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双基限时练

双基限时练
双基限时练

双基限时练(一)

1.有关正弦定理的叙述:

①正弦定理仅适用于锐角三角形;②正弦定理不适用于直角三角形;③正弦定理仅适用于钝角三角形;

④在给定三角形中,各边与它的对角的正弦的比为定值;⑤在△ABC中,sin A B C=a b c.

其中正确的个数是()

A.1B.2

C.3 D.4

解析①②③不正确,④⑤正确.

答案 B

2.在△ABC中,若A=60°,B=45°,BC=32,则AC=()

A.4 3 B.2 3

C. 3

D.

3 2

解析由正弦定理,得AC

sin B=

BC

sin A,即AC=

BC·sin B

sin A=

32×sin45°

sin60°=2 3.

答案 B

4.在△ABC中,已知3b=23a sin B,cos B=cos C,则△ABC的形状是() A.直角三角形B.等腰三角形

C.等边三角形D.等腰直角三角形

解析利用正弦定理及第一个等式,可得sin A=

3

2,A=

π

3,或

3,但由第二个等式及B与C的范围,

知B=C,故△ABC必为等腰三角形.

答案 B

5.在△ABC中,若3a=2b sin A,则B等于()

A.30°B.60°

C.30°或150°D.60°或120°解析∵3a=2b sin A,

∴3sin A=2sin B sin A.

∵sin A≠0,∴sin B=

3 2,

又0°

6.在△ABC中,已知a:b:c=4:3:5,则

2sin A-sin B

sin C=________.

解析设a=4k,b=3k,c=5k(k>0),由正弦定理,得

2sin A-sin B

sin C=

2×4k-3k

5k=1.

答案 1

8.在△ABC中,若tan A=

1

3,C=150°,BC=1,则AB=________.

解析

.

在△

∴AB

答案

9.在△ABC中,若A:B:C=1:2:3,则a b c=________.

解析由A+B+C=180°及A:B:C=1:2:3,知A=180°×

1

6=30°,B=180°×

2

6=60°,C=180°×

3

6=90°.

∴a:b:c=sin30°:sin60°:sin90°=

1

2:

3

2:1=1:3:2.

答案1:3:2

11.△ABC三边各不相等,角A,B,C的对边分别为a,b,c且a cos A=b cos B,求

a+b

c的取值范围.解∵a cos A=b cos B,∴sin A cos A=sin B cos B,

∴sin2A=sin2B.

∵2A,2B∈(0,2π),∴2A=2B,或2A+2B=π,

∴A=B,或A+B=

π

2.

如果A=B,那么a=b不合题意,∴A+B=

π

2.

a+b

c=

sin A+sin B

sin C=sin A+sin B=sin A+cos A

=2sin

?

?

?

?

?

A+

π

4.

∵a ≠b ,C =π2,∴A ∈? ????0,π2,且A ≠π4, ∴a +b

c ∈(1,2).

12.在△ABC 中,sin(C -A )=1,sin B =1

3. (1)求sin A ;

(2)设AC =6,求△ABC 的面积. 解 (1)∵sin(C -A )=1,-π

2.

∵A +B +C =π,∴A +B +A +π

2=π, ∴B =π2-2A ,∴sin B =sin ? ????

π2-2A =cos2A =13.

∴1-2sin 2

A =1

3.

∴sin 2A =13,∴sin A =3

3.

(2)由(1)知,A 为锐角,∴cos A =6

3, sin C =sin ? ??

??

π2+A =cos A =63,

由正弦定理得AB =AC ·sin C

sin B =6·6

3

13=6.

S △ABC =12AB ·AC ·sin A =12×6×6×3

3=3 2.

