第3章 补充练习题
C语言程序设计(第3版)第3章补充习题及答案
x=(a&&b)&&(c<’B’);
执行该程序段后,x的值为( )
AtureB falseC 0D 1
(9)以下程序的输出结果是( )
main()
{
int a=0,i=1;
switch(i)
{
case 0:
case 3:a+=2;
case 1:
case 2:a+=3;
default:a+=5;
(2)下面程序的输出结果是
iX=3
iX>3
。
#include "stdio.h"
main()
{intiX=3;
switch(iX)
{case 1:
case 2:printf("iX<3\n");
case 3:printf("iX=3\n");
case 4:
case 5:printf("iX>3\n");
}
}
(3)以下程序的运行结果是iM=3,iN=4,iT=3。
#include "stdio.h"
main()
{int iM,iN,iT,iA=2,iB=3;
iM=(++iA==--iB)?––iA:++iB;
iN=++iA;
iT=iB;
printf("iM=%d,iN=%d,iT=%d\n",iM,iN,iT);
(4)对switch后括号内的表达式,叙述正确的是C。
A.只能是数字B.可以是浮点数
C.只能是整型数据或字符型数据D.以上叙述都不对
第3章_资本预算补充练习及答案
第3章_资本预算补充练习及答案第三章《资本预算》补充练习⼀、单选题1、下列关于投资项⽬营业现⾦流量预计的各种作法中,不正确的是(D)。
A.营业现⾦流量等于税后净利加上折旧B.营业现⾦流量等于营业收⼊减去付现成本再减去所得税C.营业现⾦流量等于税后收⼊减去税后付现成本再加上折旧引起的税负减少额D.营业现⾦流量等于营业收⼊减去营业成本再减去所得税2、某投资⽅案,当贴现率为16%时,其净现值为338元,当贴现率为18%时,其净现值为-22元。
该⽅案的内含报酬率为(C)。
A.15.88%B.16.12%C.17.88%D.18.14%解析:内含报酬率是使净现值为0时的贴现率。
内含报酬率=16%+(18%-16%)×[338/(338+22)]=17.88%3.下列表述中,正确的是( D )。
A.净现值法可以⽤于⽐较项⽬寿命不相同的两个互斥项⽬的优劣B.使⽤净现值法评估项⽬的可⾏性与使⽤内含报酬率法的结果是⼀致的C.使⽤净现值法进⾏投资决策可能会计算出多个净现值D.内含报酬率不受设定折现率的影响4.甲公司对某投资项⽬的分析与评价资料如下:该投资项⽬适⽤的所得税税率为25%,年税后营业收⼊为1000万元,税后经营付现成本为400万元,税后经营净利润为375万元。
那么,该项⽬年营业现⾦流量为( A )万元。
A.675B.775C.1125D.1375【解析】注意题⽬给出是税后收⼊和税后付现成本。
根据题意,税后经营净利润=税前经营利润×(1-25%),所以,税前经营利润=税后经营净利润/(1-25%)=375/(1-25%)=500(万元),⽽税前经营利润=税前营业收⼊-税前经营付现成本-折旧=1000/(1-25%)-400/(1-25%)-折旧=800-折旧,所以,折旧=800-500=300(万元),营业现⾦流量=税后经营净利润+折旧=375+300=675(万元),或营业现⾦流量=税后收⼊-税后经营付现成本+折旧×税率=1000-400+300×25%=675(万元)。
语言学补充练习(1-3)
第一章语言学入门知识:I. 名词解释1.cultural transmission (as a defining property of language)Answer: While human capacity for language has a genetic basis, the details of any language system are not genetically transmitted, but instead have to be taught and learned. An English speaker and a Chinese speaker are both able to use a language, but they are not mutually intelligible. This shows that language is culturally transmitted. It is passed on from one generation to the next through teaching and learning, rather than by instinct. In contrast, animal call systems are genetically transmitted. They are born with the capacity to produce the set of calls peculiar to their species.2.descriptive linguistics vs. prescriptive linguisticsAnswer: A linguistic study is descriptive if it describes and analyses facts observed; it is prescriptive if it tries to lay down rules for "correct" behavior. Linguistic studies before the 204 century are largely prescriptive whereas modem linguistic is mostly descriptive.II. 判断正误(T for True and F for False)1. When language is used to get information, it serves an informative function.Answer: F (It serves an interrogative function).2.Most animal communication systems lack the primary level of articulation.Answer: F (The primary units in these systems cannot be further divided into elements. So what they lack is the secondary level of articulation.) 3.Descriptive linguistics are concerned with how languages work, not with how they can be improved.Answer: TIII. 填空题1.By saying that "language is arbitrary", we mean that there is no logical connection between meaning and _______.Answer: sounds2.The distinction between langue and parole is made by the Swiss linguist E de Saussure. The distinction between competence and performance is made by the American linguist__________.Answer: Noam Chomsky3.An approach to linguistic study which attempts to lay down rules of correctness as to how language should be used is _______.Answer: prescriptiveIV. 选择题1.Unlike animal communication system, human language is ______.A. stimulus freeB. stimulus boundC. under immediate stimulus controlD. stimulated by some occurrence of communal interestAnswer:A2. ____ has been widely accepted as the forefather of modem linguistics.a. Chomskyb. Saussurec. Bloomfieldd. John LyonAnswer: bV. 问答题l. Is language productive or not? Why?Answer: Firstly, Language is productive or creative. This means that language users can understand and produce sentences they have never heard before. Secondly, Productivity is unique to human language. Most animal communication systems have a limited repertoire, which is rapidly exhausted, making any novelty impossible. Thirdly, The productivity or creativity of human language originates from its duality. Because of duality, the speaker can combine the basiclinguistic units to form an infinite set of sentences. The productivity of language also means its potential to create endless sentences. This is made possible by the recursive nature of language.2. Comment on the f ollowing statement: “In linguistics, ‘language’ onlymeans what a person says or said in a given situation”.Answer: This statement is incorrect. In linguistics, "language" has several layers of meaning: firstly, the whole of a person’s language, e.g. Sha kespeare’s language; secondly, a particular variety or level of speech or writing, e.g. scientific language, literary language, colloquial language; thirdly, an abstract system underlying the totality of the speech/writing behavior of a community, e.g. the English language, the Chinese language; lastly, there is an even more abstract sense of "language", referring to the common features of all human languages that distinguish them from animal communication systems or any artificial language.3. Point out three ways in which linguistics differs from traditional grammar.Answer: Firstly, most linguistic analyses today focus on speech rather than writing. Secondly, modem linguistics is mostly descriptive while traditional grammar is largely prescriptive. Thirdly, a third difference is the priority of synchronic description over the traditional diachronic studies.4. What is the major difference between Saussure' s distinction between langue and parole and Chomsky's distinction between competence and performance?Answer: Saussure’s langue is social product, a set of conventions for a speech community. Chomsky regards competence as a property of themind of each individual. Saussure studies language more from a sociological point of view while Chomsky studies it more from a psychological point of view.