The value, V , of the claims made in one year is described byV = 100,000Ywhere Y is a random variable with density function4(1)for 01()0otherwise,k y y f y ⎧−<<=⎨⎩ where k is a constant.What is the conditional probability that V exceeds 40,000, given that V exceeds 10,000?(A) 0.08(B) 0.13(C) 0.17(D) 0.20(E) 0.5137. The lifetime of a printer costing 200 is exponentially distributed with mean 2 years.The manufacturer agrees to pay a full refund to a buyer if the printer fails during the first year following its purchase, and a one-half refund if it fails during the second year.If the manufacturer sells 100 printers, how much should it expect to pay in refunds?(A) 6,321(B) 7,358 (C) 7,869(D) 10,256(E) 12,64238. An insurance company insures a large number of homes. The insured value, X , of arandomly selected home is assumed to follow a distribution with density function43for 1()0otherwise.x x f x −⎧>=⎨⎩ Given that a randomly selected home is insured for at least 1.5, what is the probability that it is insured for less than 2 ?(A) 0.578(B) 0.684(C) 0.704(D) 0.829(E) 0.87539. A company prices its hurricane insurance using the following assumptions:(i)In any calendar year, there can be at most one hurricane. (ii)In any calendar year, the probability of a hurricane is 0.05 .(iii) The number of hurricanes in any calendar year is independent of the number of hurricanes in any other calendar year.Using the company’s assumptions, calculate the probability that there are fewer than 3 hurricanes in a 20-year period.(A) 0.06(B) 0.19(C) 0.38(D) 0.62(E) 0.9240. An insurance policy pays for a random loss X subject to a deductible of C , where01C << . The loss amount is modeled as a continuous random variable with density function()2 for 010 otherwise.x x f x <<⎧=⎨⎩Given a random loss X , the probability that the insurance payment is less than 0.5 is equal to 0.64 .CalculateC.0.1(A)0.3(B)0.4(C)0.6(D)(E) 0.841. A study is being conducted in which the health of two independent groups of tenpolicyholders is being monitored over a one-year period of time. Individual participants in the study drop out before the end of the study with probability 0.2 (independently of the other participants).What is the probability that at least 9 participants complete the study in one of the twogroups, but not in both groups?