52. What did David Rubin's research find about misnaming?
52. What did David Rubin's research find about misnaming?
根据鲁宾(Z.Rubin)对爱情与喜欢的研究,喜欢相对于爱情而言更为
根据鲁宾(Z.Rubin)对爱情与喜欢的研究,喜欢相对于爱情而言更为
Despite her lack of functional speech, Sue Rubin was able to write the narration for the documentary about her life. </p></p>
Despite her lack of functional speech, Sue Rubin was able to write the narration for the documentary about her life. </p></p>
2)__________________ at New York’s Rubin Museum, A: at the event B: at the dinner C: at the event dinner D: at the event's dinner
2)__________________ at New York’s Rubin Museum, A: at the event B: at the dinner C: at the event dinner D: at the event's dinner
被称为“Android之父”的是()。 A: ASteve Jobs B: BAndy Rubin C: CTim Cook D: DBill Gates
被称为“Android之父”的是()。 A: ASteve Jobs B: BAndy Rubin C: CTim Cook D: DBill Gates
Python语言的创始人是( )。 A: Guido van Ros B: Andy Rubin C: James Gosling D: Anders Hejlsberg
Python语言的创始人是( )。 A: Guido van Ros B: Andy Rubin C: James Gosling D: Anders Hejlsberg
The research done by psychologist Rubin centers around () A: happy and successful marriages B: friendships of men and women C: emotional problems in marriage D: interactions between men and women
The research done by psychologist Rubin centers around () A: happy and successful marriages B: friendships of men and women C: emotional problems in marriage D: interactions between men and women
The boxer Lesra finds very fascinating has, in a sense, something to do with all of the following EXCEPT A: one of the three white men. B: Rubin Carter. C: Denzell Washington. D: "Hurricane".
The boxer Lesra finds very fascinating has, in a sense, something to do with all of the following EXCEPT A: one of the three white men. B: Rubin Carter. C: Denzell Washington. D: "Hurricane".
最初对图形与背景加以区分的是 A: 布生( Gibson) B: 考夫卡( Koffka) C: 惠特海默( Wertheimer) D: 鲁宾( Rubin)
最初对图形与背景加以区分的是 A: 布生( Gibson) B: 考夫卡( Koffka) C: 惠特海默( Wertheimer) D: 鲁宾( Rubin)
若检验的假设为H0∶μ=μ0,H1∶μ≠μ0,则拒绝域为 ( )。 A: z >; zα B: z <; zα C: z >; zα/2或z <; -zα/2 D: z >; zα或z <; -zα
若检验的假设为H0∶μ=μ0,H1∶μ≠μ0,则拒绝域为 ( )。 A: z >; zα B: z <; zα C: z >; zα/2或z <; -zα/2 D: z >; zα或z <; -zα