第二章语音学和音位学I. 名词解释1.narrow transcriptionAnswer: There are two ways to transcribe speech sounds. One is the “broad transcription”----the transcription with letter-symbols only, and the other is “narrow transcription”---the transcription with letter-symbols accompanied by the diacritics which can help bring out the finer distinctions than the letters alone may possibly do.2. Illustrate the term “allophone” with at least one appropriate example. Answer: Allophones are the different members of a phoneme, sounds which are phonetically different but do not make one word different from another in meaning. For example, in English, the phoneme /l/ is pronounced differently in "let", "play" and "tell". The first /l/ is made by raising the front of the tongue to the hard palate, while the vocal cords are vibrating; the second /l/ is made with the same tongue position as the first, but the vocal cords are not vibrating; and the third /l/ is made by raisingnot only the front by also the back of the tongue while the vocal cords are vibrating.II 判断正误(T for True and F for False)1. /o/ is a mid-high front rounded vowel.Answer: F. (/o/ is a mid-high BACK rounded vowel.)2. A phoneme in one language or one dialect may be an allophone in another language or dialect.Answer: T.III. 填空题:1. The three cavities in the articulatory apparatus are _____, _______, and _____.Answer: pharynx, the nasal cavity, the oral cavity2. By the position of the ____ part of the tongue, vowels and classified as front vowels, central vowels and back vowels.Answer: highest.3. ____refers to the change of a sound as a result of the influence of an adjacent sound.Answer: Assimilation.4. You are required to fill in the blanks below abiding by the instancegiven beforehand.Example: /p/: voiced bilabial stop/s/: ________________/g/:_______________/tʃ/:______________/t/: _______________/f /: _______________Answer:/s/: voiceless alveolar fricative/g/: voiced velar stop/tʃ/: voiceless alveo-palatal/post-alveolar affricate/ t /: voiced dental fricative/f /: voiceless labiodental fricative5. Which of the following words would be treated as minimal pairs and minimal sets?pat, pen more, heat, tape, bun, fat, ban, chain, tale, bell, far, meal, vote, bet, heel, ten, men, pit, main, hit, eat, manAnswer:pat, fat; pat, pit; pit, hit;pen, ten; ten, men;heat, eat; heat, heel;tape, tale;bun, ban;chain, main;bell, bet;meal, heel;man, men, main.IV. 选择题1. All syllables contain a(n) _______.a. nucleusb. codac. onsetAnswer: a2. _____is one of the supersegmental features.a. Stopb. Voicingc. Deletiond. ToneAnswer: d3. Which of the following consonants does not exist in English?a. dental stopb. bilabial stopc. alveolar stopd. velar stopAnswer: a4. _____is not an English consonant.a. Labiodental plosiveb. Alveolar nasalc. Velar stopd. Dental fricativeAnswer: aV. 辨音选择1. What are the distinctive features that group the following sounds in these sets?1) /f, v ,s/2) /p, f, b/3) /g, z, b/4) /k, g, w/5) /m, n, ŋ/Answer: 1) fricative 2) obstruent 3) voiced 4) velar5) nasal2. There is one segment that does not belong to the natural class in each of the following groups of speech sounds. You are required to identify that segment and label the natural class, using a descriptive term asspecific as possible.a) /m/, /n/, /w/, / ŋ /b) /v/, /w/, /z/, /t/c) /n/, /f/, /l/, /s/, /t/, /d/, /z/Answer:1) /w/ is a semi-vowel, and the others are all nasals.2) /t/ is voiceless, and the others are voiced.3) /f/ is labiodental, and the rest are alveolarVI. 问答题1.Circle the words that contain a sound as required:1) a low vowel: pipe, gather, article, leave, cook2) a bilabial consonant: cool, lad, leap, bomb, push3) an approximant: luck, boots, word, once, table4) a front vowel: god, neat, pit, lush, cook5) a velar: god, fast, chat, lake, quick2.Exemplify the relationship between phone, phoneme and allophone. Answer: Firstly, a “phone” is a phonetic unit or segment. The speech sounds we hear and produce during linguistic communication are all phones. Phones may or may not distinguish meaning. Secondly, a "phoneme" is a phonological unit that is of distinctive value. As an abstract unit, a phoneme is not any particular sound. It is represented orrealized by a certain phone in a certain phonetic context. Thirdly, the phones representing a phoneme are called its "allophones". How a phoneme is represented by a phone, or which allophone is to be used, is determined by the phonetic context in which it occurs. But the choice of an allophone is not random but rule-governed in most cases.3.When we are pronouncing the following phrases, how do we actually articulate the "n" sound in the word "ten"? Do we still pronounce it as /n/?1) ten houses 2) ten teachers 3) ten colleges 4) ten pupils 5) ten buildings 6) ten classesAnswer: 1) /n/2) /n/3) / ŋ /4) /m/5) /m/6) / ŋ /4.How many functions do the vocal cords have in the production of speech sounds?Answer: They have three functions: to make a glottal stop, to produce a voiced sound and to produce a voiceless sound.