(A) 0.0960.192(B)(C)0.2350.376(D)(E) 0.46942.For Company A there is a 60% chance that no claim is made during the coming year.If one or more claims are made, the total claim amount is normally distributed withmean 10,000 and standard deviation 2,000 .For Company B there is a 70% chance that no claim is made during the coming year.If one or more claims are made, the total claim amount is normally distributed withmean 9,000 and standard deviation 2,000 .Assume that the total claim amounts of the two companies are independent.What is the probability that, in the coming year, Company B’s total claim amount will exceed Company A’s total claim amount?(A)0.1800.185(B)0.217(C)(D)0.223(E) 0.24043. A company takes out an insurance policy to cover accidents that occur at its manufacturing plant. The probability that one or more accidents will occur during any given month is 35 . The number of accidents that occur in any given monthis independent of the number of accidents that occur in all other months.Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs.(A) 0.01 (B) 0.12 (C) 0.23 (D) 0.29 (E) 0.41 44.An insurance policy pays 100 per day for up to 3 days of hospitalization and 50 per day for each day of hospitalization thereafter .The number of days of hospitalization, X , is a discrete random variable with probability function6for 1,2,3,4,5()15otherwise.k k P X k −⎧=⎪==⎨⎪⎩Determine the expected payment for hospitalization under this policy.(A) 123 (B) 210 (C) 220 (D) 270 (E) 36745. Let X be a continuous random variable with density function() for 24100 otherwise.xx f x ⎧−≤≤⎪=⎨⎪⎩Calculate the expected value of X .(A) 15 (B) 35(C)1(D) 2815(E) 12546. A device that continuously measures and records seismic activity is placed in a remote region. The time, T , to failure of this device is exponentially distributed with mean 3 years. Since the device will not be monitored during its first two years of service, the time to discovery of its failure is X = max(T , 2) .Determine E [X ]. (A) 6123e −+(B)2/34/3225e e −−−+ (C) 3(D)2/323e −+ (E) 5 47.A piece of equipment is being insured against early failure. The time from purchase until failure of the equipment is exponentially distributed with mean 10 years. The insurance will pay an amount x if the equipment fails during the first year, and it will pay 0.5x if failure occurs during the second or third year. If failure occurs after the first three years, no payment will be made.At what level must x be set if the expected payment made under this insurance is to be 1000 ?