• 2022-10-26
    甲从[tex=3.357x1.214]4bXO6/qO+ok3mTC7Qr9nBA==[/tex]中任取一数[tex=0.5x1.214]Xvgwe+yswZgMoCwmPH37UA==[/tex],乙从[tex=4.357x1.214]ghIL55221F3uKg8VExdfoUM+MzK4prWVsjM+7yIWI44=[/tex]中任取一数[tex=0.5x1.0]Oxpy3OzeKJvOuQNNMcDHuw==[/tex],求[tex=2.214x1.357]oIxZeCdF+SIXA8nz0qZ5nA==[/tex]的分布律.
  • 举一反三