第三章形态学I. 名词解释1.morphemeAnswer: The morpheme is the smallest unit in terms of relationship between expression and content, a unit which cannot be divided without destroying or drastically altering its meaning, whether it is lexical or grammatical. For instance, the word "barks" in "The dog barks" consists of two morphemes ― "bark" and "-s", neither of which can be further divided into other smaller meaningful units.2.lexemeAnswer: The term "lexeme" is postulated to reduce the ambiguity of the term "word". It is the abstract unit underlying the smallest unit in the lexical system of a language, which appears in different grammatical contexts. For example, "write" is the lexeme of the following set of words: "writes", "wrote", "writing", "written".3.inflectional morphemesAnswer: Inflectional morphemes are also called inflectional affixes. They manifest various grammatical relations or grammatical categories such as number, tense, degree and case. In English, all inflectional morphemes are suffixes, e.g. -(e)s, -ing, -(e)d, -est.II. 判断正误1.A root is not always a free form.Answer: T (There are such bound roots as “-ceive”.)III. 填空题1.Polymorphemic words other than compounds have two parts: the roots and the ____.Answer: affixes2.On, before and together are_____words ― they are words which do not take inflectional endings.Answer: grammatical (functional/form)IV.选择题1."Radar" is a/an____.a. acronymb. blendingc. coinaged. clippingAnswer: a2.Compound words consist of______ morphemes.a. boundb. freec. both bound and freeAnswer: bV. 匹配题Match each expression under A with the one statement under B that characterizes it.A B1. a noisy crow a. compound noun2. eat crow b. root morpheme plus derivational prefix3. scarecrow c. phrase consisting of an adjective plus noun4. the crow d. root morpheme plus inflection affix5. crowlike e. root morpheme plus derivational suffix6. crows f. grammatical morpheme followed by lexical morphemeg. idiomAnswer: 1. c 2. g 3. a 4. f 5. e 6. dVI. 问答题1. Divide the following words into Roots, IA (inflectional affix) and/or DA (derivational affix).1) transformations 2) looseleaves3) destructive 4) geese 5) misledAnswer:1) trans- (DA) form (Root) -ation (DA) s (IA)2) loose (Root) leave (Root) s (IA)3) de- (DA) struct (Root) -ive (DA)4) geese (IA)5) mis- (DA) led (IA)2. Label the morphological category of the morphemes underlined in each of the English expressions.a) I' ve been here.b) transformc) oxend) recurAnswer: a) bound morpheme b) derivational prefix c) inflectional suffix d) bound root3. Each of the following Persian words is poly-morphemic. You are required to match each of the notions given below with a morpheme in Persian. (Note that xar means "buy" and -id designates the past tense). xaridiYou (singular) bought.naxaridamI did not buy.namixaridandThey were not buying.xaridHe bought.naxaridimWe did not buy.mixaridHe was buying.mixarididYou (plural) were buying.xaridamI bought.Match each of the notions given below with a morpheme in Persian:a) Ib) you (singular)c) notd) was/were V-ing (continuous)Answer: a) amb) ic) nad) miVid4.It is a fact that morphological processes may be sensitive to certain phonological context. The English data given below illustrate this fact. You are required to state the phonological contexts where the addition of -en is possible.a bwhiten *bluenmadden *stupidenredden *greenenFatten *fartheren quicken *slowendeafen *difficultenLiven *abstractenharden *shallowensoften *angryendeepen *vividenAnswer: The suffix -en, which attaches to adjectives to form verbs, can only attach to monosyllabic bases ending in oral stops or fricatives.VerbAdjective-en if Adjective ends in an obstruent (oral stop or fricative).- <Φ> if Adjective ends in a sonorant (nasal, approximant, vowel) Meaning: to make (more) Adjectives5.The word uneasiness may be analyzed in either of the two ways below. You are required to find an argument to support one of the two analyses.a)NPrefixNoununAdjectiveSuffixeasinessb)NAdjectiveSuffixPrefixAdjectivenessuneasiAnswer: b) is the correct analysis, because un- only attaches to adjectivesto form other adjectives. Un- cannot be attached to a noun.。
线性代数第三章补充题及其答案
补充练习三 矩阵一、选择题:(1)设A 和B 均为n 阶方阵,则必有( )。
(A )|A+B|=|A|+|B|; (B )AB=BA (C )|AB|=|BA| (D )(A+B )-1=A -1+B -1 (2)设A 和B 均为n 阶方阵,且满足AB=0,则必有( )。
(A )A=0或B=0 (B )A+B=0 (C )|A|=0或|B|=0 (D )|A|+|B|=0 (3)设⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=333231232221131211a a a a a a a a a A ,⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+++=133312321131131211232221a a a a a a a a a a a a B , ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=1000010101P ,⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=1010100012P ,则必有( )。
