• 2022-06-04 问题

    设二维连续型随机变量(X,Y)的概率密度为f(x,y),则随机变量Z=Y-X的概率密度fz(z)=() A: ∫-∞+∞f(x,z-x)dx B: ∫-∞+∞f(x,x-z)dx C: ∫-∞+∞f(x,z+x)dx D: ∫-∞+∞f(-x,z+x)dx

    设二维连续型随机变量(X,Y)的概率密度为f(x,y),则随机变量Z=Y-X的概率密度fz(z)=() A: ∫-∞+∞f(x,z-x)dx B: ∫-∞+∞f(x,x-z)dx C: ∫-∞+∞f(x,z+x)dx D: ∫-∞+∞f(-x,z+x)dx

  • 2021-04-14 问题

    以下程序的输出结果是____?double sub(double x,double y,double z){    y=-1;    z=z+x;    return z;}main(){    double a=2.5,b=9.0;    printf("%f\n",sub(b-a,a,a));}

    以下程序的输出结果是____?double sub(double x,double y,double z){    y=-1;    z=z+x;    return z;}main(){    double a=2.5,b=9.0;    printf("%f\n",sub(b-a,a,a));}

  • 2021-04-14 问题

    【单选题】求点(x,y,z)关于原点和x轴,y轴,z轴的对称点坐标分别为 A. (-x,-y,-z),(x,-y,-z),(x,-y,z),(x,y,-z), B. (-x,-y,-z),(-x,y,z),(x,y,-z),(x,y,-z) C. (-x,-y,-z),(-x,-y,z),(x,-y,-z),(-x,y,-z) D. (-x,-y,-z),(x,-y,-z),(-x,y,-z),(-x,-y,z)

    【单选题】求点(x,y,z)关于原点和x轴,y轴,z轴的对称点坐标分别为 A. (-x,-y,-z),(x,-y,-z),(x,-y,z),(x,y,-z), B. (-x,-y,-z),(-x,y,z),(x,y,-z),(x,y,-z) C. (-x,-y,-z),(-x,-y,z),(x,-y,-z),(-x,y,-z) D. (-x,-y,-z),(x,-y,-z),(-x,y,-z),(-x,-y,z)

  • 2022-07-27 问题

    有两个优先级相同的进程P1和P2,各自执行的操作如下,信号量S1和S2初值均为0。试问P1、P2并发执行后,x、y的值各为多少? P1: P2: begin begin y:=1; x:=1; y:=y+3; x:=x+5; V(S1); P(S1); z:=y+1; x:=x+y; P(S2); V(S2); y:=z+yend z:=z+x; end

    有两个优先级相同的进程P1和P2,各自执行的操作如下,信号量S1和S2初值均为0。试问P1、P2并发执行后,x、y的值各为多少? P1: P2: begin begin y:=1; x:=1; y:=y+3; x:=x+5; V(S1); P(S1); z:=y+1; x:=x+y; P(S2); V(S2); y:=z+yend z:=z+x; end

  • 2022-05-29 问题

    ​ 若已知xy+yz+zx=xy+z,则判断等式(x+y)(y+z)(z+x)=(x+y)z是否成立,最简单方法是用对偶法则 ​ ​

    ​ 若已知xy+yz+zx=xy+z,则判断等式(x+y)(y+z)(z+x)=(x+y)z是否成立,最简单方法是用对偶法则 ​ ​

  • 2022-10-30 问题

    以下哪几项是前束范式? A: (∀x)(∀y)(P(x)∧(∃z)Q(y,z)∨R(x,z)) B: (∀z)(∀x)(∃y)(P(x)∧Q(y,z)∨R(x,z)) C: (∀x)(∀y)(P(x)∧Q(y,z)∨(∃z)R(x,z)) D: (∃x)(∃y)(∀z)(P(x)∧Q(y,z)∨R(x,z))

