设随机变量X的概率密度为 (x)=dfrac (1)(2sqrt {2pi )}(e)^-dfrac ({(x-3)^2)(8)}(-infty lt xlt infty ), 则下列服从-|||-标准正态分布的是 () .-|||-(A) dfrac (x+3)(sqrt {2)}; (B) dfrac (x+3)(2); (C) dfrac (x-3)(sqrt {2)}; (D) dfrac (X-3)(2)

qī qiǎo bǎn kàn sì píng fán què yǒu zhe wú xiàn lè qù tā kě yǐ zuò wéi wán jù-|||-七巧板看似平凡,却有着无限乐趣。它可以作为玩具,-|||-wèi shēng huó zēng tiān qù wèi yě kě yǐ zuò wéi kē pǔ yòng jù duàn liàn duō fāng miàn de-|||-为生活增添趣味;也可以作为科普用具,锻炼多方面的-|||-néng lì lìng wài yùn cáng zài qī qiǎo bǎn zhōng de shù xué zhī shi gèng shì qù wèi wú xiàn-|||-能力;另外,蕴藏在七巧板中的数学知识更是趣味无限。-|||-.xià miàn zhè gè qī qiǎo bǎn zhōng yǒu gè dān dú de zhèng fāng xíng-|||-1.下面这个七巧板中有 () 个单独的正方形。-|||-A.1 B.2 C.3 D.4-|||-.xià tú shì qī qiǎo bǎn zhōng de gè xiǎo sān jiǎo xíng hé gè zhōng sān jiǎo xíng nǐ néng-|||-2.下图是七巧板中的2个小三角形和1个中三角形,你能-|||-yòng tā men pīn chū qī qiǎo bǎn zhōng de nǎ xiē tú xíng sān kuài dōu yòng shàng-|||-用它们拼出七巧板中的哪些图形?(三块都用上)-|||-()-|||-3 1-|||-4 2-|||-1 5-|||-7 6-|||-A.1 B.1或2 C.4或6 D.1或4或6-|||-.nǐ néng yòng qī qiǎo bǎn pīn chéng shén me tú àn kuài lái pīn yi pīn-|||-3.你能用七巧板拼成什么图案?快来拼一拼!qī qiǎo bǎn kàn sì píng fán què yǒu zhe wú xiàn lè qù tā kě yǐ zuò wéi wán jù-|||-七巧板看似平凡,却有着无限乐趣。它可以作为玩具,-|||-wèi shēng huó zēng tiān qù wèi yě kě yǐ zuò wéi kē pǔ yòng jù duàn liàn duō fāng miàn de-|||-为生活增添趣味;也可以作为科普用具,锻炼多方面的-|||-néng lì lìng wài yùn cáng zài qī qiǎo bǎn zhōng de shù xué zhī shi gèng shì qù wèi wú xiàn-|||-能力;另外,蕴藏在七巧板中的数学知识更是趣味无限。-|||-.xià miàn zhè gè qī qiǎo bǎn zhōng yǒu gè dān dú de zhèng fāng xíng-|||-1.下面这个七巧板中有 () 个单独的正方形。-|||-A.1 B.2 C.3 D.4-|||-.xià tú shì qī qiǎo bǎn zhōng de gè xiǎo sān jiǎo xíng hé gè zhōng sān jiǎo xíng nǐ néng-|||-2.下图是七巧板中的2个小三角形和1个中三角形,你能-|||-yòng tā men pīn chū qī qiǎo bǎn zhōng de nǎ xiē tú xíng sān kuài dōu yòng shàng-|||-用它们拼出七巧板中的哪些图形?(三块都用上)-|||-()-|||-3 1-|||-4 2-|||-1 5-|||-7 6-|||-A.1 B.1或2 C.4或6 D.1或4或6-|||-.nǐ néng yòng qī qiǎo bǎn pīn chéng shén me tú àn kuài lái pīn yi pīn-|||-3.你能用七巧板拼成什么图案?快来拼一拼!qī qiǎo bǎn kàn sì píng fán què yǒu zhe wú xiàn lè qù tā kě yǐ zuò wéi wán jù-|||-七巧板看似平凡,却有着无限乐趣。它可以作为玩具,-|||-wèi shēng huó zēng tiān qù wèi yě kě yǐ zuò wéi kē pǔ yòng jù duàn liàn duō fāng miàn de-|||-为生活增添趣味;也可以作为科普用具,锻炼多方面的-|||-néng lì lìng wài yùn cáng zài qī qiǎo bǎn zhōng de shù xué zhī shi gèng shì qù wèi wú xiàn-|||-能力;另外,蕴藏在七巧板中的数学知识更是趣味无限。-|||-.xià miàn zhè gè qī qiǎo bǎn zhōng yǒu gè dān dú de zhèng fāng xíng-|||-1.下面这个七巧板中有 () 个单独的正方形。-|||-A.1 B.2 C.3 D.4-|||-.xià tú shì qī qiǎo bǎn zhōng de gè xiǎo sān jiǎo xíng hé gè zhōng sān jiǎo xíng nǐ néng-|||-2.下图是七巧板中的2个小三角形和1个中三角形,你能-|||-yòng tā men pīn chū qī qiǎo bǎn zhōng de nǎ xiē tú xíng sān kuài dōu yòng shàng-|||-用它们拼出七巧板中的哪些图形?(三块都用上)-|||-()-|||-3 1-|||-4 2-|||-1 5-|||-7 6-|||-A.1 B.1或2 C.4或6 D.1或4或6-|||-.nǐ néng yòng qī qiǎo bǎn pīn chéng shén me tú àn kuài lái pīn yi pīn-|||-3.你能用七巧板拼成什么图案?快来拼一拼!

