2020年电大本科《工程数学》期末考试题库及答案

2020年电大本科《工程数学》期末考试题库及答案000100a00200100a1a（=1=1.若211103c=（10）．24521A,BA,Bn(AB)BA11k0k1且kA(k)AnD）．D.n若AA1135221AA0）A,B,Cn(ACB)1（D(B)CA111A,B,CnA(AB)A2ABB222x2x4x1x1231xx0x[11,2,2]为（C.232x2x33x2x3x2123xx6（133x3x423100130,1,0,2,01410111001,,01112341230101()(A)1秩⒌A与A秩AD）．D.,,,12s量9．设Anx是nPAP1B⒈A,B(AB)BA1A与BABU且AB32⒊103131A,B，命题（CA,BA,B(0p,则在31次的概率为（D.pp)pp)pp)322X~B(n,p)E(X),D(X)与p分别是（6,0.8）．n(x)a,b(ab)E(X)（Axf(x)dx7.设fX，Bsinx,0xf(x)20,其它(x)F(x)(a,b)P(aXb)Xfb(x)dxfaX10.设XE(X),D(X)2，当（CEY)0,DY)1．C.Y是34矩阵，B是52CB24ACBAB2.设是nA）3．设A,B为A．n354.47751（D）43AAAxxa121xa0a0，x(B．aaa123223ix1xa33nA）12128.若线性方程组的增广矩阵A,D214A秩（A）=n）成立时，n1102012310.向量组B3）0037（0,0,0）1,0,0）1,2,0）1,2,3），，,,,AD．AAA．P(AB)P()P(B)）1234234A与Bx5,x,,xN(H:514．设xU=（C．1/n12n0ABU且ABA，B1C）122932D）2513~N(,)X,X,XXXX2（3123ii113(X)23ii1N(,)TB,x,,xN(,),2xxx2（112n1,x,xN(,),2（223xxx12311111x111⒉2xA为34B25为ACB为5×4C11515A．01011212006340,B(AB)A31451834A,BA,B33AB32ABA1,B3AB)－3．121aAa0．012124022。033,AA12AO1A11O1OA．OA122x0x１12．x0x120,0,0,1,1,1121,2,3,1,2,0,1,0,0,0,0,0300xxx1121,0,233123,,1230,1,0,0,．21231,,,,,,．12s12sAX05(A)3２bXAX0X,Xb0123XkXkX．011229．若0IAA1A32.已知P(),P(B)A,BP(AB)，P(AB)．．PAA,BBAP(AB)P(AB)P(AB),P(A)pP(B)1P．A,BP()p,P(B)qP(AB)．pqP(),P(B)A,BP(AB)，P(AB)．0x0~U(0,F(x)XX01．xx1x18.若X9.若X~B(20,E(X)6．~N(,)P(X)2．2E[(XE(X))(YEY))](X,Y)的．．和和，．,x,,xN(,)2（24．设x12nxHH;U0．/0010n|xu（u011121x2A01，-1，2，-2．A22x1422．设A,B3AB3(B)8．13则设A,B3AB33B1则设A,B3AB32AB15．设4=B（）=1，那么=B3XXA6．设A为n阶方阵，若存在数nXAX为相应于特征值A,BP()0P(BA)0P(A)0.8,P(AB)0.5,P(AB)_____.X~（，（）=．11n,x,,x~(0,XNxxx~N)x．12nnini112n,x,,xX~N(,)xxD(x)11．设x=12nnini112．若P(A)0.8,P(AB)0.5P(AB)0.3．(X)2E(X)9D(2X)XE，20．24设XD(X)=3,则012若X则0.20.5aˆ设ˆ是_E().AB1．设,Bn()．112．设,B)．n(设,B(AB)Bn为4．设A,Bn为5．设A，BP(AB)P()P(B)A，B6．设A是mn是stBC是(snBA是ns是Bms）11A0，2，则3A(0，6)．11311120121AA)．1120113,x,,xN(,)2xxx10．设x11．设x55512n123x5,x,,xN(H:5U=（1/n12n0aaa1a1aa23332312．设bbb2ababab2（1ccc21c12c23c3121230123~P(X2)13．设X0.10.30.40.2,x,,xN(,)(,x14．设x2212n115．若AAA16．若（r(A)nnAXOABUAB且A，BX与YD(2XY)=（4D(X)9DY)12X19若X、X=B、AX=OX=B33121212X~N(Y3X2~（N()2A与BP(AB)P(A)P(B)X21112500x（3若X~N(2,Y，Y~N(0,．若（201x3若A,BP(AB)P()P(B)A与B5E(X)D(X)D(X)E(X2)[E(X)]2X和0AXb若AX0r(A)nnA,BABA与B123412341252若AA（1若A()A*（351234311234111111102022210,1,2,3,,,,3的秩是（4）．00037000412,[224],12],[23,1234121,0,2,2,3,5,1,2,11,3,2123123（N(,2)(x,x,,x)，t212n133~N(,)X,X,XXXX2（i．