英语周报2020一2021七下仁爱版51答案

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18.(本小题满分12分)(1)由已知,学生为优秀的概率为记优质学生数为X,由题意知,X的所有可能取值为0,1,2,3则P(X=0)=C3(0.4)=0064,P(X=1)=C1(0.4)20.6=0.288P(X=2)=C0.4(0.6)2=0.432,P(X=3)=C3(0.6)=0.216故X的分布列为0.0640.2880.430.216所以X的数学期望为E(X)=3×0.6=1.86分(2)填写列联表如下优秀学生非优秀学生合计甲方案乙方案合计120计算k2=12040×28-20×32)。22252.70660×60×72×48所以不能在犯错误的概率不超过0.1的前提下认为学生综合测试评分优秀与教学方案有关12分19.(本小题满分12分)(1)取AB中点为O,连结OD,OB,因为△ABA中BB=BA,所以OB⊥AB因为AB⊥B1D,OB∩BD=B,所以AB⊥平面BOD因为ODc平面BOD,所以AB⊥OD3分由已知,BC⊥BB,又OD∥BC,所以OD⊥BB1因为AB∩BB1=B,所以OD⊥平面ABBA又ODc平面ABC,所以平面ABC⊥平面ABBA(2)由(1)知,OB,OD,OB两两垂直.以O为坐标原点,OB的方向为x轴的方向,OD的方向为y轴的方向,OB的方向为2轴的方向,1OB为单位长度1,建立如图所示的空间直角坐标系O-xyz可知B.00,B10、3),D0.0),A-10.0),c(.2.0),c1(02,)则BB1=(-10,√3),BD=(0,-√3),B1A=(-10.-√3),CC=(-10.3)龙江考试-x+√3z=0B有即m=(√3,3,1)设平面BA的法向量为n=(x,y,z),则Bn·B1A=0x+√3z=0有{y=1,即n=(0.10)则cos(m,n)=11分所以二面DBB1一4的余弦值为y?712分0.(本小题满分12分)(1)因为焦点F以点A,B的坐标分别为所以S△AOB2p.=2,故P=2故抛物线C的方程为y2=4x.5分(2)由题设P(x,y),Q(x2y),易知直线的斜率存在,记为k则设直线l:y=kx+m,与y2=4x联立得ky2-4y+4m=0得y+y2=,H1y2=42mr,+xVy2黑龙江考试VI Jx1x2y1·y2mk+=互+卫2=52(6+m)+x(6+m)2+m(+)=4又知kop=tana,ko=tan4tan a+tantan(a+B)OAOBm -= tan -=lI-tan a tan B 1-kokor1 4k解得m=4k+4,所以直线l:y=kx+4k+4=k(x+4)+4,恒过定点(4,4)12分

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英语周报2020一2021七下仁爱版51答案

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