Nối tiếp nội dung phần 1 Tài liệu Thiết kế bài giảng Đại số 10 nâng cao (Tập 2), phần 2 giới thiệu tới người đọc cách thiết kế bài giảng Đại số nâng cao về chủ đề thống kê, góc lượng giác và công thức lượng giác. Mời các bạn cùng tham khảo.
Chi/dNq V THONG KE PHAN1 OTlClVG VAIV D E CUA CHUW^TG L NOIDUNG Ndi dung chinh cua chuong gdm nhiing va'n de sau : Khai niem md dau: Miu so lieu, kich thudc miu Miu sd lieu: Bang phan bd tin so, tan suat Bieu dd : Bieu tin so : Hinh cot, dudng gap khiic Bieu dd tan suit: Hinh cot, hinh quat, dudng gap khiic So trung binh cdng, sd trung vi va mdt • Phuong sai va lech chuin IL MUC TIEU Kien thufc Nim dugc toan bd kien thiic co ban chuong da neu tren Nim dugc cac khai niem : Sd lieu thdng ke, tin sd, tin suit, bang phan bd tin so, tin suat ghep ldp Cac bieu dd tin sd va tin sua't, md ta, dgc va ve dugc bieu dd Nim dugc sd trung binh cdng, sd trung vi va mdt cua day hoac bang so lieu 180 Nim va tinh dugc phuong sai va lech chuin, tir dd cd the danh gia, du doan dugc cac tieu chi cin thiet Kl nang Kl nang tinh toan cac sd lieu thdng ke Kl nang nhan biet bao chinh xac danh gia cac sd lieu thdng ke, tir dd cd nhirng du Kl nang dgc va ve dugc cac bieu dd thdng ke Kl nang tinh phuong sai va lech chuan Thai Hgc sinh ren luyen tinh cin than, kien tri va khoa hgc tinh toan cac sd lieu thdng ke Hgc sinh thay dugc y nghia va tam quan trgng ciia thdng ke va khoa hgc ciia nd ddi sdng 181 PHAN CAC B A I SOAl^ §1 Mot vai khai niem md dau (1 tiet) I MUC TIEU Kien thiifc Giiip HS nim dugc : Khai niem thdng ke la gi? Miu sd lieu va kich thudc miu KT nang Ren luyen ki nang nhan biet khai niem thdng ke Kl nang tim kich thudc miu Thai Thdng qua khai niem thdng ke, miu sd lieu va kich thudc miu HS lien he vdi thuc te' va tir thuc te' cd the thie't lap mdt bai toan thdng ke Hieu ro hon vai trd ciia thdng ke ddi sdng II CHUAN BI CUA GV VA HS Chuan hi ciia GV : GV : Chuin bi mot sd bang SGK Chuan bi mot sd cau hdi nhim din dit HS thao tac day hgc Chuan bj ciia HS : HS : Cin dn lai mdt sd kien thiic ve ham sd da hgc d ldp IIL PHAN PHOI THOil LUONG Bai chia lam tiet 182 Bdi tap phdn ndy chii ye'u Id hudng ddn ve nhd IV TIEN TRINH DAY HOC A OAT VAN DE Cau hdi 1) Em hay thdng ke diem cac mdn hgc cua em 10 tuan dau tien Cau hdi Em hay sip xep cac diem so theo thii tu tang din B BAI IVl6l HOAT DONG 1 Thdng ke la gi? • GV neu mdt so vi du ve thdng ke: Thdng ke dan so ciia mdt dia phuang, thdng ke ke't qua hgc tap ciia hgc sinh, thdng ke tang trudng kinh te' ciia mot don vi sin xuat, HI Em hay neu mdt vi du thdng ke ma em biet H2 Hay neu ddi tugng dieu tra thdng ke em vira neu HOAT DONG 2 Mau sd lieu • GV treo bang sach, sau dd neu cac cau hdi sau: HI Da'u hieu dieu tra d day la gi? H2 Don vi dieu tra d day la gi? Sau dd GV neu khai niem kich thudc miu, Mot tap hihi han cdc dan vi dieu tra dugc ggi Id mot mau Sd phdn tic ciia mot mau dugc ggi Id kich thudc mdu Day cdc gid tri ciia ddu hieu thu dugc tren mdu dugc ggi Id mot mdu so lieu 183 H3 Neu kich thudc mau vi du tren • Ne'u thuc hien dieu tra tren mgi don vi dieu tra thi la dieu tra todn bo Ne'u chi dieu tra tren mdt miu thi dd la dieu tra mdu Thuc hien H1 Hoat ddng cua GV Hoat ddng ciia HS Cau hdi Mot nha may thirdng san xuat vdi so lirgng hop siia nhidu hay it? Cau hdi Cd the dieu tra dirac toan bd hay khdng? Ggi y tra Idi cau hdi Thuang la nhieu va khdng dem thii cdng dirge Ggi y tra Idi cau hdi Khdng the dieu tra toan bd dirge GV neu kha nang dieu tra: Chi dieu tra miu TOM TAT BAI HOC Thdng ke la khoa hgc ve cac phuong phap thu thap, td chiic, trinh bay, phan tich