2123ii113(X)（3ii1,BP(AB)P()P(B)P(AB)ABBABAAA1113241616α,α,α,α(α,α,α,α)2r1234123410210111A,B10ABa（10a2311121~N(2,2)b~N(a,b1X22xxa121xaaaa0a0，x(232123ix1xa33xx112xx023AXbr()r(Ab)）n93个红球，2）25377517（）．48．1~B(3,)P(X2)（X）45284361．设A,B3A3(B)B8．132．设A,B3AB33B—8．1设A,B3AB32AB．1设A,B是3A3,B22AB．1A,BP()0P(BA)设A,BnA,B0．(BA)(A)B．11111设A，BP(AB)P()P(B)A与B．X8．设A为n阶方阵，若存在数nXAX9．设A为n阶方阵，若存在数nXAX，则称为AXX为A相应于特征值设A,B,CAC(BC).设A为34B52为24BC设,C,DnB,CABXCDXB(DA)C1．1012~Xa=．0.20.5aX~（，（）=np．X~BE(X)．k,0x14(x)1x2k=f101．0,其它~a0.45X．0.3a0.25012~P(XX．30111x2xXf(x)P(X)．0其它28XE(XE(X))0．(X)2,E(X)5E(X)3．XD22．设XD(X)3D(3X2)2．ˆˆˆ)的23．设Eˆ24．设ˆE()．1AA2A1=2．,,,,,,．12n12n27．设4=B（）=1，那么=B314,x,,xN(,4)x~N(,)．设x1012ii11n2n,x,,xN(,)xx()Dx设x212nini17111230．设f(x)1x2()02．2fx2x14211211x2031．设A2A1，-1，2，-2．2x142111040()_________________rA．232.设A07033．若P(A)0.8,P(AB)0.5P(AB)．11n,x,,x~(0,XNxxx~N)x12nnini12000132，，Rk．123212k1[0,2]D(X)X~U．3121A2142若AXr(A)n．n0若P(AB)0.9,P(AB)0.1,P(AB)0.5P(AB)．ˆˆˆˆˆˆ)D)比更．和D121221BP()P(B)，BA（A-）=．AAA(X)2E(X)9D(2X)XEXE，．．2(X)2E(X)9D(2X)，20),(0,k)110,0,0,231,0,0,1,2,31,2,0,,,1234234．P()0.8,P(AB)0.2P(AB)．1025~~E(X)X．0.30.10.10.51025XE(X)3．0.50.50.50.5386512107的值为=．aA1）.12N(,)2T81x111x111xx2．41121．52354AXA是r(Ab)=3．b450AX1123x2xx0102(x,xA134x2x34000024xx1当=112xx112112215235,BB．A01X3241121011210023501001121032400012301121011210001121001121000151001511109221002010107210107210015100151201即A1721512012011XAB72111113151562110200121,B050AA1B．22300511010011010012101001111022300104320191101100110100100105310110016410016410043101053100164431即A1531644312008155AB5310501015516400512205231123011,B112AA，求：（1）；（2）1．001012231A0112001解：（1）因为12301111B112112112012012AB2．231100AI0110100010012301011001/23/210100110100110010010010011/23/21A011．1001100111()AA．A11011001111111101013210101112210111101111001320100211101220001110100201100201100201001112011101010011011101001112001112201()011即1112112235A，求（1），（2）1．AA3241121121122350110111324012001（1）A1121001121002350100112103240010123011121011210001121001121000151001511109221002010107210107210015100151201721即A15101011B11120XA，B．X1035(I)XB110100110100(IAI)101010011110102001012101101010100021011110010121001011001011102(I)121即1010211113(I)B1212024X1．0153312323B357,58AB．