(A) 3858(B) 4449 (C) 5382 (D) 5644(E) 723548. An insurance policy on an electrical device pays a benefit of 4000 if the device fails during the first year. The amount of the benefit decreases by 1000 each successive year until it reaches 0 . If the device has not failed by the beginning of any given year, the probability of failure during that year is 0.4 .What is the expected benefit under this policy?(A) 2234 (B) 2400 (C) 2500 (D) 2667 (E) 2694 49.An insurance policy pays an individual 100 per day for up to 3 days of hospitalization and 25 per day for each day of hospitalization thereafter .The number of days of hospitalization, X , is a discrete random variable with probability function6for 1,2,3,4,5()15otherwise.k k P X k −⎧=⎪==⎨⎪⎩Calculate the expected payment for hospitalization under this policy.(A) 85163(B)168(C)213(D)(E) 25550. A company buys a policy to insure its revenue in the event of major snowstorms that shutdown business. The policy pays nothing for the first such snowstorm of the year and10,000 for each one thereafter, until the end of the year. The number of majorsnowstorms per year that shut down business is assumed to have a Poisson distributionwith mean 1.5 .What is the expected amount paid to the company under this policy during a one-yearperiod?(A) 2,769(B) 5,000(C) 7,231(D) 8,347(E) 10,57851. A manufacturer’s annual losses follow a distribution with density function2.53.52.5(0.6)for 0.6()0otherwise.x f x x ⎧>⎪=⎨⎪⎩To cover its losses, the manufacturer purchases an insurance policy with an annual deductible of 2.What is the mean of the manufacturer’s annual losses not paid by the insurance policy?(A) 0.84(B) 0.88 (C) 0.93 (D) 0.95 (E) 1.0052.An insurance company sells a one-year automobile policy with a deductible of 2 . The probability that the insured will incur a loss is 0.05 . If there is a loss, the probability of a loss of amount N isKN, for N = 1, . . . , 5 and K a constant. These are the only possible loss amounts and no more than one loss can occur. Determine the net premium for this policy.(A) 0.031 (B) 0.066 (C) 0.072 (D) 0.110 (E) 0.150 53.An insurance policy reimburses a loss up to a benefit limit of 10 . The policyholder’s loss, Y , follows a distribution with density function:32for 1f ()0,otherwise.y y y ⎧>⎪=⎨⎪⎩What is the expected value of the benefit paid under the insurance policy?