双基限时练(二)

1.在△ABC 中,a 2+b 2

D .等边三角形

解析 由a 2

+b 2

,知cos C =a 2+b 2-c 2

2ab

<0,

又0

2.在△ABC 中,已知a 2+b 2-c 2=ab ,则C =( ) A .60° B .120° C .30°

D .45°或135°

解析 由cos C =a 2+b 2-c 22ab =ab 2ab =12, 又0°

3.在△ABC 中,a :b :c =3:5:7,则△ABC 的最大角是( ) A .30° B .60° C .90°

D .120° 解析 由a :b :c =3:5:7,知最大边为c ,

∴最大角为C ,设a =3k ,b =5k ,c =7k (k >0),则cos C =a 2+b 2-c 22ab =-1

2,又0°

4.在△ABC 中,B =60°,b 2=ac ,则这个三角形是( ) A .不等边三角形 B .等边三角形 C .等腰三角形

D .直角三角形

解析 由b 2=ac 及余弦定理,得 b 2=a 2+c 2-2ac cos60°, 即ac =a 2+c 2-ac ,

∴(a -c )2=0,∴a =c ,又B =60°, ∴△ABC 为等边三角形. 答案 B

5.△ABC 的三边长分别为AB =7,BC =5,CA =6,则AB →·BC →

的值为( ) A .19 B .14 C .-18

D .-19

解析 由余弦定理,得cos B =AB 2+BC 2-CA 2

2·AB ·BC

=72+52-622·

7·5=19

35.

∴AB →·BC →=|AB →||BC →|cos 〈AB →,BC →

〉=7×5×? ????

-1935=-19.

答案 D

6.在△ABC 中,已知a ,b 是方程x 2-5x +2=0的两根,C =120°,则边c =____________. 解析 由韦达定理,得a +b =5,ab =2. 由(a +b )2=a 2+b 2+2ab , 得a 2+b 2=52-2×2=21. ∴c 2=a 2+b 2-2ab cos120°=23. ∴c =23. 答案

23

7.在△ABC 中,若a =7,b =8,cos C =13

14,则最大角的余弦值为____________. 解析 c 2=a 2+b 2-2ab cos C =72+82-2×7×8×13

14=9. ∴c =3,因此最大角为B ,由余弦定理,得 cos B =a 2+c 2-b 22ac =-17. 答案 -17

8.在△ABC 中,角A ,B ,C 所对的边分别为a ,b ,c ,若a =1,b

解析 由余弦定理,得

cos B =a 2+c 2-b 22ac =1+3-72×1×3=-32,∴B =5π6.

答案 5π6

10.在△ABC 中,已知a =7,b =10,c =6,判断△ABC 的形状. 解 由余弦定理,知cos B =a 2+c 2-b 22ac =72+62-1022×7×6=-5

28.

在△ABC 中,0°

11.在△ABC 中,角A ,B ,C 所对的边分别为a ,b ,c ,且2b ·cos A =c ·cos A +a ·cos C . (1)求角A 的大小;

(2)若a =7,b +c =4,求bc 的值.

解 (1)根据正弦定理及2b ·cos A =c ·cos A +a ·cos C , 得2sin B cos A =sin C cos A +sin A cos C =sin(A +C )=sin B . ∵sin B ≠0,∴cos A =1

2. ∵0

3. (2)根据余弦定理得

7=a 2=b 2+c 2-2bc cos π

3=(b +c )2-3bc , ∵b +

12???sin C 2, n =? ?的夹角为π3.

(1)求(2)已知c =72,三角形面积S =33

2,求a +b . 解 (1)∵m =(cos C 2,sin C

2), n =(cos C 2,-sin C

2), ∴m ·n =cos 2C 2-sin 2C

2=cos C . 又m ·n =|m |·|n |cos π3=12, ∴cos C =1

2.又0

3.

(2)∵c 2=a 2+b 2-2ab cos C ,c =7

2, ∴49

4=a 2+b 2-ab =(a +b )2-3ab . ∵S =12ab sin C =12ab sin π3=3

4ab , 而S =33

2,∴ab =6.

∴(a+b)2=49

4+3ab=

49

4+18=

121

4.

∴a+b=11

2.

双基限时练(三)

1.在△ABC中,角A,B,C的对边分别为a,b,c,若a2+c2-b2=3ac,则角B的值为()

A.π

6 B.π3

C.π

6,或

6 D.

π

3,或

3

解析由余弦定理,得cos B=a2+c2-b2

2ac=

3ac

2ac=

3

2,又0

π

6.