    以下哪几项是前束范式? A: (∀x)(∀y)(P(x)∧(∃z)Q(y,z)∨R(x,z)) B: (∀z)(∀x)(∃y)(P(x)∧Q(y,z)∨R(x,z)) C: (∀x)(∀y)(P(x)∧Q(y,z)∨(∃z)R(x,z)) D: (∃x)(∃y)(∀z)(P(x)∧Q(y,z)∨R(x,z))

  • 2022-06-07 问题

    若有以下程序: #include<stdio.h> void sub(double x, double*y, double*z) { *y=*y-1.0; *z=*z+x; } main() {double a=2.5, b=9.0, *pa, *pb; pa=&a; pb=&b; sub(b-a, pa, pa); printf("%f\n", a); } 程序的输出结果是 A: 9.000000 B: 1.500000 C: 8.000000 D: 10.500000

    若有以下程序: #include<stdio.h> void sub(double x, double*y, double*z) { *y=*y-1.0; *z=*z+x; } main() {double a=2.5, b=9.0, *pa, *pb; pa=&a; pb=&b; sub(b-a, pa, pa); printf("%f\n", a); } 程序的输出结果是 A: 9.000000 B: 1.500000 C: 8.000000 D: 10.500000

  • 2021-04-14 问题

    【单选题】已知序列x(n)的双边z变换为X(z),收敛域|z|>|a|,则x(-n)的双边z变换和收敛域为() A. X(-z),|z|>|a B. X(1/z),|z|>1/|a| C. X(1/z),|z|<1/|a| D. X(-z),|z|

    【单选题】已知序列x(n)的双边z变换为X(z),收敛域|z|>|a|,则x(-n)的双边z变换和收敛域为() A. X(-z),|z|>|a B. X(1/z),|z|>1/|a| C. X(1/z),|z|<1/|a| D. X(-z),|z|

  • 2021-04-14 问题

    已知()x()(()n())()的()z()变换是()X()(()z())(),()ROC()是()|()z()|()>()a(),则()x(()-()n()-()5())()的()z()变换和()ROC()是()()A.()()z()-()5()X()(1/()z()),z()>()1/()a()B.()()z()5()X()(1/()z()),z()>()1/()a()C.()()z()-()5()X()(1/()z()),z()<()1/()a()D.()()z()5()X()(1/()z()),z()<()1/()a

    已知()x()(()n())()的()z()变换是()X()(()z())(),()ROC()是()|()z()|()>()a(),则()x(()-()n()-()5())()的()z()变换和()ROC()是()()A.()()z()-()5()X()(1/()z()),z()>()1/()a()B.()()z()5()X()(1/()z()),z()>()1/()a()C.()()z()-()5()X()(1/()z()),z()<()1/()a()D.()()z()5()X()(1/()z()),z()<()1/()a

  • 2022-06-08 问题

    设方程\({x^2} + {y^2} + {z^2} = 2Rx\)确定函数\(z=z(x,y)\),则\( { { \partial z} \over {\partial x}}=\) A: \( { { \partial z} \over {\partial x}} = { { R +x} \over z}\) B: \( { { \partial z} \over {\partial x}} =- { { R +x} \over z}\) C: \( { { \partial z} \over {\partial x}} = { { R - x} \over z}\) D: \( { { \partial z} \over {\partial x}} =- { { R - x} \over z}\)

    设方程\({x^2} + {y^2} + {z^2} = 2Rx\)确定函数\(z=z(x,y)\),则\( { { \partial z} \over {\partial x}}=\) A: \( { { \partial z} \over {\partial x}} = { { R +x} \over z}\) B: \( { { \partial z} \over {\partial x}} =- { { R +x} \over z}\) C: \( { { \partial z} \over {\partial x}} = { { R - x} \over z}\) D: \( { { \partial z} \over {\partial x}} =- { { R - x} \over z}\)

  • 1 2 3 4 5 6 7 8 9 10