(A) 1.0 (B) 1.3 (C) 1.8 (D) 1.9 (E) 2.054. An auto insurance company insures an automobile worth 15,000 for one year under a policy with a 1,000 deductible. During the policy year there is a 0.04 chance of partial damage to the car and a 0.02 chance of a total loss of the car. If there is partial damage to the car, the amount X of damage (in thousands) follows a distribution with density function/20.5003 for 015()0otherwise.x e x f x −⎧<<=⎨⎩ What is the expected claim payment? (A) 320 (B) 328 (C) 352 (D) 380(E) 54055. An insurance company’s monthly claims are modeled by a continuous, positiverandom variable X , whose probability density function is proportional to (1 + x )−4, where 0 < x < ∞ .Determine the company’s expected monthly claims. (A) 16 (B) 13 (C) 12 (D) 1 (E) 356. An insurance policy is written to cover a loss, X , where X has a uniform distribution on [0, 1000] .At what level must a deductible be set in order for the expected payment to be 25% of what it would be with no deductible? (A) 250 (B) 375 (C) 500 (D) 625 (E) 750 57.An actuary determines that the claim size for a certain class of accidents is a random variable, X , with moment generating functionM X (t ) =1125004()−t .Determine the standard deviation of the claim size for this class of accidents.(A) 1,340 (B) 5,000 (C)8,660(D) 10,000(E) 11,18058. A company insures homes in three cities, J, K, and L . Since sufficient distanceseparates the cities, it is reasonable to assume that the losses occurring in these cities are independent.The moment generating functions for the loss distributions of the cities are:M J(t) = (1 – 2t)−3M K(t) = (1 – 2t)−2.5M L(t) = (1 – 2t)−4.5X represent the combined losses from the three cities.LetCalculateE(X3) .(A) 1,320(B) 2,082(C) 5,760(D) 8,000(E) 10,560。
(完整word版)移动云计算导论复习资料整理
移动云计算导论复习资料1选择题1。
云计算是对（ D ）技术的发展与运用A. 并行计算B网格计算C分布式计算D三个选项都是2。
将平台作为服务的云计算服务类型是( B ）A。
IaaS B.PaaS C。
SaaS D。
三个选项都不是3。
将基础设施作为服务的云计算服务类型是（ A ）A. IaaSB.PaaSC.SaaSD.三个选项都不是4. IaaS计算实现机制中，系统管理模块的核心功能是（ A ）A。
负载均衡 B 监视节点的运行状态C应用API D. 节点环境配置5. 云计算体系结构的（ C )负责资源管理、任务管理用户管理和安全管理等工作A。
物理资源层 B. 资源池层C。
管理中间件层 D. SOA构建层6。
云计算按照服务类型大致可分为以下类（A、B、C )A。
IaaS B。
PaaS C. SaaS D。
效用计算7. 下列不属于Google云计算平台技术架构的是（ D ）A. 并行数据处理MapReduce B。
分布式锁ChubbyC。
结构化数据表BigTable D.弹性云计算EC28。
（ B ）是Google提出的用于处理海量数据的并行编程模式和大规模数据集的并行运算的软件架构.A. GFSB.MapReduce C。
Chubby D.BitTable9。
Mapreduce适用于（ D ）A。
任意应用程序B。
任意可在windows servet2008上运行的程序C。
可以串行处理的应用程序 D. 可以并行处理的应用程序10。
MapReduce通常把输入文件按照（ C ）MB来划分A. 16 B32 C64 D12811. 与传统的分布式程序设计相比，Mapreduce封装了( ABCD ）等细节，还提供了一个简单而强大的接口.A。
并行处理B。
容错处理C。
本地化计算 D. 负载均衡12。
( D ）是Google的分布式数据存储于管理系统A。
