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下表给出了某班12名同学两次考试的成绩。要求:(1)计算两次考试成绩(X,Y)的相关(2)求Y对X的线性回归方程;(3)对所求方程进行方差分析,以检验其显著性提示: [tex=32.643x1.286]I/NcS0LDveLL2lm9KWY5LD58OUbW3mOKab/gXVs2biQhukfjqQyMsYbqdVyHNSWU+KBYz10rIPPQhNoxSo15XjYyflE4OL/kQH8ndOyWBWBybmZzuzE9E+2AwiLjTgev[/tex][img=1004x134]17cc057c9c476c6.png[/img]
下表给出了某班12名同学两次考试的成绩。要求:(1)计算两次考试成绩(X,Y)的相关(2)求Y对X的线性回归方程;(3)对所求方程进行方差分析,以检验其显著性提示: [tex=32.643x1.286]I/NcS0LDveLL2lm9KWY5LD58OUbW3mOKab/gXVs2biQhukfjqQyMsYbqdVyHNSWU+KBYz10rIPPQhNoxSo15XjYyflE4OL/kQH8ndOyWBWBybmZzuzE9E+2AwiLjTgev[/tex][img=1004x134]17cc057c9c476c6.png[/img]
下列选项为某变速器一、二、三、四挡的传动比,则三挡的传动比为()。 A: 0.969 B: 1.286 C: 1.944 D: 3.455
下列选项为某变速器一、二、三、四挡的传动比,则三挡的传动比为()。 A: 0.969 B: 1.286 C: 1.944 D: 3.455
下列变量组()是一个闭回路。 A: {x,x,x,x,x,x} B: {x,x,x,x,x} C: {x,x,x,x,x,x} D: {x,x,x,x,x,x}
下列变量组()是一个闭回路。 A: {x,x,x,x,x,x} B: {x,x,x,x,x} C: {x,x,x,x,x,x} D: {x,x,x,x,x,x}
以下谓词蕴含式正确的是(): (∀x) (A(x)→B(x))=>( ∀x)A(x)→(∀x)B(x)|(∀x) (A(x)↔B(x))=>( ∀x)A(x)↔(∀x)B(x)|(∀x)A(x)∨(∀x)B(x)=>( ∀x) (A(x)∨B(x))|(∃x) (A(x)∧B(x))=>(∃x)A(x)∧(∃x)B(x)
以下谓词蕴含式正确的是(): (∀x) (A(x)→B(x))=>( ∀x)A(x)→(∀x)B(x)|(∀x) (A(x)↔B(x))=>( ∀x)A(x)↔(∀x)B(x)|(∀x)A(x)∨(∀x)B(x)=>( ∀x) (A(x)∨B(x))|(∃x) (A(x)∧B(x))=>(∃x)A(x)∧(∃x)B(x)
以下谓词蕴含式正确的是(): (?x) (A(x)→B(x))=>( ?x)A(x)→(?x)B(x)|(?x) (A(x)?B(x))=>( ?x)A(x)?(?x)B(x)|(?x)A(x)∨(?x)B(x)=>( ?x) (A(x)∨B(x))|(?x) (A(x)∧B(x))=>(?x)A(x)∧(?x)B(x)
以下谓词蕴含式正确的是(): (?x) (A(x)→B(x))=>( ?x)A(x)→(?x)B(x)|(?x) (A(x)?B(x))=>( ?x)A(x)?(?x)B(x)|(?x)A(x)∨(?x)B(x)=>( ?x) (A(x)∨B(x))|(?x) (A(x)∧B(x))=>(?x)A(x)∧(?x)B(x)
下列式中错误的是: A: (∀x)(A(x)Úp(x)) Û (∀x)A(x)Ú (∀x)p(x) B: ($x)A(x) Ù p Û ($x)(A(x) Ù p ) C: (∀x)(A(x)ÚB(x)) Þ (∀x)A(x)Ú( ∀x)B(x) D: ($x)(A(x)ÙB(x)) Þ ($x)A(x)Ù( $x)B(x)
下列式中错误的是: A: (∀x)(A(x)Úp(x)) Û (∀x)A(x)Ú (∀x)p(x) B: ($x)A(x) Ù p Û ($x)(A(x) Ù p ) C: (∀x)(A(x)ÚB(x)) Þ (∀x)A(x)Ú( ∀x)B(x) D: ($x)(A(x)ÙB(x)) Þ ($x)A(x)Ù( $x)B(x)
判断下列推证是否正确。 (∀x)(A(x)→B(x))⇔(∀x)(¬A(x)∨B(x)) ⇔(∀x)¬( A(x)∧¬B(x) ) ⇔¬(∃x) ( A(x)∧¬B(x) ) ⇔¬( (∃x)A(x)∧(∃x)¬B(x) ) ⇔¬(∃x)A(x)∨¬(∃x)¬B(x) ⇔¬(∃x)A(x)∨(∀x)B(x) ⇔(∃x)A(x)→(∀x)B(x)
判断下列推证是否正确。 (∀x)(A(x)→B(x))⇔(∀x)(¬A(x)∨B(x)) ⇔(∀x)¬( A(x)∧¬B(x) ) ⇔¬(∃x) ( A(x)∧¬B(x) ) ⇔¬( (∃x)A(x)∧(∃x)¬B(x) ) ⇔¬(∃x)A(x)∨¬(∃x)¬B(x) ⇔¬(∃x)A(x)∨(∀x)B(x) ⇔(∃x)A(x)→(∀x)B(x)