X58100112310012310035701001231058100010255011231001204630123100105520011210011210064101055200112641即A15521264123813XAB552581523112018123xxx1x12342x7x2xx21234x4x3x2x112342x4x8x2x212341311113111131111311127212010100101001010143221012300022000220248202640006600000015xx14xxx244xx3412令xX4000).0x5x14xxx424xx34令xX111.4XkXXk01x3x3x2xx0123452x6x9x5x3x012345x3x3x2x0123513321133212695300311=1330200621332113010003110031000000000x3xx1241xxx，x334240x5令x=1，x=0，得X=(0)；241(0)x=0，x=3，得X=242{X，X}．12kXkX1k，k12212x3x2x01232x5x3x0，1233x8xx0123132132101125301101A=38016005当50即5r()3xx13xxx323令x=1得X={X}．311kXk111833概率；（2）取到3A3AAB123到3CC38(A)1P(A)1P(A)110.2550.745．（1）P11231213CC343(B)P(AA)P(A)P(A)0.2550.2550.0180.273．（2）P=2323122x4x5x3x512343x6x5x2x512344x8x15xx1234解2453536525481115245351201000555120100055500555120100011100000x2xx124x，xxx12434xx0X(00)令．240x2xx124x，x44xx23x1，x0X(20)令令；241x0，x1X(11)．242XkXkX(00)k(20)k(11)，X0112212k，k12当xxx21242xx4x3x12342x3xx5x21234110121101211214301132315201132101211012101131011310000000033x2xx,x1344（xxx3x34231,x0x0,x1x及343414X1110,X2301120,x0X1100令x340XkXkXk,k1X01122213．当xx2xx212342xx7x3x612349x7x4xx1123411212112122173601597410219112121094801115100111500000000119x4xx134xx5x2341,x0x0,x1x及3434X91110,X4501120,x0X81000令x340XkXkXk,kX0112212(2)(2，(4)，，1234124328131）=451234142401140357213102070100700200110000,,,所以，r()=3．1234,,,,134234P(X7)15．设X~N(3,，试求:PX9)．0.8413,(2)0.9772,0.9987）53X393X3解：(1)PX9)P()P22220.9987X373P(X7)P()2215X3X3P(2)1P(2)221(2)116．设X~N(3,4)，试求：(1)P(XPX7)．(2)）X313解：(1)P(XP()22X3P((21153X373X3PX7)P()P2)2222(2)X~N(4,．（1）求P(X4；（2）若P(Xk)k(2)0.9332解：（1）P(X4＝1－P(X4=1－P(2X42)＝1－（(2)(2)）=2（1－(2)）＝0.045．（2）P(Xk)P(X4k4)＝1－P(X4k4)＝1－(k(k4)1(即－4=，＝2.5．X~（1<X<（X<a．(已知，0.9(2.0)0.9773)．，13X373X3解：（1）X<7）=P()P(1=2)2222=(2)(=+–1=X3a3)(a3)（X<）=P(==222a31.28，a=3+21.28=2~N2)P(XP(X1(0.5)和19．设X(2)0.9772,）X3Y~N(X3532P(XP()0.84132203X323P(X1P(0X2)P()222=P(1.5Y((=(0.5)0.93320.691516（x=21，求知u1.96)xN(0,1)1.963，n=64，且u~n31.96ux=21，un2295%的[xu,xu][20.265,．nn22（x=99%的置信区间.(已知u2.576)x2，n=625，且uN(0,1)~n0.0110.995u2.576x=2.5，，，222un299%的[xu,xu][2.294,2.706].nn22X~N(32.5,9,u6强度（单位：kg／cm0.975H:210xU~N(0,nxx，393n6，u0.975x36u0.975n915.1mm，若0.062(u)．0.975x~N(0,2Un0.0630.02x9[xu,xu]1.96u990.9750.975,15.1392]170.15cm.从一4cm）10.4，10.6，10.1，10.40.05u1.96)？(,:10.50.15H0xU～N(0,n410.375，，xn10.37510.5x0.075n1.67x1.961.96.u2n2X~N,0.09)914.9，,u615（0.975:150.09H20xU~N(0,nx0.3x14.90.1，130.19nu6，0.975x11.96u0.975n15．