答案 A

4.已知三角形的三边之比为a:b:c=2:3:4,则此三角形的形状为() A.锐角三角形B.钝角三角形

C.直角三角形D.等腰直角三角形

解析设三边长为2a,3a,4a(a>0),它们所对的三角形内角依次为A,B,C.

则cos C=(2a)2+(3a)2-(4a)2

2×2a×3a

=-

1

4<0,

∴C为钝角.故该三角形为钝角三角形.

答案 B

6.△ABC中,已知2A=B+C,且a2=bc,则△ABC的形状是() A.两直角边不等的直角三角形

B.顶角不等于90°,或60°的等腰三角形

C.等边三角形

D.等腰直角三角形

解析解法1:由2A=B+C,知A=60°.

又cos A=b2+c2-a2

2bc,∴

1

2=

b2+c2-bc

2bc

∴b2+c2-2bc=0.即(b-c)2=0,∴b=c. 故△ABC为等边三角形.

解法2:验证四个选项知C成立.

答案 C

8.△ABC中,已知a=2,c=3,B=45°,则b=________.

解析由余弦定理,得b2=a2+c2-2ac cos B=2+9-2×2×3×

2

2=5,∴b= 5.

答案 5

9.在△ABC中,a=23,cos C=

1

3,S△ABC=43,则b=________.

解析∵cos C=

1

,∴sin C=

22

.又S△ABC=

1

2ab sin C,

∴43b=3 2.

答案

10cos C

是方程2x2-3x-2=0的一个根,求△ABC周长的最小值.

x1=-

1

2,x2=2,而cos C为方程2x

2-3x-2=0的一个根,∴cos C=-c2=a2+b2-2ab cos C,得c2=a2+b2+ab.∴c2=(a+b)2-ab=100-ab=100-a(10-a)=a2-10a+100=(a-5)2+75≥75,∴当a=b=5时,c min=5 3.从而三角形周长的最小值为10+5 3.

11.在△ABC中,如果lg a-lg c=lgsin B=-lg2,且B为锐角,试判断此三角形的形状.

解∵lgsin B=-lg2,∴sin B=

2

2.又∵B为锐角,∴B=45°.∵lg a-lg c=-lg2,∴

a

c=

2

2.

由正弦定理,得

sin A

sin C=

2

2.

即2sin(135°-C)=2sin C.

∴2(sin135°cos C-cos135°sin C)=2sin C.

∴cos C=0,∴C=90°,∴A=B=45°.

∴△ABC是等腰直角三角形.

12.a,b,c分别是△ABC中角A,B,C的对边,且(sin B+sin C+sin A)(sin B+sin C-sin A)=

18

5sin B sin C,边b和c是关于x的方程x2-9x+25cos A=0的两根(b>c).

(1)求角A的正弦值;

(2)求边a,b,c;

(3)判断△ABC的形状.

解 (1)∵(sin B +sin C +sin A )(sin B +sin C -sin A )=18

5sin B sin C , 由正弦定理,得(b +c +a )(b +c -a )=18

5bc , 整理,得b 2+c 2-a 2=8

5bc .

由余弦定理,得cos A =b 2+c 2-a 22bc =45,∴sin A =3

5. (2)由(1)知方程x 2-9x +25cos A =0可化为x 2-9x +20=0, 解之得x =5或x =4,∵b >c ,∴b =5,c =4. 由余弦定理a 2=b 2+c 2-2bc cos A ,∴a =3. (3)∵a 2+c 2=b 2,∴△ABC 为直角三角形.