GFS B. MapReduce C。
Chubby D.Bigtable13. 在Bigtable中，（ A ）主要用来存储子表数据以及一些日志文件A。
(完整word版)软件测试复习题
一、名词解释题①软件生命周期：软件从产生到报废的过程， 1.问题定义及规划2.需求分析3.软件设计4.程序编码5.软件测试6.软件维护②软件测试：使用人工或者自动手段来运行或测试某个系统的过程③CMM：能力成熟度模型，是对于软件组织在定义、实施、度量、控制和改善其软件过程的实践中各个发展阶段的描述④软件质量：软件与明确的和隐含的定义的需求相一致的程度⑤等价类划分：分步骤地把无限的测试用例减的很少，但过程同样等效⑥V&V：Verification和Validation，验证与确认⑦灰盒测试：边看代码、边利用代码的信息帮助测试的一种测试方法⑧回归测试：是在软件维护阶段，对软件进行修改之后进行的测试⑨驱动模块(Drive)：用来模拟被测试模块的上一级模块，相当于被测模块的主程序⑩QA：（软件）质量保证，检查和评价当前软件开发的过程，找出改进过程的方法，以达到防止软件缺陷的出现的目标11需求：产品为向涉众提供价值而必须具备的特性12特别测试：是一种没有实际计划下执行的测试13集成测试：把多模块按照一定的集成方法和策略，逐步组装成子系统，进而组装成整个系统的测试14黑盒测试：软件测试人员只需知道软件运行的结果而无需知道软件的内部是如何运行的。
又称功能行测试或行为测试15回归测试：回归测试是在软件维护阶段，对软件进行修改之后进行的测试16评审：对软件元素或者项目状态的一种评估手段，以确定其是否与计划的结果保持一致，并使其得到改进。
17软件缺陷：计算机软件或程序中存在的某种破坏正常运行能力的问题、错误，或者隐藏的功能缺陷。
18SQA：软件质量保证，建立一套有计划，有系统的方法，来向管理层保证拟定出的标准、步骤、实践和方法能够正确地被所有项目所采用。
19单元测试：对软件中的最小可测试单元进行检查和验证。
二、判断题(√)1、在千年虫例子中，Dave有错吗？有错(√)2、在没有产品说明书和需求文档的条件下可以进行动态黑盒测试。
(完整word版)软件设计模式复习题
软件设计模式复习题1、熟悉每个设计模式的定义、模式UML图解、模式使用情形以及模式优缺点。
2、设计模式的两大主题是什么？3、设计模式分成几大类？每大类各包含哪些模式？4、为什么要使用设计模式？使用设计模式有哪些好处？5、比较抽象工厂模式和工厂方法模式相似性和差异性。
什么情况下使用抽象工厂模式？什么情形下使用工厂方法模式？6、简述原型模式中浅层克隆和深度克隆的区别？分别给出代码说明。
7、模板方法和普通的实现类继承抽象类方式有何区别？8、是比较和分析适配器模式和桥接模式之间的共性和差异性。
9、请用组合模式实现学校人事管理模式。
10、综合应用装饰模式、命令模式和状态模式实现工具条命令按钮鼠标进入时高亮显示状态以及鼠标单击按钮后呈现凹陷状态，表明当前按钮为选中状态。
11、请阐述享元模式是如何节省系统内存的？试举例分析使用享元模式前后的内存节约之比。
12、使用解析器模式实现对学生成绩表的查询输入语句进行解析并执行查询，成绩表结14、试比较和分析中介者模式和观察者模式之间的相似性和差异性？两者是否能够相互转化？15、模板方法和普通的抽象类继承有什么区别？16、使用访问者模式对12题中的学生信息进行报到。
1、熟悉每个设计模式的定义、模式UML图解、模式使用情形以及模式优缺点。
(见书)答：单件模式：①单例模式：class Sin glet on {private static Sin gleton in sta nee;private Sin glet on(){}public static Sin glet on Get In sta nce(){if (in sta nee == n ull) {in sta nee = new Sin glet on();}retur n in sta nee; }客户端代码：class Program{static void Main(string[] args) {Sin glet on s1 = Sin glet on .Getl nsta nce();Sin glet on s2 = Sin glet on .Getl nsta nce();if (s1 = = s2) {Con sole.WriteL in e("Objects are the same in sta nee");}Co nsole.Read();}}②多线程时的单例(Lock是确保当一个线程位于代码的临界区时，另一个线程不进入临界区。
亚马逊aws-sap考试题库
1.问题 1 一家公司正在将业务关键型应用程序迁移到 AWS上。
它是使用 Oracle 数据库的传统三层Web应用程序。
数据在传输和静止时都必须加密。
该数据库承载12TB 的数据。
允许通过内部网络与源 Oracle 数据库建立网络连接，该公司希望通过使用AWS托管服务来降低运营成本。
仅所有主键；但是，它包含许多二进制大对象（ BLOB）字段。
由于许可限制，无法使用数据库的本机复制工具。
哪种数据库迁移解决方案将对应用程序的可用性产生最小的影响？A.为 Amazon实例设置 Amazon RDS。
在虚拟私有云中托管 RDS数据库（ VPC）子网具有 Internet访问权限，并将 RDS数据库设置为源数据库的加密只读副本。
使用SSL 加密两个数据库之间的连接。
通过观察RDSReplicaLag 指标来监视复制性能。
在应用程序维护窗口期间，在没有更多复制滞后时，请关闭本地数据库，并将应用程序连接切换到 RDS实例。
将只读副本提升为独立的数据库实例。
B.设置一个 Amazon EC2实例并安装相同的 Oracle 数据库软件。