三、计算题(本大题共5个小题,每小题6分,共-|||-30分)-|||-1.已知连续型随机变量X的概率密度为-|||-f(x)= ) 2x,xin (0,A) 0,

3.求下列极限:-|||-(1) lim _(xarrow 2)dfrac ({x)^2-3x-4}({x)^2-4} =-|||-(3) lim _(xarrow infty )dfrac (3{x)^2+5x+1}({x)^2+3x+4} =-|||-(5) lim _(xarrow infty )(1+dfrac (1)(x))(2-dfrac (1)({x)^2}) ;-|||-(7) lim _(xarrow 1)dfrac (sqrt {3-x)-sqrt (1+x)}({x)^2-1} ;-|||-(9) lim _(xarrow 1)dfrac ({x)^m-1}({x)^n-1} (m,n是正整数);-|||-(11) lim _(xarrow 0)dfrac ((1+x)(1+2x)(1+3x)-1)(x) ;-|||-(2) lim _(harrow 0)dfrac ({(x+h))^3-(x)^3}(h) ;-|||-(4) lim _(xarrow infty )dfrac ({(2x-3))^20((3x+2))^30}({(5x+1))^50} :-|||-(6) lim _(xarrow infty )dfrac (2{x)^3+3x+1}(4{x)^5+2x+7} =-|||-(8)lim _(xarrow 1)(dfrac (1)(1-x)-dfrac (3)(1-{x)^3});-|||-(10) lim _(xarrow 1)dfrac (sqrt [3]{x)-1}(sqrt {x)-1} =-|||-(12) lim _(xarrow +infty )(sqrt ((x+2)(x-1))-x) .

设某班车起点站上客人数X服从参数为λ(λ＞0)的泊松分布，每位乘客在中途下车的概率为p(0＜p＜1)，且中途下车与否相互独立．以Y表示在中途下车的人数，求：(1)在发车时有n个乘客的条件下，中途有m人下车的概率；(2)二维随机变量(X，Y)的概率分布．

5.(计算题，22分)设A=(}4&2&31&1&0-1&2&3)，且AB=A+2B，求B.

【单选题】设m×n矩阵A的秩等于n,则必有A. m=nB. mC. m>nD. m≥n

用3台机床加工同样的零件，零件由各机床加工的概率分别为0.5，0.3，0.2，各机床加工的零件为合格品的概率分别为0.94，0.9，0.95，求：(1)任取一个零件，其为合格品的概率；(2)任取一个零件，若是次品，其为第二台机床加工的概率.

1.用洛必达法则求下列极限:-|||-(1) lim _(xarrow 0)dfrac (ln (1+x))(x)-|||-(4) lim _(xarrow pi )dfrac (sin 3x)(tan 5x) =-|||-(7) lim _(xarrow 0)dfrac (ln tan 7x)(ln tan 2x) ;-|||-(10) lim _(xarrow 0)dfrac (ln (1+{x)^2)}(sec x-cos x) ;-|||-(13) lim _(xarrow 1)(dfrac (2)({x)^2-1}-dfrac (1)(x-1)) ;-|||-(16) lim _(xarrow {0)^+}((dfrac {1)(x))}^tan x --|||-(2) lim _(xarrow 0)dfrac ({e)^x-(e)^-x}(sin x) ;-|||-(5) lim _(xarrow dfrac {pi )(2)}dfrac (ln sin x)({(pi -2x))^2} :-|||-(8) lim _(xarrow dfrac {pi )(2)}dfrac (tan x)(tan 3x) ;-|||-(11) lim _(xarrow 0)xcot 2x ;-|||-(14) lim _(xarrow infty )((1+dfrac {a)(x))}^x-|||-(3) lim _(xarrow a)dfrac (sin x-sin a)(x-a) ;-|||-(6) lim _(xarrow a)dfrac ({x)^m-(a)^m}({x)^n-(a)^n} ;-|||-(9) lim _(xarrow +infty )dfrac (ln (1+dfrac {1)(x))}(arctan x) ;-|||-(12) lim _(xarrow 0)(x)^2(e)^dfrac (1{{x)^2}} ;-|||-(15) lim _(xarrow {0)^+}(x)^sin x ;