双基限时练(四)

1.在△ABC 中,若sin B :sin C =3:4,则边c b 等于( ) A .4:3,或16:9 B .3:4 C .16:9

D .4:3

解析 由正弦定理c sin C =b sin B ,得c b =sin C sin B =4

3. 答案 D

3.在△ABC 中,若a cos A =b cos B =c

cos C ,则△ABC 是( ) A .直角三角形 B .等边三角形 C .钝角三角形

D .等腰直角三角形

解析 由正弦定理及题设条件,知sin A cos A =sin B cos B =sin C cos C .由sin A cos A =sin B

cos B ,得sin(A -B )=0.∵0

答案 B

4.在△ABC 中,如果BC =6,AB =4,cos B =1

3,那么AC 等于( ) A .6 B .2 6 C .3 6

D .4 6

解析 由余弦定理,得 AC 2=BC 2+AB 2-2·AB ·BC ·cos B

=62+42-2×6×4×1

3 =36,∴AC =6. 答案 A

双基限时练(五)

6.某人向正东方向走x km 后,向右转150°,然后朝旋转后的方向走3 km 后他离最开始的出发点恰好为 3 km ,那么x 的值为________.

解析 AB =x ,BC =3,AC =3,∠ABC =30°. 2×3×x cos30°,

即x 2-=3,x 2=23,经检验都适合题意. 答案 3或2 3

7.某海岛周围38海里有暗礁,一轮船由西向东航行,初测此岛在北偏东60°方向,航行30海里后测得此岛在东北方向,若不改变航向,则此船________触礁的危险(填“有”或“无”).

双基限时练(六)

1.在△ABC 中,已知BC =6,A =30°,B =120°,则△ABC 的面积等于( ) A .9 B .18 C .9 3

D .18 3

解析 由正弦定理得

AC sin B =BC

sin A

, ∴AC =BC ·sin B sin A =6×sin120°

sin30°=6 3. 又∠ACB =180°-120°-30°=30°, ∴S △ABC =12×63×6×1

2=9 3. 答案 C

2.在△ABC 中,若a 2+b 2+ab

B .锐角三角形

C .直角三角形

D .形状无法判定

解析 由a 2+b 2+ab

2.

又cos120°=-1

2,∴C >120°,故△ABC 为钝角三角形. 答案 A

3.在△ABC 中,BC =2,B =π3,若△ABC 的面积为3

2,则tan C 为( ) A. 3 B .1 C.33

D.32

解析 由S △ABC =12BC ·BA sin B =3

2,得BA =1, 由余弦定理,得

AC 2

=AB 2

+BC 2

-2AB ·BC cos B . ∴AC =3,∴AC 2

+BA 2

=BC 2

.

∴△ABC 为直角三角形,其中A 为直角. ∴tan C =AB AC =33. 答案 C

4.三角形的两边长为3和5,其夹角的余弦值是方程5x 2

-7x -6=A .6 B.152 C .8

D .10

解析 由5x 2-7x -6=0,得x =-35,或x =2(舍去).∴cos α=-35,答案 A

5.△ABC 中,A =60°,b =16,此三角形的面积S =2203,则a 的值为( ) A .7 B .25 C .55

D .49

解析 由S =220 3,得1

2bc sin A =220 3. 即12×16×c ×3

2=220 3,∴c =55.

∴a 2=b 2+c 2-2bc cos60°

=162+552-2×16×55×1

2=2401. ∴a =49. 答案 D

6.在△ABC 中,a ,b ,c 分别是角A ,B ,C 所对的边,已知a =3,b =3,C =30°,则A =________. 解析 c 2=a 2+b 2-2ab cos C =3+9-2×3×3×3

2=3, =3·123=12,

所对的边分别为a ,b ,c ,若(3b -c )cos A =a cos C ,则cos A =______. , (3sin B -sin C )cos A =sin A cos C .

∴3sin B cos A =sin(A +C )=sin B .∴cos A =33.

答案 3

3

8.在△ABC 中,a 2-b 2+bc ·cos A -ac ·cos B =________.

解析 由余弦定理cos A =b 2+c 2-a 22bc ,得bc ·cos A =12(b 2+c 2-a 2),同理ac ·cos B =12(a 2+c 2-b 2). ∴a 2-b 2+bc ·cos A -ac ·cos B

=a 2-b 2+12(b 2+c 2-a 2)-1

2(a 2+c 2-b 2) =a 2-b 2+b 2-a 2=0. 答案 0

9.在△ABC 中,A =60°,b =1,c =4,则a +b +c

sin A +sin B +sin C 的值为________.