(A )AP 1P 2=B ; (B )AP 2P 1=B ; (C )P 1P 2A=B ; (D )P 2P 1A=B (4)设n 维行向量⎪⎭⎫⎝⎛=21,0,,0,21α,矩阵ααT E A -=,ααT E B 2+=,其中E 为n 阶单位矩阵,则AB=( )。
(A )0; (B )E ; (C )-E (D )ααTE + (5)设n 阶方阵A 非奇异(n ≥2),A *是A 的伴随矩阵,则( )。
(A )(A *)*=|A|n-1A ; (B )(A *)*=|A|n+1A ; (C )(A *)*=|A|n-2A ; (D )(A *)*=|A|n+2A(6)设n 阶方阵A 、B 、C 满足ABC=E ,其中E 是n 阶单位矩阵,则必有( )。
(A )ACB=E ; (B )CBA=E ; (C )BAC=E ; (D )BCA=E(7)设⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=44434241343332312423222114131211a a a a a a a a a a a a a a a a A ,⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=41424344313233342122232411121314a a a a a a a a a a a a a a a a B ,⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=00010100001010001P ,⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=10000010010000012P ,其中A 可逆,则B -1等于( )。
第三章知识题及答案解析
.六.补充练习题(一)单项选择题1.以下各项目属于会计科目的有()A.欠供应单位料款B.所有者投资C.银行存款D.未分配利润2.对会计要素进行分类核算的工具是()A.会计科目B.账户C.会计科目和账户D.会计科目或账户3.会计科目是()A.会计报表的名称B.会计账户的名称C.会计账簿的名称D.会计要素的名称4.每一项经济业务的发生,都会影响()账户发生增减变化。
A.一个B.两个C.两个或更多D.全部5.对每个账户来说,在某一时刻其期末余额()A.只能在借方B.只能在账户的一方C.只能在贷方D.可能在借方和贷方6.按照借贷记账法的记账规则记账,资产类账户的借方发生额和贷方发生额()A.总是相等的B.总是不等的C.借方发生额一定大于贷方发生额D.两者之间没有必然的对应关系7.被世界各国普遍采用的复式记账法是()A.增减记账法B.收付记账法C.单式记账法D.借贷记账法8.采用复式记账的方法,主要为了()A.便于登记账簿B.全面地、相互联系地反映资金运动的来龙去脉C.提高会计工作效率D.便于会计人员的分工协作9.在借贷记账法下,应付账款账户的贷方表明()A.企业债权的增加B.企业债务的减少C.企业债务的增加D.企业债权的减少10.在借贷记账法下,负债类账户的借方表明()A.企业债权的增加B.企业债务的减少C.企业债务的增加D.企业债权的减少11.下列账户中本期发生额减少登记在借方的是()A.管理费用B.财务费用C.短期借款D.制造费用12.本月共购入材料5 000元,各车间部门领用3 500元,退回200元,月末结存为2 000元,则材料上月结存应为()A.300元B.700元C.500元D.900元13.账户发生额试算平衡公式存在的依据是()A.借贷记账法的记账规则B.经济业务的内容C.静态会计方程式D.动态会计方程式14.采用借贷记账法,哪方记增加、哪方记减少,是根据()A.借方记增加、贷方记减少的规则决定B.贷方记增加、借方记减少的规则决定C.企业会计准则决定D.每个账户的基本性质决定15.某单位会计员小赵、小钱、小孙、小李在一起讨论账户借贷双方登记的内容,其正确者是()A.小赵说:借方反映资产减少,负债、所有者权益增加,费用减少,收入增加。
马原,第三章习题及答案
第三章人类社会及其发展规律(课后练习题)一、单项选择题1 .人类社会历史发展的决定力量是( )A .生产方式B .地理条件C .社会意识D .人口因素2 .社会意识相对独立性的最突出表现是它( )A .同社会存在发展的不同步性B .具有历史的继承性C .对社会存在具有能动的反作用D .同社会经济的发展具有不平衡性3 .在生产关系中起决定作用的是( )A .生产资料所有制B .产品的分配和交换C .在生产中人与人的关系D .管理者和生产者的不同地位4 . “手推磨产生的是封建主的社会,蒸汽磨产生的是工业资本家的社会” , 这句话揭示了( )A .生产工具是衡量生产力水平的重要尺度B .科学技术是第一生产力C .社会形态的更替有其一定的顺序性D .物质生产的发展需要建立相应的生产关系5 .十一届三中全会以来,我党制定的一系列正确的路线、方针、政策促进了我国经济的迅猛发展,这说明( )A .经济基础发展的道路是由上层建筑决定的B .上层建筑的发展决定经济基础的发展方向C .上层建筑对经济基础具有积极的能动作用D .社会主义社会的发展不受经济基础决定上层建筑规律的制约6 .一定社会形态的经济基础是( )A .生产力B .该社会的各种生产关系C .政治制度和法律制度D .与一定生产力发展阶段相适应的生产关系的总和7 .上层建筑是指( )A .社会的经济制度B .科学技术C .社会生产关系D .建立在一定社会经济基础之上的意识形态及相应的制度和设施8 .社会形态是( )A .生产力和生产关系的统一B .同生产力发展一定阶段相适应的经济基础和上层建筑的统一体C .社会存在和社会意识的统一D .物质世界和精神世界的统一9 .人类社会发展的一般规律是( )A .生产方式内部的矛盾规律B .生产力和生产关系、经济基础和上层建筑之间的矛盾运动规律C .社会存在和社会意识的矛盾规律D .物质生产和精神生产的矛盾规律10 .阶级斗争对阶级社会发展的推动作用突出表现在( )A .生产力的发展B .生产关系的变革C .社会形态的更替D .科技的进步11 .社会革命根源于( )A .人口太多B .少数英雄人物组织暴动C .先进思想和革命理论的传播D .社会基本矛盾的尖锐化12 .社会主义改革的根本目的在于( )A .改变社会主义制度B .完善社会主义制度C .解放和发展生产力D .实现社会公平13 . “蒸汽、电力和自动纺织机甚至是比巴尔贝斯、拉斯拜尔和布朗基诸位公民更危险万分的革命家。
热学各章补充练习题
2021/4/9
43
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a 3
T
1 c 2
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6
9、一摩尔双原子分子理想气体,经历如图所示
的循环
,其中ab是等温过程,
0
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2021/4/9
24
解:(3)
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分别表示氧气和氢
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28
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2022-2022现代文理学院分子生物学章节练习题第3章练习题
2022-2022现代文理学院分子生物学章节练习题第3章练习题第三章生物信息的传递-从DNA到RNA练习题一、【单项选择】1.转录终止因子为A.σ因子B.α因子C.β因子D.ρ因子E.γ因子2.转录的含义是A.以DNA为模板合成DNA的过程B.以DNA为模板合成RNA的过程C.以RNA为模板合成RNA的过程D.以RNA为模板合成DNA的过程E.以DNA为模板合成蛋白质的过程3.关于DNA指导的RNA聚合酶,下列说法错误的是A.以DNA为模板合成RNAB.是DNA合成的酶C.以四种NTP为底物D.催化3‘,5‘–磷酸二酯键的形成E.没有DNA时,不能发挥作用4.关于DNA聚合酶和RNA聚合酶,下列说法正确的是A.都以dNTP 为底物B.都需要RNA引物C.都有3‘→5‘核酸外切酶活性D.都有5‘→3‘聚合酶活性E.都有5‘→3‘核酸内切酶活性5.关于DNA复制和转录,下列说法错误的是A.都以DNA为模板B.都需核苷酸作原料C.遵从A—T配对,G—C配对D.都需依赖DNA的聚合酶E.产物都是多核苷酸链10.下列对转录的描述错误的是A.RNA链延伸方向5‘→3‘B.转录多以一条DNA链为模板C.合成的RNA都是前体D.转录延长过程中RNA聚合酶是全酶E.真核生物的结构基因是断裂的,有些基因的顺序不表达在相应的mRNA中15.关于真核生物的RNA聚合酶,下列说法错误的是A.RNA聚合酶Ⅰ的转录产物是45S–rRNAB.RNA聚合酶Ⅱ转录生成hnRNAC.利福平是其特异性抑制剂D.真核生物的RNA聚合酶是由多个亚基组成E.RNA聚合酶催化转录时,还需要多种蛋白质因子16.在真核生物中,RNA聚合酶Ⅲ催化的转录产物是A.tRNA、5–rRNA和nRNAB.hnRNAC.28–rRNAD.5.8–rRNAE.nRNA21.关于真核生物mRNA的聚腺苷酸尾巴,错误的说法是A.是在细胞核内加工接上的B.其出现不依赖DNA模板C.维持mRNA作为翻译模板的活性D.先切除3‘末端的部分核苷酸然后加上去的E.直接在转录初级产物的3‘末端加上去的22.关于外显子和内含子叙述错误的是A.外显子是基因中编码序列,并表达为成熟RNA的核酸序列B.外显子能转录,内含子不能转录C.去除内含子,连接外显子的过程叫拼接D.基因中外显子加内含子的长度相当于hnRNA的长度E.基因中外显子和内含子相互间隔排列37.大肠杆菌RNA聚合酶的亚基中Aα亚基用于识别不同的启动子Bβ亚基用于识别不同的启动子Cβ'亚基用于识别不同的启动子Dω亚基用于识别不同的启动子Eσ亚基用于识别不同的启动子38.大肠杆菌RNA聚合酶的亚基中Aα亚基执行聚合反应,催化磷酸二脂键形成Bβ亚基执行聚合反应,催化磷酸二脂键形成Cβ'亚基执行聚合反应,催化磷酸二脂键形成Dω亚基执行聚合反应,催化磷酸二脂键形成Eσ亚基执行聚合反应,催化磷酸二脂键形成39.转录是A.以DNA双链中一股单链为模板B.以DNA双链为模板C.以RNA链为模板D.以编码链为模板E.以前导链为模板49.下列关于mRNA的叙述正确的是A.在三类RNA中分子量最小B.由大小两个亚基组成C.更新最快D.占RNA总量的85%E.含大量稀有碱基50.比较RNA转录与DNA复制,叙述错误的是A.都以DNA为模板B.都依赖DNA的聚合酶C.原料都是dNTPD.新链延伸方向都是5′→3′E.都遵从碱基配对规律51.RNA的剪接作用A.仅在真核发生B.仅在原核发生C.真核原核均可发生D.仅在rRNA发生E.以上都不是52.原核生物经转录作用生成的mRNA是A.内含子B.单顺反子C.多顺反子D.插入序列E.间隔区序列53.真核mRNA后加工的顺序是A.带帽.运输出细胞核、加尾、剪接B.带帽、剪接、加尾、运输出细胞核C.剪接、带帽、加尾、运输出细胞核D.带帽、加尾、剪接、运输出细胞核E.运输出细胞核、带帽、剪接、加尾54.转录真核细胞rRNA的酶是A.RNA聚合酶ⅠB.RNA聚合酶ⅡC.RNA聚合酶ⅢD.RNA聚合酶Ⅰ、ⅢE.RNA聚合酶Ⅱ、Ⅲ55.RNA病毒的复制由哪一种酶催化A.RNA聚合酶B.RNA复制酶C.DNA聚合酶D.反转录酶E.核酸酶56.DNA复制与RNA转录中的不同点是A.合成体系均需要酶和多种蛋白因子B.新生子链合成方向均为5′→3′C.聚合过程都是核苷酸间生成磷酸二酯键D.RNA聚合酶缺乏校正功能E.遵从碱基配对规律57.以下反应属于RNA编辑的是A.转录后碱基的甲基化B.转录后产物的剪接C.转录后产物的剪切D.转录产物中核苷酸残基的插入、删除和取代E.以上反应都不是58.以下对tRNA合成的描述,错误的是A.RNA聚合酶Ⅲ参与tRNA前体的生成B.tRNA前体加工除去5′和3′端多余核苷酸C.tRNA前体中含有内含子D.tRNA3′端需添加ACC-OHE.tRNA上有些碱基还需进行特征性修饰【单项选择】参考答案1.D2.B3.B4.D5.C6.C7.B8.C9.A10.D11.C这个题目的问题是一个同学提出来的,值得表扬。