X~N(4((已知u1.96)H:20xU~N(0,n14.95xx14.95，0.50.24n6，u0.975x0.51.96u0.975n千克.18,BnAB1．设A是,BA(AB)B,BAA,BBA(AB)ABBA3(I)1IAA2．A是n=0，则A(I)(IAA2)223AAA=IA=3I=IAA(I)1IAA23．设nA(AI)(AI)0A(AI)(AI)AI0AI22所以，A,,,,，，线1231122233311230kkk112233k)k(3)k(4)0112223331(kk(2kk(2k4k0131122233kk013,,230,,kkkkk123121231232k4k023,BA,BAP()P(B)P(AB)P(B)P(A)P(B)P(BP(A))P(A)P(B),BA,BP(A)P(AB)P(AB)AAUA(BB)ABAB(AB)AB而(AB)ABP(A)P(AB)P(AB)A,BP(B)P(A)P(BA)P(A)P(BA)．BUB(AA)ABABB而(AB)(AB)P(B)P(AB)P(AB)P(A)P(BA)P(A)P(BA)A,BP(AB)P()P(AB)AAUA(BB)ABABAB(AB)而(AB)(AB)P(A)P(AB)P(AB)即P(AB)P()P(AB)设A是AB0A0．,BnB00BABB111AIA0B0ABB110A设A,B()(AB)()BB19AAA则A()(A)A1A111,,,,也线性无关.设,12k3122313,k,kk)k)k)0123112223313(kk(kk(kk0,,即131122233123kk013k0k12kk023kk0,,k123122313设A,BP(B)P(A)P(BA)P(A)P(BA)．BUB(AA)ABABB(AB)(AB)而P(B)P(AB)P(AB)P(A)P(BA)P(A)P(BA)BP(AB)P()P(B)．A,BABA(AB)BA(AB)BP()P(AB)P(B)而P(AB)P()P(B)即()1()PABPB．A,BABABABBPB()()PAB由ABP(B)1P(B)P(AB)1P(B)XE(X)D(X)X(X)0Y的均值、方差都存在,且D0．[E(X)E(X)]0XE(X)D(X)11EY)E()E(XE(X))D(X)D(X)121,C54,BABAC2ACA5BABA3543(AB)C．0366ABAC2AC04183777A5B(AB)C01280114121103A,BC321BC．AC012210021140246410ACBC(ABC3212210201002310102121,B11132AXBX．A3422113A2XB解：2034183225112X(3AB)25211227115711522241020143602533110,aa414142020120a(4360(421364505341a422531234100011001222122312；⑶．⑴；⑵11111101111221102解：（1）132001212100122100102231A|I2rr2rr21320102120362100362rr32rr12322100106320100922113231220110210092919r3221212rrr012103101032rr9323232929900001192199999122991922A192929199922626171000201751100(过程略)（2）A1A110210110011415301011011110110010121021132021101101111011010110111011011011011rr120110111rr13242rrrr141012102113200001110000111001112210001110()3RA10110110110111rr34000111000000003x2xx6x12343x8xx5x012342xx4xx1234432xxxx123413216132161019233rr0183rr01718381501272182rr5rrA1323rrrr141421410581000273914132013480010122610101003rr19rr413017801780101317rrrr322345rr43300330011400114005600560001001000242rr2x01041010011115rrr42143rrx400114001012x1300013000133x4111x11y112z?111111122rrA111101112rrrr21313111110111223］112011)rr2300(2))212()()3当且RARA1当RARA()()1,,1232283235756,,,710312310321x11xx331232223581030137756341,,,A1037001011712332110000571R(A)R()123131117392,8,0,612343933641331311131117390112,,,2806000181234393300000000413363xx2x0x12345xx2x3x01234xx2x5x012343540xxx1245