解析 在△ABC 中,由正弦定理得

a sin A =

b sin B =

c sin C

=2R ,得a +b +c =2R (sin A +sin B +sin C ).

又a 2

=b 2

+c 2

-2bc cos A =1+16-2×1×4×1

2=13,

∴a =13,

∴a +b +c sin A +sin B +sin C =2R =a sin A =13sin60°=2393. 答案

2393

10.在锐角△ABC 中,a ,b ,c 分别为角A ,B ,C 所对的边,又c =21,b =4,且BC 边上的高h =2 3.

(1)求角C ; (2)求边a 的长.

解 (1)由于△ABC 为锐角三角形,过A 作

AD ⊥BC 于D 点,

sin C =234=3

2,则C =60°. (2)由余弦定理,可知 c 2=a 2+b 2-2ab cos C ,

则(21)2=42+a 2-2×4×a ×1

2,即a 2-4a -5=0. 所以a =5,或a =-1(舍). 因此所求角C =60°,边a 长为5.

11.在△ABC 中,内角A ,B ,C 对边的边长分别是a ,b ,c ,已知

(1)若△ABC 的面积等于3,求a ,b ;

(2)若sin C +sin(B -A )=2sin2A ,求△ABC 的面积. 解 (1)由余弦定理及已知条件,得 a 2+b 2-ab =4.

又因为△ABC 的面积等于3, 所以1

2ab sin C =3得ab =4,

联立方程组???

a 2+

b 2

-ab =4,

ab =4,

解得a =2,b =2.

(2)由题意,得sin(B +A )+sin(B -A )=4sin A cos A , 即sin B cos A =2sin A cos A . 当cos A =0时,A =π2,B =π

6, ∴a =433,b =233.

∴△

b =2

3 3. 当cos 解得???

??

a =233,

b =433.

∴△ABC 的面积S =12ab sin C =23

3.

12.△ABC 的面积是30,内角A ,B ,C 所对边长分别为a ,b ,c ,cos A =12

13. (1)求AB →·AC →;

(2)若c -b =1,求a 的值.

解 (1)在△ABC 中,∵cos A =1213,∴sin A =513. 又S △ABC =1

2bc sin A =30,∴bc =12×13. ∴AB →·AC →=|AB →||AC →|cos A =bc cos A =144. (2)由(1)知bc =12×13,又c -b =1, ∴b =12,c =13.

在△ABC 中,由余弦定理,得 a 2=b 2+c 2-2bc cos A

=122+132-2×12×13×12

13=25, ∴a =5.

人教新课标版语文高一人教版必修三双基限时练 12动物游戏之谜

双基限时练(十二)动物游戏之谜 一、基础测试 1.下列词语中,加点字的注音完全正确的一项是() A.缅.甸(miàn)聒.噪(ɡuō) 潜.游(qián) 嬉.闹(xī) B.脊.背(jǐ) 挨.挤(ái) 倒.立(dào) 消耗.(hào) C.汲.取(jí) 安抚.(fǔ) 尾鳍.(qí) 格.斗(ɡé) D.反馈.(kuì) 模.式(mó) 捉.摸(zuó) 嚼.烂(jiáo) 解析A.缅miǎn;B.挨āi;D.捉zhuō。 答案 C 2.下列词语中,没有错别字的一项是() A.伙伴兴高彩烈平衡史无前例 B.默契得意扬扬厮打销声匿迹 C.跳越猩猩相惜消融铁打金刚 D.元气自娱自乐调剂杂草老藤 解析A.兴高彩.烈—采;B.得意扬扬 ..—洋洋;C.跳越.—跃,猩猩 ..相惜—惺惺。 答案 D 3.下列加点的成语使用不正确的一项是() A.在动物身上,无论从形态结构、生理过程,还是行为方面去分析,尽可能节省能量 的例子几乎俯拾皆是 ....。 B.单独游戏时,动物常常兴高采烈 ....地独自奔跑、跳跃,在原地打圈子。 C.每当刮起大风时,成群的露脊鲸把尾鳍高高举出水面,正对着大风,以便像船帆似 的,让大风推着它们,得意洋洋 ....地“驶”向海岸。 D.研究者们各执己见,众说纷纭 ....。 解析A项中应是“比比皆是”。“比比皆是”和“俯拾皆是”都表示相同的事物很多,到处都是。前者侧重表示多得很,到处都是;后者侧重表示容易得到。 答案 A 4.下列各句没有语病的一句是() A.动物的游戏行为,是动物行为研究中被认为最复杂、最难以捉摸、引起争论最多的