使用支持的工具创建源数据库的备份。
在应用程序维护窗口期间，将备份还原到在EC2 实例中运行的 Oracle 数据库中。
设置Amazon RDS forOracle 实例，并在 AWS托管的数据库之间创建导入作业。
作业完成后，关闭源数据库并将数据库连接切换到 RDS实例。
C.使用 AWS DMS在本地 Oracle 数据库和 AWS上托管的复制实例之间加载和复制数据集。
提供一个AmazonRDSO racle 实例与透明数据加密（ TDE）启用，其配置为目标的复制实例。
创建一个客户管理的AWSK MS 主密钥，将其设置为复制实例的加密密钥。
使用AWSD MS任务将数据加载到目标 RDS实例中。
在应用程序维护窗口期间以及加载任务到达正在进行的复制阶段之后，将数据库连接切换到新数据库。
D.在应用程序维护窗口期间，在本地 Oracle 数据库上创建压缩的完整数据库备份。
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SOA 基础复习题题型：选择题20*1分，简答题8*5分，问答题4*10分一、选择题（每题1分）注意：有些题目可能有不止一个正确答案C 应用集成D. 改善的系统管理C．Web services基于JAX-RPC标准，并通过远程方法调用提供一种松耦合方法。
D．Web services 通过使用BPEL应用程序定义的接口，实现服务功能。
3.下面哪一种新方案有助于实现SOA业务的敏捷性。
A.为服务的提供者找合适的用户B.购买预置对象快速实现业务功能C.编排业务流程以有效地满足业务功能需求D.重新设计和开发新的业务流程以替代旧的业务流程C.建立能够执行多种任务的一般服务D.在一个连贯的软硬件平台上从新实施所有的服务C.由于医生和执业医师不懂得信息技术，所以对于他们来讲SOA可以屏蔽底层的技术实现D.使用.Net的小型企业的医疗服务供应商可以通过Web services的互操作性，和采用Java的较大的母公司进行合作。
C管理方法，基础设施和连通性D互操作性，标准和应用集成E高可用性的硬件，操作系统和服务7.最适合SOA实施的时间是什么时候A实时性能是关键B直接投资回报率是关键C应用程序接口需要高度定制D业务的功能对于许多组织部门是十分必要的8.有一个零售公司很难改变其业务流程，许多业务需要维护他们的数据库和相应的接口，下面一些包括重构数据在内的手工步骤。
零售公司应该怎样改变这些流程进而提高流程的效率A用来减少成本的航运功能外包B开发具有整合应用程序和信息资源的服务C创建公共服务以减少流程的冗余D用最新的可用的技术来整合应用程序和资源9.SOA的哪一个基本特征可以提高业务敏捷性。
A重用和安全机制B松耦合和重用C管理和整合D点对点的整合和扩展10.如果整体系统的一个功能组件能够独立于其他组件正常运行，这是SOA架构C松耦合D关注点分离E综合执行.C.在ESB曾采用一种通用的数据模型D.业务分析师创建集成逻辑之间进行谈判的细节服务12.一个公司已经采用SOA服务构建了一些业务流程。
他们怎样确定新的业务流程帮助公司实现了业务目标B.他们可以联系第三方等专门机构来分析和评估业务目标的新流程C.他们通过设定和监控Key Performance Indicators来跟踪业务业务目标的新流程D.他们可以通过商业模式模拟器来模拟新的业务流程以确定瓶颈因素和潜在的性能问题13.在什么情况下，生产工厂最可能为他们的客户订单分析和跟踪系统选用SOA 的方法？A．公司希望开发和操作尽可能的简单。
B．公司希望打破部门并鼓励所有人使用同一套工具。
C．公司的流程大部分是手工的或者以文档为中心的，很少用到自动化。
D．遗留系统已经发展为不易更新、无法处理新兴的业务需求。
C．互联网通信协议D．网络服务信息的负载容量15.在SOA管理中，企业服务总线（ESB）的作用是什么？A.通过限制谁可以访问服务来提供服务的管理B.提供用户和服务提供者之间点对点的连接C.提供灵活性使业务流程效率快速得到纠正D.在任何地方任何时候使用不同的协议时确保信息的无缝传递C．重命名一个操作。
D．改变操作参数。
17.企业服务总线比点对点解决方案有哪些优点？A．灵活性和成本B．冗余和灵活性C．性能和可伸缩性D．多功能性和适应性C．服务实现和服务的访问控制D．服务路由和服务端点注册19.某组织计划自动化其业务流程并以服务的方式对外开放，请问在SOA环境下应怎样与上述服务交互？A.可能有某种Web应用程序扮演服务消费者的角色，它允许自动调用服务流程，而且任何一种服务流程中的活动都可能是如此被调用B.上述业务流程不适宜有人工干预，人的反应过于迟钝不适于同步方式C.可能有某种Web应用程序扮演服务消费者的角色，它允许从浏览器中自动调用服务流程，所有服务流程中的活动都是如此被调用D.可能有某种Web应用程序扮演服务消费者的角色，它允许自动调用服务流程，只有特殊的步骤可能采取上述方式E.所有活动都应该以自动服务的方式对外开放20.在SOA中Web services扮演了哪两种角色？A.它提供了服务请求的调解机制B.它提供了一种信息框架的服务C.它支持不同技术间的可操作性D.它提供了描述服务接口的开放式的标准E.它为企业服务总线(ESB)提供了信息机制C封装，关注，高内聚，松耦合，正确的粒度和抽象D业务灵活性，业务敏捷性，成本效益，目标相关的，可追溯的要求和持续逐步改善C. 侧重于Web services的企业服务总线D. 侧重于XML消息机制的企业服务总线23.从IT的角度看，转变为SOA的主要目标是什么？A使IT也更具责任性B使IT更灵活C平台和技术独立性D减少短期开发和维护成本24.为在企业层面确保共享服务生命周期的资金支持，哪种方法是最有效的第一步？A．退款机制到位，以防止非法的服务B．使用一种“税”的方法，这种方法建立在每条业务线都IT化的基础上C．建立一个SOA治理组织，并且提供相应的资金支持D．使用“第一需要”的方法，需要服务的工程第一个得到资金支持25.客户对一个公司介绍引入的复杂新产品的满意度比较低，这一点会影响到产品的销售。