人教新版初三上学期《第3章物质构成的奥秘》练习题(综合)
人教新版初三上学期《第3章物质构成的奥秘》练习题(综合)单元练习(综合)一.选择题(共14小题)1.金鱼在凉开水中不能存活,是因为在凉开水中含量不足的是()A.氧元素B.氧原子C.氧分子D.氧离子2.我国已启动“嫦娥工程”探月打算,拟于2021年前发射绕月飞行的探测卫星﹣﹣“嫦娥一号”.人类探月的重要目的之一是勘察、猎取地球上蕴藏量专门小而月球上却极为丰富的核聚变燃料﹣﹣“C﹣13”是指原子核内含有6个质子、7个中子的碳原子,则“He﹣3”所指的氦原子核内()A.含有3个质子,没有中子 B.含有2个质子,1个中子C.含有1个质子,2个中子D.含有3个中子,没有质子3.能证明分子在化学变化中能够再分的变化是()A.碘受热变碘蒸汽B.一定条件下水制成冰块C.块状的硫碾磨成硫粉D.加热氧化汞产生氧气4.据报导,科学家发觉了新的氢微粒。
这种氢微粒由3个氢原子核(只含质子)和2个电子构成。
下列说法中正确的是()A.该微粒不显电性B.该微粒比一个一般H2分子相对分子质量多3C.该微粒可用H3﹣表示D.该微粒比一个一般H2分子多一个氢原子核5.下列操作或现象与分子对应的性质不一致的是()选项操作或现象分子的特性A给篮球充气分子间有间隔B在花园中可闻到花的香味分子是运动的C5mL酒精和5mL水混合在一起,体积小于分子是有质量的10mLD蜡烛燃烧生成二氧化碳和水分子是能够再分的A.A B.B C.C D.D6.如图所示,在a烧杯中盛有酚酞溶液,在b烧杯中盛有浓氨水,有关现象和说明正确的是()A.现象:a烧杯溶液变红色,说明:b中氨分子运动到a中,氨气溶于水B.现象:a烧杯溶液变红色,说明:氨气与酚酞反应生成红色物质C.现象:b烧杯溶液为无色,说明:b中氨分子没有运动D.现象:b烧杯溶液为无色,说明:a中酚酞分子运动到b中,氨水呈中性7.科学家最近制造出第112号新元素,其原子的相对原子质量为277.关于该元素的下列叙述正确的是()A.其原子核内中子数和质子数差不多上112B.其原子核内中子数为277,核外电子数为112C.其原子质量是碳﹣12原子质量的277倍D.其原子质量与碳﹣12原子质量之比为277:128.青青年正处于成长期,需要摄入足量钙,那个地点的“钙”是指()A.元素 B.分子C.原子D.单质9.每100 克黑木耳中含铁185 毫克,它比绿叶蔬菜中含铁量最高的菠菜高出20 倍,比动物性食品中含铁量最高的猪肝还高出约7 倍,是各种荤素食品中含铁量最多的。
资料:第三章 补充练习题
第三章补充题库一、Choice Questions1. The principle function of the International Monetary Fund was originally toA. lend to member nations experiencing a shortage of foreign exchange reserves.B. finance postwar reconstruction, particularly in Europe and Japan.C. reduce trade barriers and settle disputes among countries relating to currencynegotiations.D. act as a supranational regulatory agency for domestic central banks.2. The U.S. dollar today is an example of aA. commodity money.B. fiat money.C. commodity-backed money.D. currency basket.3. The gold standard was in place for most major economics of the world during theperiodA. from the beginning of the Great Depression until World War II.B. from 1973 until the present.C. from the mid-1870s until World War I.D. since the end of World War II.4. Under the gold standard, if the mint parity condition for the French franc was set atFfr107.1 per ounce of gold, and the German mark was set at DM88.7 per ounce of gold, then it is possible to compute the exchange rate between the German mark and the French franc (DM/f) as approximatelyA. 0.83.B. 1.21.C. 9490.9D. It is not possible to compute the exchange rate between the mark and the francwith these values, because these values are relative to the price of gold.5. Under the Bretton Woods system, most of the major currencies of the system, otherthan the U.S. dollar,A. pegged their values against the value of an ounce of gold.B. pegged their values against the value of the dollar.C. allowed their currencies to float.D. pegged their values against the value of the Euro.6. Which of the following is not an institution that arose under the Bretton WoodsAgreement?A. The International Monetary Fund.B. The International Bank for Reconstruction and Development.C. The League of Nations.D. The General Agreement on Tariffs and Trade.7. In the Plaza Agreement of September, 1985, the "Group of Five" or "G5" countriesannounced that they believed thatA. the "G5" needed to be expanded to include an additional five major industrializedcountries to make up what is now referred to as the "G10".B. the Bretton Woods system would no longer be sustainable.C. it was necessary to "float" the dollar relative to gold.D. the exchange value of the dollar was too strong and that they would coordinatetheir central bank interventions in order to drive down the value of the dollar.8. Which of the following is not a member of the "Group of Ten" industrializedcountries?A. JapanB. BelgiumC. SwitzerlandD. Sweden9. An example of a country that maintained a crawling-peg exchange-rate system duringthe early 2000s isA. the United States.B. Canada.C. France.D. Nicaragua.10. A currency board is anA. exchange market in which the major currencies of the world are exchanged on theopen market among private banks at prevailing rates.B. independent monetary agency that substitutes for a central bank by pegging thegrowth of the domestic money stock to the foreign-exchange holdings of the board.C. independent monetary agency which is responsible for setting bank reserverequirements for the domestic currency.D. exchange market in which the notes and bills issued by the domestic governmentare traded on the open market among private banks.11. Which of the following arrangements places the greatest restriction on policymakersand requires the greatest sacrifice of policy autonomy?A. a free float.B. a currency-basket peg.C. dollarization.D. a currency board.12. If the Chinese renminbi, whose unit of currency is the yuan, is revalued relative to theU.S. dollar, thenA. for each yuan, one can expect to buy fewer dollars.B. for each yuan, one can expect to buy more dollars.C. the exchange rate between the renminbi and the dollar will remain constant.D. it is impossible to tell what will happen to the exchange rate or the number ofdollars that one can buy with each yuan, since this depends on the supply and the demand for the yuan relative to the dollar.13. Currency basket pegs usually involve pegging the domestic currency toA. each of the major currencies of the world.B. the relative price of a chosen basket of consumer