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双基限时练26

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双基限时练(三) 1.设f ′(x 0)=0,则曲线y =f (x )在点(x 0,f (x 0))处的切线( ) A .不存在 B .与x 轴垂直 C .与x 轴平行 D .与x 轴平行或重合 答案 D 2.一木块沿某一斜面自由下滑,测得下滑的水平距离s 与时间t 之间的函数关系为s =18t 2,则当t =2时,此木块在水平方向的瞬时速 度为( ) A. 2 B. 1 C.12 D .14 解析 s ′=lim Δt →0 Δs Δt =lim Δt →0 18(t +Δt )2-18t 2Δt =lim Δt →0 14tΔt +18(Δt )2Δt =lim Δt →0 (14t +18Δt )=14t . ∴当t =2时,s ′=12. 答案 C 3.若曲线y =h (x )在点P (a ,h (a ))处切线方程为2x +y +1=0,则 ( ) A .h ′(a )<0 B .h ′(a )>0 C .h ′(a )=0 D .h ′(a )的符号不定 解析 由2x +y +1=0,得h ′(a )=-2<0. ∴h ′(a )<0. 答案 A

4.曲线y =9x 在点(3,3)处的切线方程的倾斜角α等于( ) A .45° B .60° C .135° D .120° 解析 k =y ′=lim Δx →0 Δy Δx =lim Δx →0 9x +Δx -9x Δx =lim Δx →0 -9x (x +Δx )=-9x 2. ∴当x =3时,tan α=-1.∴α=135°. 答案 C 5.在曲线y =x 2 上切线倾斜角为π4的点是( ) A .(0,0) B .(2,4) C .(14,116) D .(12,14) 解析 y ′=lim Δx →0 Δy Δx =lim Δx →0 (x +Δx )2-x 2Δx =lim Δx →0 2xΔx +(Δx )2Δx =lim Δx →0 (2x +Δx )=2x . 令2x =tan π4=1,∴x =12,y =14. 故所求的点是(12,14). 答案 D 6.已知曲线y =2x 2上一点A (2,8),则过点A 的切线的斜率为________. 解析 k =f ′(2)=lim Δx →0 2(2+Δx )2-2×22Δx

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解析 分层抽样应按比例抽取,所以应抽取三年级的学生人数为200×2 10=40. 答案 B 6.一个单位共有职工200人,其中不超过45岁的有120人,超过45岁的有80人.为了调查职工的健康状况,用分层抽样的方法从全体职工中抽取一个容量为25的样本,应抽取超过45岁的职工________人. 解析 依题意得,抽取超过45岁的职工人数为25 200×80=10. 答案 10 7.某工厂生产A ,B ,C 三种不同型号的产品,产品数量之比依次为 现用分层 抽样方法抽出一个容量为n 的样本,样本中A 种型号产品有16件,那么此样本的容量n =________. 解析 由题意得n =16×10 2=80. 答案 80 8.课题组进行城市空气质量调查,按地域把24个城市分成甲、乙、丙三组,对应城市数分别为4,12,8.若用分层抽样抽取6个城市,则丙组中应抽取的城市数为________. 答案 2 9.某企业有三个车间,第一车间有x 人,第二车间有300人,第三车间有y 人,采用分层抽样的方法抽取一个容量为45人的样本,第一车间被抽取20人,第三车间被抽取10人,问:这个企业第一车间、第三车间各有多少人? 解 x =20×30045-20-10=400(人),y =10×300 45-20-10 =200(人). 10.某单位有工程师6 人,技术员12 人,技工18 人,要从这些人中抽取一个容量为 n 的样本.如果采用系统抽样和分层抽样方法抽取,都不用剔除个体;如果样本容量增加1 个,则在采用系统抽样时,需要在总体中先剔除1个个体,求样本容量n . 解 解法1:总体容量为6+12+18=36(人).当样本容量是n 时,由题意知,系统抽样的间隔为36n ,分层抽样的比例是n 36,抽取工程师人数为n 36×6=n 6人,技术人员人数为n 36×12 =n 3人,技工人数为n 36×18=n 2 人,所以n 应是6的倍数,36的约数,即n =6,12,18. 当样本容量为(n +1)时,总体容量是35 人,系统抽样的间隔为35n +1,因为35 n +1 必须是整数,所以n 只能取6,即样本容量n =6. 解法2:总体容量为6+12+18=36(人). 当抽取n 个个体时,不论是系统抽样还是分层抽样,都不用剔除个体,所以n 应为