客服中心花费很多时间处理产品问题的方式是不可取的。
要想达到快速减少用户抱怨的目标，下面哪几种方法可取？A．连接所有涉及到的系统B．收集所有产品的信息放置于一个信息库中C．为客服中心提供所有可能的信息和工具，以便于他们能向专家询问详细的信息D．将呼叫过程建模，以确定的瓶颈，然后处理自动化工具，可用于获取所有必要的产品信息，包括人工任务26.最近一个公司的一项战略指令，要求业务预算中减少航空货运的2%。
首席信息官（CIO）表示转向SOA在开发成本上可以帮助提高效率。
该公司刚刚与包裹邮递公司签了战略合作伙伴关系，这需要用到两个公司系统的集成。
准备采用SOA的时候，什么对于显示转向SOA后的投资回报率最重要？A．目前公司及合作伙伴公司的预算B．当前公司及伙伴公司的IT治理模块重叠C．当前公司级伙伴公司的关键业绩指标D．运行现在的信息系统实现与伙伴公司的点对点集成的预算27.某公司已经决定采用SOA改进企业系统，以加快集成，减少应用冗余，并提供灵活地改变服务实现。
应为该公司建议下面那个项目？A.开发一个通用的源数据模型，以应用于所有工程开发B.实现企业服务总线，并在SOA架构中注册一个连接层C.文档化现存的体系结构、存在的系统和接口。
使这些在普通的库中便能被所有开发者使用D.使用业务建模工具对应用的开发流程建模、模仿，找出瓶颈，并标明优先级较高的瓶颈。
28.一个企业使用SOA已经一段时间了并且对结果大失所望，他们即使创建了数百个服务也没有看到应有的投资回报，哪一个最可能是这个问题的最主要原因？A．服务是基于语言和地理需要被复制的。
B．对于如何创建可重用的服务，开发者需要更好的训练。
C．管理过程没有合适的审查服务。
D．控制环境对执行如此大的SOA来说是个挑战。
29.哪一个是组件和服务的主要差别？A．一个服务只配置一次而一个组件会配置多次。
B．一个组件只配置一次而一个服务会配置多次。
C．一个组件有一个接口而一个服务实现这个接口。
D．一个服务有一个接口而一个组件实现这个接口。
E．一个服务没有一个定义良好的接口而一个组件有。
30.哪一个标准描述了Web服务的接口？A. WS-IB. BPELC. SOAPD. WSDL31.在SOA设计中XML扮演什么角色？A．X ML用于执行UML设计。
B．X ML模式驱动信息的有效荷载。
C．X ML用于SOA所需要的Web服务。
D．XML用于描述接口和信息绑定。
32.为什么定义一个合适的KPIs对于SOA的成功十分重要？A．这样可以提供IT执行步骤。
B．这样可以决定SOA的可销售性。
C．这样可以为SOA的移植提供资金。
D．为指标性商业投资回报提供反馈。
E．使商业领导层了解商业处理过程的障碍和瓶颈。
33.SOA的哪一个方面能够降低交叠和复制？A．模块化和封装。
B．S OA管理和服务重用。
C．服务重用和处理建模。
D．松耦合和关系分离。
34.关于SOA中的业务流程下面哪种说法正确？A．它几乎不包括手工步骤。
B．它可以跨越人，系统和信息。
C．它完成需要相对较短的时间。
D．所有的服务通过企业服务总线（ESB）与流程连接。
35.下列哪三种说法描述了在SOA中的服务？A. 服务是用Java或C#写的。
B. 服务可以兼容不同的技术。
C．服务可以改变位置并对客户端没有影响。
D．服务根据SOAP文件提供了接口。
E．服务可以停止并对客户没有影响。
F．服务的实现可以被改变而不改变用户。
36.为什么重用对推动SOA的灵活性是一个重要的方面？A．服务往往代表SOA的功能。
B．现有的应用可作为服务接口的基础。
C．服务接口允许它不用改变地集成到其它进程。
D．服务接口细节根据预期的业务需求可被定义在较高的水平。
C．识别黄金客户传入的请求D．改变业务流程为黄金客户提供增强的功能38.下列哪一个最准确地描述了业务流程？A．以最佳的方式完成一个工作单元的步骤B．用来获取业务需求的工具C．必须由系统或人在特定的顺序下完成的任务流D．BPEL代码运行在业务流程引擎驱动人和系统工作的环境下39.下列哪两个描述了SOA的特点？A．多个业务单元使用相同的服务以确保最大程度的利用和可用性。
B．通过确保服务都是通用的来实现最大程度的利用，这样相同的服务可以服务许多不同的业务任务。
C．诸如Web服务和企业服务总线（ESB）的新技术展示了现有的API来提高业务灵活性和敏捷性。
D．通过合约来编程保证了服务的使用者不知道服务提供者促成低耦合的实现细节。
E．通过对实施和运行细节的分层抽象，提供在重用和特异性之间取得适当平衡的软件资源成为可能。
40.下列哪项标准使得服务可以和任何相关数据一起被请求？A. XHTMLB. ESBC. UDDID. SOAP41.一个SOA的业务分析人员需要描述一个业务流程。
这个SOA的分析人员必须把这个业务流程描述为：A．一个企业服务总线（ESB）B．一个软件部署路线C．实现业务任务必须的web serviceD．一个或多个业务服务的组合42.下面哪种方式是最被接受的流程执行方式？A．ebXMLB．XML ParserC．WS-CoordinationD．SOAP4JE．BPEL43.基于以下哪两种组件的支持，SOA中的松耦合是可能的？A．业务流程模型B．SOA治理C．企业服务总线（ESB）D．JDBC适配器E．Web Service接口44.一个银行的贷款手续从顾客填写纸质的贷款申请并由职员将这些信息录入到安全数据库中开始，银行希望能够将这个过程升级为由顾客直接在线的填写贷款申请并自动的将信息录入到安全数据库中。