goods.C. a weighted average of only a small selected number of different currencies.D. within an upper and lower limit of a band relative to either the U.S. dollar or theJapanese yen.14. A "dirty float" exchange rate system refers toA. an exchange rate system wherein policymakers allows the value of the domesticcurrency to be determine only by the forces of supply and demand.B. an exchange rate system whereby each of the members of the system peg theircurrency against one of the major currencies, such as the U.S. dollar, which is in turn pegged against a commodity, such as gold.C. an exchange rate arrangement in which the domestic currency is primarilymanaged by the central bank of a foreign country, which is typically the major trading partner.D. an exchange rate arrangement in which a nation allows the international value ofits currency to be primarily determined by market forces, but intervenes occasionally to stabilize its currency.15. The G7/G8 economic summits were started by:A. Russian President Bois YeltsinB. U.S. President Gerald FordC. U.S President William Clinton.D. French President Valery Giscard d’Estaing.16. The Smithsonian agreement refers toA. the understanding that most of the major industrialized countries came to at the end ofthe depression era regarding the non-sustainability of the gold standard as it was currently practiced.B. an agreement that was made to establish new par values for the G10 countries thathad participated in the Bretton Woods system.C. the agreement that President Nixon came to vis a vis the other G10 members tosuspend temporarily the convertibility of the dollar into gold or other reserve assets.D. the initial agreement that set forth future economic cooperation under the NorthAmerican Free Trade Association.17. Which of the following did not contribute to the eventual collapse of the BrettonWoods system?A. Increased federal spending for social programs termed the "Great Society" underthe Johnson administration.B. Heightened U.S. involvement in Vietnam.C. The conditions set forth in the Louvre Accord.D. U.S. balance of payments deficits with Germany and Japan.18. A Monetary Order isA. a set of rules that determine the international value of a currency.B. a set of laws and regulations that establishes the framework within whichindividuals conduct and settle transactions.C. an exchange rate arrangement in which a country pegs the international value ofthe domestic currency relative to the currency of another nation.D. a specialized form of a currency board.19. Which of the following is an example of a commodity money that has been usedextensively?A. Credit cards.B. Demand deposits and certificates of deposit.C. Gold.D. The Euro.20. A key challenge to the exchange rate system of the leading industrialized countries inthe 1970s came as a consequence ofA. a rapid increase in the price of petroleum.B. the formation of the Bundesbank.C. the heightening of Cold War tensions.D. the establishment of many currency baskets worldwide.21. The IMF constitution was amended to allow member nations to determine their ownexchange rate arrangements under theA. Smoot-Hawley Act.B. Jamaica Accord.C. Smithsonian Agreement.D. Treaty of Rome.22. An advantage of the gold standard system was thatA. it promoted conditions that helped to avoid periodic financial and bankinginstability.B. it did not require countries to have central banks.C. it ensured short-run stability of the monetary stock.D. minting and transportation costs were relatively low as compared to otherexchange rate systems.23. The gold window for the U.S. dollar was closed byA. President Ronald Reagan.B. President James Carter.C. Federal Reserve Chairman Paul V olker.D. President Richard Nixon.24. An advantage of a flexible exchange rate system is that itA. can help a country overcome external shocks such as an unusual inflow of capitalfrom abroad.B. can reduce the volatility of nominal exchange rate over time.C. eliminates the need for the central bank to target interest rates.D. ensures a greater volume of trade in goods and services among member countries.25. Which of the following is a principal function of the International Monetary Fund?A. To act as a forum for international monetary cooperation.B. To provide central banks with a range of financial services for managing theirexternal reserves.C. T o act as a lender of last resort for countries facing temporary external balance ofpayments problems.D. To act as an agent or trustee which facilitates the implementation of variousinternational financial agreements.二、本章重要概念Exchange-Rate Systemmint parity rateThe Bretton Woods SystemTriffin DilemmaJamaica AccordsPlaza AgreementCurrency Board DollarizationSoft PegCurrency Basket PegFree FloatingFixed Exchange Rates system Floating Exchange Rates system。
人教版数学七年级上册 第3章 3.1 --3.3基础练习题含答案