113121312310rr5rr014371414321421512312rrrr07A13233rrrr1424111250143700000035040143100351251251010100121414r14r131r3131311201r01rr0102rr332334140231402114000000000000010000002355143xx14133xx令x1140142330x415x2x3xx12343xx4x2x51234x9x4x1245361xxxx123497121523111523111015rr3142rr0142728142112rrrr0142728A134235rr2rr14219041701427280000061028414565300000971210171x1xx11134120129121147200000xxx27223400000令，，xkxkkk31421271712kk11x929121x11112kk2kk27122712012x30kx1204k01,a,a,aa123411110111，，，00112340001010001000010121324300011000a1a0100))a)aaaaaaa20010a12334112213324433a00014(aa(aaaa(a121223434424AXBn()()RARAn()AXBRAn0R()nAXB010A1AIA1AA1AA1A1()(()1A11即A1xxx2xx2xx2xx2xxfx2122232412242334f(xx)2x2x22xx2xx2xx(xx)2x22xxx)x22xx1234242334123324424(xx)(xxx)x222212324令yxx，yxxx，yx，xy11223243244xyy113yx即23xyyy3234xy44fy2yy222311.设A,B,CA,B,C⑴A,B,C⑵A,B,C⑶A,B,C⑷A,B,C⑸A,B,C⑹A,B,CCABCABCABCABCCABCABCABCABCABC3个红球，22⑴2⑵21解:设A=“2B=“21C2C2312C1C1C2639P()2P(B)3332C25C255i道工序出正品”（i=1,2）AiP(AA)P(A)P(A|A)12121A""A""A""12325B""P(B)P(A)P(B|A)P(A)P(B|A)P(A)P(B|A)112233XpP(XPP(X2)P)PP(XP)2PP(Xk)P)1Pk故X123kpp)pp)2pp)1pkX0123456P(X4),P(2XP(X．(X(X(X(X(X(XP(2XP(X2)P(XP(X4)P(X0.20.30.120.10.72P(X1P(X10.30.7X2x,0x1f(x)0,其它11(X),P(X2)P．241121114P(X)f(x)dx2xdxx2222001151621P(X2)f(x)dx2xdxx2141114442x,0x1~f(x)E(X),D(X)．设X0,其它2310231E(X)xf(x)dxx2xdxx302224121E(X)xf(x)dxx2xdxx41200121D(X)E(X2)[E(x)]2()22318(X)P(X．设~.6)PXN2X1X121X1P(X0)P(1.67)110.95250.04750.61n,X,,XE(X),D(X)2XX10.设X12n11nii1E(X),D(X)．26111n()(EXE)(EXXXEXEXEX)[()()()]Xin12nn12nni11nn(X)(1X)1D(XXX)1[(X)(X)(X)]nnin212nn212ni111n22n2nX2s．x11xi3.6xi111s(xx)25.92219ii1X,01xxf(x;)0,其它．3121n2x1ˆnE(X)x(x1xx,i1x0i1nL(x,x,,x;)x)(xxx)n12ni12ni1dLnnˆnnLnx,x01，d1iinxi1i1ii15m）：N(,)2与222求1155ˆˆ()xx2s2xx551iii1i1)10.975（1）当21－α＝0.95，(2ss[x,x]nnN(,)24H:2003x/|U||，0n4/4()10.975由2|U>H0278度为（单位：cm）：520.0125s20.0671xx|T||0s/n/8t(nt2.62∵|T|<∴H04AAA(A')'(')'AA'AAA是nIA11或．证明：A是nI2AA1I1或A1AAA证明：AAA1)()即A()(A1A11AA100111(AA)A1．101100111111111010132101011122111101111001320100211101220001110100201100201100201001112011101010011011101001112001112201()011即111228110200121,B050AA1B2230051101001101001210100111102230010432011101100110100011100105310016410016410043101053100164431即A15316443120085AB531050516400552341113.设A123,B111,那么A.231230234111123123解：AB123111012,AB01210,231230001001所以A可逆。