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双基限时练(十二) 用样本的频率分布估计总体的分布 基础强化 1、下列关于频率分布直方图的说法,正确的就是() A、直方图的高表示取某数的频率 B、直方图的高表示该组上的个体在样本中出现频数与组距的比值 C、直方图的高表示该组上的个体在样本中出现的频率 D、直方图的高表示该组上的个体在样本中出现频率与组距的比值 解析频率分布直方图的纵坐标表示错误!,故选D、 答案D 2、从一群学生中抽取一个一定容量的样本对她们的学习成绩进 行分析,前三组就是不超过80分的人,其频数之与为20人,其频率之与(又称累积频率)为0、4,则所抽取的样本的容量就是() A、100 B、80 C、40 D、50 解析样本容量为亠 = 50、故选D、 0、4 答案D 3、下列说法不正确的就是() A、频率分布直方图中每个小矩形的高就就是该组的频率 B、频率分布直方图中各个小矩形的面积之与等于1

C、频率分布直方图中各个小矩形的宽一样大 D、频率分布折线图就是依次连接频率分布直方图的每个小矩形 上端中点得到的 解析在频率分布直方图中各个小矩形的高就就是该组的错误!、 答案A 4、某市教育行政部门为了对2012届高中毕业生的学生水平进行评价,从该市高中毕业生中抽取1000名学生的数学成绩作为样木进行统计,其频率分布直方图如图所示、则这1000名学生的数学平均成绩的最大可能值为() 频率 O 40 50 60 70 80 90 100分数(分) A、67、50 B、72、05 C、76、50 D、77、50 解析由题意得平均成绩的最大可能值为0、05X50 + 0、1X60 + 0、25X70 + 0、35X80 + 0、15X90 + 0、1X100 = 77、50、 答案D 5、一个社会调查机构就某地居民的月收入调查了10000人,并根据所得数据画了样本的频率分布直方图(如下图)、为了分析居民的收入与年龄、学历、职业等方而的关系,要从这10000人中再用分?层抽样方法抽出100人作进一步调查,则在[2500, 3000)(元)月收入段应

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双基限时练(二十一) 1.已知a =(-3,4),b =(5,2),则a ·b =( ) A .23 B .7 C .-23 D .-7 解析 a ·b =-3×5+4×2=-7,故选D. 答案 D 2.已知向量a =(1,-1),b =(2,x ).若a·b =1,则x =( ) A .-1 B .-12 C.12 D .1 解析 由a =(1,-1),b =(2,x )可得a·b =2-x =1,故x =1. 答案 D 3.若非零向量a ,b ,满足|a |=|b |,(2a +b )·b =0,则a 与b 的夹角为( ) A .30° B .60° C .120° D .150° 答案 C 4.已知A ,B ,C 是坐标平面上的三点,其坐标分别为A (1,2),B (4,1),C (0,-1),则△ABC 的形状为( ) A .直角三角形 B .等腰三角形 C .等腰直角三角形 D .以上均不正确 解析 AB →=(3,-1),AC →=(-1,-3),BC → =(-4,-2), ∴|AB →|=10,|AC →|=10,|BC → |=20.

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