人教版数学七年级上册第3章 3.1 --3.3基础练习题含答案3.1从算式到方程一.选择题1.若关于x的方程(k﹣2020)x﹣2019=7﹣2020(x+1)的解是整数,则整数k的取值个数是()A.6B.8C.9D.102.已知k位非负整数,且关于x的方程3(x﹣3)=kx的解为正整数,则k的所有可能取值为()A.4,6,12B.4,6C.2,0D.2,0,﹣6 3.下列四组变形中,变形正确的是()A.由x=2,得x=B.由2x﹣3=0得2x﹣3+3=0C.由5x=7得x=35D.由5x+7=0得5x=﹣74.关于x的一元一次方程2x a﹣1+m=2的解为x=1,则a﹣m的值为()A.5B.4C.3D.25.下列等式变形正确的是()A.若4x=2,则x=2B.若4x﹣2=2﹣3x,则4x+3x=2﹣2C.若4(x+1)﹣3=2(x+1),则4(x+1)+2(x+1)=3D.若=1,则3(3x+1)﹣2(1﹣2x)=66.下列等式变形不正确的是()A.由x+2=y﹣2,可得x﹣y=4B.由2x=y,可得x=yC.由﹣x=y,可得x=﹣5y D.由y﹣x=﹣2,可得x=y+27.如图,两个天平都平衡,则六个球体的重量等于()个正方体的重量.A.7B.8C.9D.108.已知(a≠0,b≠0),下列变形错误的是()A.B.3a=4b C.D.4a=3b9.运用等式性质进行的变形,正确的是()A.若x=y,则=B.若=,则x=yC.由4x﹣5=3x+2,得到4x﹣3x=﹣5+2D.若a2=3a,则a=310.下面是一个被墨水污染过的方程:3x﹣2=x﹣,答案显示此方程的解是x=2,被墨水遮盖的是一个常数,则这个常数是()A.2B.﹣2C.D.二.填空题11.已知关于x的方程2﹣(a﹣1)x|a|=0是一元一次方程,则a=.12.已知方程(m﹣1)x|m|﹣5=0是关于x的一元一次方程,则m的值为.13.已知关于x的一元一次方程+3=2020x+m的解为x=2,那么关于y的一元一次方程+3=2020(1﹣y)+m的解y=.14.设“●■▲”分别表示三种不同的物体,如图所示,前两架天平保持平衡,如果要使第三架天平也平衡,那么“?”处应该放“■”的个数为.15.如果(a+3)x|a|﹣2=3是一元一次方程,那么a=.三.解答题16.关于x的方程x﹣2m=﹣3x+4与2﹣x=m的解互为相反数.(1)求m的值;(2)求这两个方程的解.17.已知x=﹣2是关于x的方程a(x+3)=a+x的解,求代数式a2﹣2a+1的值.18.【定义】若关于x的一元一次方程ax=b的解满足x=b+a,则称该方程为“友好方程”,例如:方程2x=﹣4的解为x=﹣2,而﹣2=﹣4+2,则方程2x=﹣4为“友好方程”.【运用】(1)①﹣2x =,②x =﹣1两个方程中为“友好方程”的是 (填写序号); (2)若关于x 的一元一次方程3x =b 是“友好方程”,求b 的值;(3)若关于x 的一元一次方程﹣2x =mn +n (n ≠0)是“友好方程”,且它的解为x =n ,则m = ,n = .19.我们规定,若关于x 的一元一次方程ax =b 的解为a +b ,则称该方程为“合并式方程”,例如:3x =﹣的解为﹣,且﹣,则该方程3x =﹣是合并式方程.(1)判断x =1是否是合并式方程并说明理由;(2)若关于x 的一元一次方程5x =m +1是合并式方程,求m 的值.3.2解一元一次方程(一)—合并同类项与移项一、选择题1.下列各方程中,合并同类项正确的是( )A .由3x -x =-1+3,得2x =4B .由23x +x =-7-4,得53x =-3 C .由52-13=-x +23x ,得136=13x D .由6x -4x =-1+1,得2x =0 2.下列变形一定正确的是( )。
第3章习题解答
《数字通信系统原理》教材习题解答第三章练习题33-1 填空(1)模拟信号在数字通信系统中的传输,首先必须把模拟信号转变为 数字信号 ,转换的方法有 脉冲编码调制 和增量调制等。
(2)衡量量化性能好坏的常用指标是 量化信噪比。
此值越大,说明量化性能越 好。
(3)非均匀量化的PCM 中,信号功率小时,量化噪声功率 小,适用于动态范围较宽的信号。
(4)目前,数字通信系统中采用两种压扩特性:一种是A 律压扩特性:另一种是 μ律压扩特性 。
(5)采用增量调制的目的是 简化模拟信号的数字化方法 :采用自适应增量调制的目的是 提高小信号的量化信噪比。
补充题:1.线性PCM 的量化噪声与信号功率大小有关吗?无关,它适用于动态范围小的信号。
2.在对数PCM 中,量化噪声功率与信号功率的定性关系是信号功率小,量化噪声功率就小,适用于动态范围大的信号。
在对数M 中,信号在某一段落内变化时,量化噪声功率是否变化?不变。
3.在对数PCM 和自适应增量调制中,抗噪声能力强的是自适应增量调制,量化噪声小的是对数PCM 。
4.均匀量化器的量化信噪比与编码位数的关系是编码增加1位,量化信噪比增大6dB ,非均匀量化器可以提高小信号的量化信噪比。
5.若A 律13折线PCM 编码器输入信号为直流且幅度等于最小量化间隔的1.5倍,则编码器的输出为10000001。
6.线性PCM 编码器的抽样信号频率为8kHz ,当信息速率由80kbit/s 下降到56kbit/s 时,量化信噪比增大18dB 。
3-2 试画出PCM 通信的原理图,并简述PCM 通信的过程。
3-3 PAM 信号、量化信号和PCM 信号属于什么类型的信号?3-4 对基带信号t t t g ππ4cos 3cos 2)(+=进行理想抽样。
(1)为了在接收端不失真地从已抽样信号中恢复出,怎样选取抽样间隔?(2)若抽样间隔为0.2s ,试画出已抽样信号的频谱。
解:(1)基带信号可以看成是低通信号,由于Hz f m 2=根据抽样定理,得Hz f f m s 42=≥(2)由已知得,抽样频率为Hz f s 52.01==。
八年级物理上册第三章第一节习题精选试题
第三章第一节?运动与静止?补充练习题制卷人:打自企;成别使;而都那。
审核人:众闪壹;春壹阑;各厅……日期:2022年二月八日。
一、填空题:1、“月亮在云里穿行〞,这句话中,月亮是______,云是______.〔填“研究对象〞或者“参照物〞〕2、组织同学们乘车去华山游览。
车快开到华山时,甲同学大声说“我终于来到了华山!〞乙同学那么说:“华山终于来到了我面前!〞这两种说法中,甲同学是以__________为参照物(选填“华山〞或者“汽车〞),甲、乙同学不同说法表达了运动和静止具有____________性.3、公路上,甲、乙、丙三个同学骑自行车向东行驶,甲感觉顺风,乙感觉无风,丙感觉逆风,由此可断定当时刮的是 ______风.假如以乙为参照物,甲向______运动,丙向______运动.4、坐在运动着的直升飞机中的人,看到楼房顶部竖直向上运动,此时这人是以 ______为参照物来描绘运动的.假设以地面为参照物,飞机做的是_________的运动〔填“向上〞或者“向下〞〕5、“小小竹排江中游,巍巍青山两岸走〞这句歌词中,竹排“游〞是以为参照物,青山“走〞又是以为参照物。
6、小明同学放学回家,正碰上刮风下雨,他以18km/h的速度由西向东快跑。
此时他发现了奇怪的现象,雨滴成竖直下落状态。
请你确定这时刮的是风_____〔填“东〞、“西〞、“南〞或者“北〞〕,风速是________m/s。
7、甲、乙两辆汽车分别在平直的公路上作匀速直线运动且方向一样,以甲车为参照物,乙车向东运动,假设以地面为参照物,那么乙车向________运动,且v Z_____v甲(选填“大于〞、“等于〞或者“小于)。
8、一千多年前,唐朝的大诗人李白曾在感慨长江的壮美景观:“中断楚江开,碧水东流至此回。
两岸青山相对出,孤帆一片日边来〞。
从物理学的角度看,“两岸青山相对出〞和“孤帆一片日边来〞所选的参照物分别是和。
9、小明在马路边上拍街景照片,先后连续拍了两张,如图甲、乙所示,假如以小轿车为参照物,大卡车向运动〔选填“左〞或者“右〞〕;假设以地面为参照物,运动速度最慢的物体是〔选填“小轿车〞、“大卡车〞或者“自行车〞〕二、选择题1、夜晚看见月亮在云层里以很快的速度向东运动,这是因为我们选定的参照物是( )A地球; B.向东运动的月亮; C.向东运动的云层; D.向西运动的云层.2、在无风的雨中,坐在自东向西行驶的汽车中的乘客看到雨点的运动方向是〔〕A.向下的 B.向下偏东 C.向下偏西 D.不知车速无法确定3、寒假期间,新版?西游记?在许多电视台热播,我们从电视画面上看到的孙悟空腾云驾雾,其实在拍摄过程中,是演员在原地做着各种姿势,快速挪动后面的背景形成的。
第三章练习题
第三章《营养学基础》习题第一部分:营养学概论、能量及宏量营养素一、单项选择题,请将最佳答案填入题后的括号内。
1.各种体力活动所消耗的能量约占人体总能量消耗的()A.10~15% B. 10~20% C. 10~30% D. 15~30%2.随机个人的营养素摄入水平达到RNI时,发生摄入不足的概率在()以下:A.10% B. 6% C. 3% D. 1%3. 明确提出食疗概念的医学家是()A.孙思邈 B. 张仲景 C. 忽思慧 D. 崔昊4. 营养化学奠基人之一()最早确立了食物组成与物质代谢的概念。
A.拉瓦锡 B. 李比希 C. 芬克 D. 卡尔宏邦5. 到目前为止,我国共开展了()次全国性膳食营养调查。
A.6次 B. 5次 C. 4次 D. 3次6. 我国第一部《膳食营养素参考摄入量》是于()年公布的。
A.1989 B. 1997 C. 2000 D. 20027.人体食物脂肪的主要运输形式为()A.高密度脂蛋白 B. 低密度脂蛋白 C. 极低密度脂蛋白 D.乳糜微粒8.碳水化合物占总能量的适宜百分比为()A、12-14%B、25-30%C、55-65%D、30-50%9.一成年男性,身高175cm,体重70kg,其身体质量指数为()A、 B、 C、D、10.蛋白质的基本组成单位是()A、脂肪酸B、氨基酸C、肽D、葡萄糖醛酸11.优质蛋白质主要来源于()A、谷类B、肉蛋奶类C、蔬菜D、水果12.酪氨酸在体内可由哪种必需氨基酸转化而成()A、色氨酸B、癞氨酸C、苯丙按酸D、蛋氨酸13.蛋白质功效比值是表示蛋白质的()A表现消化率 B利用率 C蛋白质含量 D净利用率14.下列属于半完全蛋白的是()A小麦中的麦谷蛋白 B小麦中的麦胶蛋白 C玉米中的谷蛋白 D玉米中的玉米胶蛋白15.食物特别动力作用是由于()而产生的。
A.咀嚼动作引起咀嚼肌运动产热B.食物本身所具有的热能和温度C.摄食吞咽动作产生的D.是机体由于摄取食物而引起体内能量消耗增加的现象。
第3章 汇编语言程序设计补充习题
MCS—51单片机指令系统及程序设计一.单项选择题,从供选择的答案中选出正确的选项,并将标号填写到答题册中,每小题1分。
1.MCS-51汇编语言指令格式中,唯一不可缺少的部分是B。
A.标号 B.操作码 C.操作数 D.注释2.MCS-51的立即寻址方式中,立即数前面D。
A.应加前缀“/:”号B.不加前缀号C.应加前缀“@”号D.应加前缀“#”号3.下列完成8031单片机内部RAM数据传送的指令是D。
A.MOVX A,@DPTR B.MOVC A, @A+PCC.MOV A,#data D.MOV direct,direct4.MCS-51的立即寻址的指令中,立即数就是B。
A.放在寄存器R0中的内容 B.放在程序中的常数C.放在A中的内容D.放在B中的内容5.单片机中PUSH和POP指令常用来C。
A.保护断点B.保护现场C.保护现场,恢复现场 D.保护断点,恢复断点6.MCS-51寻址方式中,操作数Ri加前缀“@”号的寻址方式是A。
A.寄存器间接寻址 B.寄存器寻址 C.基址加变址寻址 D.立即寻址7.MCS-51寻址方式中,位寻址的寻址空间是 D 。
A.工作寄存器R0~R7 B.专用寄存器SFR C.程序存贮器ROMD.片内RAM的20H~2FH字节中的所有位和地址可被8整除的SFR的有效位8.MCS-51寻址方式中,直接寻址的寻址空间是D。
A.工作寄存器R0~R7 B.专用寄存器SFR C.程序存贮器ROM D.片内RAM 00H--7FH的128字节和80H--FFH中的SFR9.执行指令MOVX A,@DPTR时,、脚的电平为C。
A.高电平,高电平 B.低电平,高电平C.高电平,低电平 D.低电平,低电平10.主程序执行完ACALL后返回主程序后,堆栈指针SP的值A。
A.不变 B.加2 C.加4 D.减211.单片机中使用MOVX A,@R1指令 C 寻址数据存贮器1050H单元。
A.能直接B.不能C.与P2口配合能 D.与P1口配合能12.下列指令判断若P1口最低位为高电平就转LP,否则就执行下一句的是B。
第三章练习题
第三章练习题第3章资本和剩余价值一、填空题1.资本总公式的矛盾是指资本总公式在形式上与存在的矛盾。
2.资本主义生产过程是和的统一。
3.生产资本划分为不变资本和可变资本的依据是。
4.准确反映资本家对工人剥削程度的是。
5.剩余价值生产的两种基本方法是和。
6.工人的劳动时间分为和两部分。
7.资本主义工资的本质是的转化形式。
8.剩余价值既不能,又不能,这就是解决资本总公式矛盾的条件。
9. 是货币转化为资本的前提。
10. 的生产是以整个社会劳动生产率的提高为条件的。