123100120103100121012010010012010012，001001001001001001121故(AB)01210012x4x5x3x512343x6x5x2x512344x8x15xx123解424535365254824535120100055529120100055500555120100011100000x2xx124x，xxx12434xx0X(00)令．240x2xx124x，x44xx23x1，x0X(20)令令；241x0，x1X(11)．242(00)k(20)k(11)，XXkXkX0112212k，k12xx2xx212342xx7x3x612349x7x4xx1123411212112122173601597410219112121094801115100111500000000119x4xx134xx5x2341,x0x0,x1x及3434X91110,X4501120,x0令x34X810000XXkXkXk,k0112212x3x2x01235.2x5x3x0,1233x8xx012330132132132解：A253011011,38016005时，方程组有非零解；132101=xx011011，方程组一般解为：其中x是自由未知量），1x=x3300000023其基础解系为X(1,1,1),通解为XkX其中k为任意常数）1111(1)，，X~（3，4）．求：（1）（5<X<9）,（2）（X>7）,(已知)．253解：(1)P(5<X<9)=()()(3)(1)273(2)P(X1P(X1()1(2)122x11(1)X~（3，（X<（2P.，,)．5-3解：(1)P(X<5)=()(1)0.8413;223)(03)(2)P(X11)P(1X11)P(0X2)(22(0.5)(1.5)1(0.5)1(1.5)(1.5)(0.5)0.93320.69150.2417X~（1<X<（X<a．(已知，0.9(2.0)0.9773)．，13X373X3解：（1）X<7）=P()P(1=2)2222=(2)(=+–1=X3a3a3（X<）=P()(=)=222a31.28，a=3+21.28=2X~N(3,4)，试求：(1)P(XPX7)．(2)）X313解：(1)P(XP()22X3P((21153X373X3PX7)P()P2)2222(2)（=99%的置信区间.(已x知u2.576)31x2，n=625，且uN~n1，u，x=2.5，2122un1299%的u,xu].nn1122915.1mm，2．2xU~Nnx39u,xuu99XN92kg0.975解：零假设H:于0x-知，故选取样本函数2u=N(0,1);/n由样本观测值计算统计量值x-31.1232.531.1232.5/n故拒绝零假设，即认为这批砖的抗断强度不合格。u3.73u1.96,1.1/90.37XN(414.7，0.975解：零假设H:20x-u=N(0,1);/n1由样本观测值计算统计量值:x=(14.715.114.815.2)14.954x-/n故接受零假设，即认为这批零件的平均重量为15千克。14.95150.05u0.5u1.96,40.1每根标准直径IOOmm，随机取出9根测得直径的平均值为99.9mm，样本标准差sO.，?0.,32x解：零假设H:由于未知，故选取样本函数t2099.9100snX0123P是n13．设nA(AI)(AI)0A(AI)(AI)AI0A2I2所以，A,,,,，，123112223331123kkk112233k)kk(4)0112223331(kk(2kk(2k4k0131122233013,,kk1231223,,kkk123123,BA,BA33P()P(B)P(AB)P(B)P(A)P(B)P(BP(A))P(A)P(B),BAB,P(A)P(AB)P(AB)AAUA(BB)ABAB(AB)AB(AB)而P(A)P(AB)P(AB)设证明：因为A=AU=A(B+B)=AB+AB=AB+(A-B)，而(A-B)AB=，故由概率性质知：P(A)=P(A-B)+P(AB),即P(A-B)=P(A)-P(AB),证毕。一、单项选择题AB1．设,B都是n阶方阵，则下列命题正确的是(．11,B,Bn2．设3.设均为阶可逆矩阵，则下列等式成立的是(．(AB)Bn为阶矩阵，则下列等式成立的是（）．）．A,Bn为阶矩阵，则下列等式成立的是（4．设5．设A，B是两事件，则下列等式中（P(AB)P()P(B)，其中，B互不相容）是不正确的．6．设A是mn矩阵，B是st矩阵，且ACB有意义，则C是(sn)矩阵．7．设A是ns矩阵，是ms矩阵，则下列运算中有意义的是（）B18．设矩阵A的特征值为0，2，则3A的特征值为(0，6)．1131119.设矩阵A201，则的对应于特征值的一个特征向量=(A21)．112011310．设x,x,,x是来自正态总体N(,)的样本，则（xxx）是无偏估计．255512n123x511．设x,x,,x是来自正态总体N(的样本，则检验假设H:5采用统计量U（）．1/n12n0aaa1a1aa232312．设bbb2，则ababab（2）．1ccc2331c12c23c312123012313．设X~，则P(X2)（0.4）．0.10.30.40.214．设x,x,,x是来自正态总体N(,)(,均未知）的样本，则（