• 2021-04-14 问题

    能正确表示“当x的取值在[-58,-40]和[40,58]范围内为真,否则为假”的表达式是: (x>= -58) && (x<= -40) && (x>=40) && (x<=58)|(x>= -58) && (x<= -40) || (x>=40) && (x<=58)|(x>= -58) | |(x<= -40) && (x>=40) || (x<=58)|(x>= -58) || (x<= -40) || (x>=40) || (x<=58)

    能正确表示“当x的取值在[-58,-40]和[40,58]范围内为真,否则为假”的表达式是: (x>= -58) && (x<= -40) && (x>=40) && (x<=58)|(x>= -58) && (x<= -40) || (x>=40) && (x<=58)|(x>= -58) | |(x<= -40) && (x>=40) || (x<=58)|(x>= -58) || (x<= -40) || (x>=40) || (x<=58)

  • 2022-07-01 问题

    设随机变量X服从正态分布N(2,1),其概率密度函数为f(x),分布函数为F(x),则有 A: P(X≥0)=P(X≤0)=0.5 B: P(X≥2)=P(X≤2)=0.5 C: [img=398x40]1803b3bad5a359e.png[/img] D: [img=475x43]1803b3bae0a2852.png[/img]

    设随机变量X服从正态分布N(2,1),其概率密度函数为f(x),分布函数为F(x),则有 A: P(X≥0)=P(X≤0)=0.5 B: P(X≥2)=P(X≤2)=0.5 C: [img=398x40]1803b3bad5a359e.png[/img] D: [img=475x43]1803b3bae0a2852.png[/img]

  • 2022-06-17 问题

    398宽带续订手机宽带需在398独立宽带到期当月进行续约操作。()

    398宽带续订手机宽带需在398独立宽带到期当月进行续约操作。()

  • 2022-07-26 问题

    能正确表示“当x的取值在&#91;-58,-40&#93;和&#91;40,58&#93;范围内为真,否则为假”的表达式是()。 A: (x>=-58)&&(x<=-40)&&(x>=40)&&(x<=58) B: (x>=-58)∥(x<=-40)∥(x>=40)∥(x<=58) C: (x>=-58)&&(x<=-40)∥(x>=40)&&(x<=58)

    能正确表示“当x的取值在&#91;-58,-40&#93;和&#91;40,58&#93;范围内为真,否则为假”的表达式是()。 A: (x>=-58)&&(x<=-40)&&(x>=40)&&(x<=58) B: (x>=-58)∥(x<=-40)∥(x>=40)∥(x<=58) C: (x>=-58)&&(x<=-40)∥(x>=40)&&(x<=58)

  • 2022-10-31 问题

    假定16<X≤40,那么用边界值分析法,X在测试中应该取的边界值是: A: X=16,X=17,X=40,X=41 B: X=15,X=16,X=40,X=41 C: X=16,X=17,X=39,X=40 D: X=15,X=16,X=39,X=40

    假定16<X≤40,那么用边界值分析法,X在测试中应该取的边界值是: A: X=16,X=17,X=40,X=41 B: X=15,X=16,X=40,X=41 C: X=16,X=17,X=39,X=40 D: X=15,X=16,X=39,X=40

  • 2022-06-14 问题

    设有变量定义:x=(10,20,30)和y=&#91;10,20,30&#93;,以下正确的赋值语句是_______。 A: x(2)=40 B: x[2]=40 C: y(2)=40 D: y[2]=40

    设有变量定义:x=(10,20,30)和y=&#91;10,20,30&#93;,以下正确的赋值语句是_______。 A: x(2)=40 B: x[2]=40 C: y(2)=40 D: y[2]=40

  • 2022-06-09 问题

    “int x=40;”,声明了一个字符型类的x变量,其默认值为40。

    “int x=40;”,声明了一个字符型类的x变量,其默认值为40。

  • 2021-04-14 问题

    398 . “灰”就是画面上没有最亮和最深的颜色,缺少

    398 . “灰”就是画面上没有最亮和最深的颜色,缺少

  • 2021-04-14 问题

    【单选题】#include <stdio.h> int x1 = 30, x2 = 40; void sub(int x, int y) { x1 = x; x = y; y = x1; } int main() { int x3 = 10, x4 = 20; sub(x3, x4); sub(x2, x1); printf(" %d, %d, %d, %d ", x3, x4, x1, x2); return 0; } A. 10, 20, 40, 40 B. 10,40, 40, 40 C. 20, 20, 40, 40 D. 10, 10, 40, 40

    【单选题】#include <stdio.h> int x1 = 30, x2 = 40; void sub(int x, int y) { x1 = x; x = y; y = x1; } int main() { int x3 = 10, x4 = 20; sub(x3, x4); sub(x2, x1); printf(" %d, %d, %d, %d ", x3, x4, x1, x2); return 0; } A. 10, 20, 40, 40 B. 10,40, 40, 40 C. 20, 20, 40, 40 D. 10, 10, 40, 40

  • 2022-05-28 问题

    以4,9,1为为插值节点,求\(\sqrt x \)的lagrange的插值多项式 A: \( {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) + {1 \over {24}}(x - 4)(x - 9)\) B: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) + {1 \over {24}}(x - 4)(x - 9)\) C: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x +1) + {1 \over {24}}(x - 4)(x - 9)\) D: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) - {1 \over {24}}(x - 4)(x - 9)\)

    以4,9,1为为插值节点,求\(\sqrt x \)的lagrange的插值多项式 A: \( {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) + {1 \over {24}}(x - 4)(x - 9)\) B: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) + {1 \over {24}}(x - 4)(x - 9)\) C: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x +1) + {1 \over {24}}(x - 4)(x - 9)\) D: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) - {1 \over {24}}(x - 4)(x - 9)\)

  • 1 2 3 4 5 6 7 8 9 10