二、单项选择题1.资本总公式是()A G-W-GB G-W-G’C W-G-WD W-G-W’2.劳动力商品使用价值的重要特点是()A 劳动力的使用价值可以还原为生活资料价值B 它包括历史和道德的因素C 它是价值的源泉并且是大于自身价值的价值源泉D 它会随着社会经济的发展而不断扩大物质内容3.超额剩余价值的源泉是()A 工人的必要劳动B 工人的剩余劳动C 绝对价值D 相对剩余价值4.相对剩余价值的获得是()A 绝对延长工作日的结果B 提高工人劳动强度的结果C 个别企业提高劳动生产率的结果D 整个社会劳动生产率提高的结果5.绝对剩余价值生产是()A 个别资本所有者首先改进技术的结果B 全社会劳动生产率提高的结果C在必要劳动时间不变时延长工作日的结果D 缩短必要劳动时间相应地延长剩余劳动时间的结果6.机器设备等物成为资本是因为它们是( )。
A生产劳动产品的条件 B生产商品的条件C生产剩余价值的条件 D生产物质财富的条件7.个别资本家提高劳动生产率的目的和直接动机是( )。
A追逐相对剩余价值 B追逐绝对剩余价值C降低劳动力价值 D追逐超额剩余价值8.剩余价值率反映的是( )。
A预付资本的增殖程度 B可变资本的增殖程度C资本家对工人的剥削程度 D不变资本的增殖程度9.“剩余价值不能从流通领域中产生,但又不能离开流通领域而产生”,这句话的意思是()A剩余价值可以在流通中产生,也可以在生产中产生B剩余价值既不在流通中产生,也不在生产中产生C剩余价值在生产中产生,但要以流通为条件D剩余价值在流通中产生,但要以生产为条件10.分析资本主义社会工资水平及其变动趋势主要看()A货币工资 B实际工资 C相对工资 D国民工资11.在商品生产过程中,生产资料的价值是借助于生产者的()A具体劳动而增殖大 B具体劳动而转移到新产品中去C抽象劳动而增殖 D抽象劳动而转移到新产品中去12.个别企业加强工人劳动强度而多得的那部分价值属于()A剩余价值外负担 B绝对剩余价值C相对剩余价值 D超额剩余价值13.某资本家经营的企业通过改进技术、提高劳动生产率,使其生产商品花费的劳动时间比社会必要劳动时间少10%,由此形成商品个别价值低于社会价值的那部分是()A超额剩余价值 B绝对剩余价值 C必要剩余价值 D个别剩余价值14.某资本家经营的企业通过强化管理,在工作日不变的情况下,工人的必要劳动时间缩短了20%,假定其他条件不变,则该企业的剩余价值率将由100%()。
第三章补充练习题
1.如下图所示二极管环形电桥检波测量电路,UP 为恒压信号源,C1和C2是差动式电容传感器,C0是固定电容,其值C0>>C1、C0>>C2,设二极管D1~D4正向电阻为零,反向电阻为无穷大,信号输出经低通滤波器取出直流信号ABe。
要求:⑴分析检波电路测量原理;⑵求桥路输出信号e AB=f(C1,C2)的表达式;⑶画出桥路中VA 、VB、e AB各处在=在C1=C2,C1>C2,C1<C2三种情况下电压波形图。
(提示:画出UP正半周等效电路图,标出工作电流流程即可求出e AB表达式)2. 如下图为二极管环形检波测量电路。
C1和C2为差动式电容传感器,C3为滤波电容,RL 为负载电阻。
R为限流电阻。
UP是正弦波信号源。
设RL很大,并且C 3>>C1,C3>>C2。
⑴试分析此电路工作原理;⑵画出输出端电压UAB在C1=C2、C1>C2、C1<C2三种情况下波形图;⑶推导),(21CCfU AB 的数学表达式。
3.1 、L 1是差动电感,D 1 ~D 4是检波二极管(设其正向电阻为零,反向电阻为无穷大),C 1是滤波电容,其阻抗很大,输出端电阻R 1 =R2 = R ,输出端电压由C 、D 引出e CD ,U p 为正弦波信号源。
求:(1)分析电路工作原理(即指出铁心移动方向与输出电压e CD 极性的关系)。
(2)分别画出铁心上移及下移时流经电阻R 1和R 2的电流i R1和i R2及输出端电压e CD 的波形图。
4. 用一涡流式测振仪测量某机器主轴的轴向振动。
已知传感器的灵敏度为20mV/mm ,最大线性范围为5mm 。
现将传感器安装在主轴两侧,如下图(a )所示,所记录的振动波形如图(b )所示。
请问:(1)传感器与被测金属的安装距离L 为多少时测量效果较好? (2)轴向振动的最大值A 为多少? (3)主轴振动的基频f是多少?。
概率论第三章补充练习答案
《概率论》第三章 练习答案一、填空题:1.设随机变量ξ与η相互独立且具有同一分布律:则随机变量ηξζ+=的分布律为: 。
2.随机变量ξ服从(0,2)上均匀分布,则随机变量ξη2=在(0,4)的密度函数为⎪⎩⎪⎨⎧=041)(yy f η 其他4〈〈y o )()()()()()()()()(,0)20(,21)(),2,0(~2y F y F y p y p y y p y p y p y F f U --=-≤-≤=≤≤-=≤=≤=⎪⎩⎪⎨⎧<<=ξξηξξξξηξξξ其他yyO yy F y f 41212121)()(/=∙+∙==ηη3.设x 表示10次独立重复射击命中目标的次数,每次射中的概率为0.4,则x 2的数学期望E (x 2) = DX+(EX )2=2.4+16=18.4 。
4.2,4),4.0,10(~===npq DX EX b X 则4.设随机变量x 服从 [1, 3 ] 上的均匀分布,则E (X1)=⎰=∙32121113Ln dx x5.设DX =4,DY =9,P XY =0.5,则D (2x – 3y) =4Dx+9Dy-2cov(2x,3y)=61 。
3),cov(,32),cov(5.0=∴⨯==Y X Y X ρ6.若X 与Y 独立,其方差分别为6和3,则D(2X -Y)=___27_______。
),cov(44)2(Y X DY DX Y X D -+=-二、单项选择:1.设离散型随机变量(ηξ,)的联合分布律为:若ξ与η独立,则α与β的值为: ( A ) A .α=92,β=91 B .α=91,β=92C .α=61,β=61D .α=185,β=18131)311819161(1=+++-=+βα还原为(ηξ,):2. 设(X ,Y )是一个二元随机变量,则X 与Y 独立的充要条件是:( D ) A 、 cov (X,Y )= 0 B 、)()(i j i ij X Y P X P P = C 、 P = 0 D 、j i ij P P P ⨯=3.已知(X ,Y )的联合密度为=)(x ϕ 04xy其它1,0≤≤y x ,则F (0.5,2)=( B )A 、0B 、0.25C 、0.5D 、0.1{})(41442,5.025.015.005.001利用图像),(===≤≤=⎰⎰⎰⎰ydy xdx xydxdy Y X P F4.如果X 与Y 满足D (X +Y )=D (X -Y ),则必有 ( )A .X 与Y 独立B .X 与Y 不相关C .D (Y )=0 D .D (X )D (Y )=0BEY Y EX X E 故选),())((00cov 0=⇒=⇒=--ρηξ5.对任意两个随机变量X 和Y ,若E (X ,Y )=E (X )E (Y ),则( B )A .D (XY )=D (X )D (Y )B .D (X +Y )=DX +DYC .X 和Y 独立D .X 与Y 不独立6.设DX =4,DY =9,P XY =0.5,则D (2X -3Y )=____。
第3章第1-7节《程序设计初步》补充习题
第3章程序设计初步第1-7节一、选择题1、下列有关算法的说法错误的是( B )。
A) 为解决一个问题而采用的方法和步骤称为算法。
B) 排序算法为数值运算算法。
C) 算法的每个步骤应具有确定性和有效性。
D) 一个算法应有一个或多个输出。
2、流程控制语句的基本控制结构有三种,不属于这一种结构的是( D )。
A) 顺序结构 B) 选择结构 C) 循环结构 D) 计算结构3、输入输出格式控制是在哪个头文件中定义的?( B )。
A) iostream B) iomanip C) istream D) ostream4、输入输出时需要包含哪个头文件?( A )。
A) iostream B) iomanip C) istream D) ostream5、对于语句cout<<endl<<x;中的各个组成部分,下列叙述中错误的是( D )。
A)”cout”是一个输出流对象 B) ”endl”的作用是输出回车换行C)”x”是一个变量 D) ”<<”称作提取运算符6、下列有关输入、输出的说法错误的是( C )。
A) C++语言本身没有提供输入、输出语句B) C++语言中输入输出功能是通过输入输出流库中的流对象cin和cout实现的C) 函数putchar()只能用来输出字符型变量的内容D) 函数getchar()一次只能输入一个字符7、在ios中提供控制格式的标志位中,哪个是转换为十六进制形式的标志位?(A )。
A) hex B) oct C) dec D) left8、关系表达式的求值结果为( B )。
A)1或-1 B)1或0 C)任意正整数或0 D)任意负整数或09、设变量a和b的值满足a<b,判断变量x的值是否位于闭区间[a,b]的表达式是(C )。
A)a<x<y B)a<=x<=b C)x>=a&&x<=b D)a<x&&x<b10、能正确表示逻辑关系:“a≥10或a≤0”的C语言表达式是( D )。
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第三章
1 在SQL中,关系模式称为_____,子模式称为_______,元组称为_______,属性称为_______.
2 在SQL中,表有3种: _________、_________和________.
2在基本表中,“主键“概念应该体现其值的________和_________两个特性。
3 操作“元组IN (集合)“的语义是________________.
4 表达式中的通配符“%“表示___________,” _”表示________________.
5 操作“元组>some (集合)”的语义是______________________.
6操作“元组<all (集合)”的语义是______________________.
7 操作“Not exists (集合)”的语义是______________________.
8操作“Not unique (集合)”的语义是______________________.
9 SQL有两种使用方式:____________和________________.
10 在SQL中,聚合函数COUNT(列名)用于
A计算元组的个数B计算属性的个数
C对一列中的非空值计算个数D对一列中的非空值何空值计算个数
11在SQL中,与“NOTIN”等价的操作符是
A=SOMEB<>SOME
C=ALLD<>ALL
12在SQL中,谓词EXISTS可用来测试一个集合是否
A有重复元组B有重复的列名
C为非空集合D有空值
13对基本表EMP(ENO,ENAME,SALARY,DNO),其属性表示职工的工号、姓名、工资和所在部门编号。
对基本表DEPT(DNO,DNAME),其属性表示部门的编号和名称。
有SQL语句:SELECT COUNT(DISTINCT DNO)
FROM EMP ;
其等价的查询语句是:
A 统计职工的总人数
B 统计每一部门的职工人数
C 统计职工服务的部门数目
D 统计每一职工服务的部门数目
14 SQL语句:
UPDA TE EMP
SET SALARY=SALARY* 1。
05
WHERE DNO=‘D6‘
AND SALARY< (SELECT A VG(SALARY)
FROM EMP);
其等价的修改语句为:
A 为工资低于D6部门平均工资的所有职工加薪5%
B 为工资低于整个企业平均工资的职工加薪5%
C 为在D6部门工作、工资低于整个企业平均工资的职工加薪5%
D为在D6部门工作、工资低于本部门平均工资的职工加薪5%
答案
1 基本表视图行列
2 基表视图临时表
3 唯一非空
4 与零个或多个字符组成的字符串匹配与单个字符匹配
5 若元组值大于集合中某一个值,则其值为true,否则为false
6 若元组值小于集合中每一个值,则其值为true,否则为false
7 若集合为空,则其值为true,否则为false
8 若集合中存在重复元组,则其值为true,否则为false
9 交互式SQL 嵌入式SQL
10 C
11 D
12 C
13 C
14 C。