va xir li sd lieu Mot tap hitu han cdc dan vi dieu tra dugc ggi Id mot mdu So phdn tit ciia mot mdu dugc ggi Id kich thudc mdu Day cac gia tri ciia dau hieu thu dugc tren miu dugc ggi la mot mdu sd'liiu MOT SO CAU HOI T R A C NGHIEM Cdu Khi dieu tra dan so, mdt can bd dieu tra dua ket luan sau: (a) Ket qua dieu tra ludn ludn dung tai mgi thdi diem (b) Ket qua dieu tra ludn ludn diing tai mgi thdi diem trudc dieu tra (c) Ket qua dieu tra ludn ludn dung tai thdi diem ket thiic dieu tra (d) Ket qua dieu tra chi de tham khao de phan doan mdt sd lieu cin thie't nao dd Hay chgn khing dinh diing 184 Trd Idi Chgn(d) Cdu Khi dieu tra chieu cao cua mdt khdi hgc tai mdt trudng phd thdng, ngudi ta chgn 30 em HS bat ki ciia khdi do: a) Hay chgn khing dinh diing cac khing dinh sau: (a) Miu sd lieu la tat ca HS cua khdi (b) Miu sd lieu la tat ca HS toan trudng (c) Miu sd lieu la HS ciia khdi (d) Miu sd lieu la 30 HS cua khdi Trd Idi Chgn (d) b) Hay chgn hing dinh diing cac khang dinh sau: (a) Kich thudc miu la 30 (b) Kich thudc miu khdng xac dinh (c) Kich thudc miu la (d) Kich thudc miu la mdt so khac 30 va Trd Idi Chgn (a) HUdNG DAN B A I TAP SGK Bai Deldm bdi tap ndy, HS edn dn tap lai: - Ddu hieu dieu tra -Mdu sdlieu - Kich thudc mdu Hudng ddn a) Dau hieu la sd mdt gia dinh Kich thudc miu la 80 b) Cd gia tri khac miu so lieu tren l a O ; l ; ; ; ; ; ; Bai De ldm bdi tap ndy, HS can dn tap lai: - Ddu hieu dieu tra -Mdu sdlieu - Kich thudc mdu 185 Hudng ddn Dan vi dieu tra la gia dinh Dau hieu la sd dien nang tieu thu mdt thang cua mdt gia dinh Kich thudc miu la 30 Cd 18 gia tri khac miu sd lieu tren la 40 ; 42 ; 45 ; 50 ; 53 ; 57 ; 59 ; 65 ; 70 ; 75 ; 84 ; 85 ; 90 ; 100 ; 133 ; 141 ; 150; 165 186 §2 Trinh bay mot m a u so' l i e u (tiet 2, 3) L MUC TIEU Kien thirc Giiip HS nim dugc : Khai niem tin sd, tin suit, bang phan bd tan so, tin suit Cach tim tan so va tin suit cua mot bang sd lieu thdng ke KT nang Ren luyen ki nang tinh toan thdng qua viec tim tin sd, tin sua't KT nang dgc va thiet lap bang phan bd tan so, tin sua't ghep ldp KT nang du bao cac tieu chi, thdng qua so lieu thdng ke Thai Thdng qua khai niem tin sd, tan suat, HS lien he vdi nhieu bai toan thuc te va til' thuc te cd the thiet lap mdt bai toan thdng ke Hieu rd hon vai trd cua toan hgc ddi sdng IL CHUAN BI CUA GV VA HS Chuan bi ciia GV : GV : Chuin bi mot sd bang SGK nhu : bang Iden bang Chuin bi mdt sd hinh SGK nhu : hinh 5.1, 5.2, 5.3, 5.4 Chuin bi mdt so cau hdi nhim din dit HS thao tac diiy hgc Chuin bi pha'n mau, mot sd dung cu day hgc khac nhu thudc, compa, may tinh bd tiii Chuan hi ciia HS : HS : Cin dn lai mdt sd kien thiic ve ham so da hgc d ldp IIL PHAN PHOI T H I LUONG Bai chia lam tiet 187 Tiit 1: Tit ddu din hit phdn Tiit 2: Phdn Iqi vd hudng ddn bdi tap ve nhd IV TIEN TRINH DAY HOC A BAI CCl Cau hdi 1) Em hay thdng ke diem cac mdn hgc cua em 10 tuin diu tien 2) Xac dinh xem diem sd nao xuat hien nhieu nhat Tinh ti le phin tram mdi diem sd xuat hien Cau hdi Em hay tu lam mdt dieu tra nhd va cho biet: Miu so lieu va kich thudc miu B BAI Mdi HOAT DONG 1 Bang phan bd tan sd - tan suat • GV neu vidu Sau dd dat cac cau hdi sau: HI Trong cac so lieu tren cd bao nhieu gia tri H2 Hay neu sd lin xuat hien cua tiing gia tri • GV neu khai niem tin sd So lan xudt hien ciia moi gid tri mdu so' lieu dugc ggi Id tdn sd ciia gid tri • GV treo bang va gidi thieu ve bang phan bd tin sd Sau dd neu nhan xet: - Thudng bang phan bd tin sd gdm hang : Gia tri va tin so - Sd cot thudng la sd gia tri (tap hgp cac gia tri) Neu cac cau hdi sau: 188 HI Tdng sd cac tan sd bing bao nhieu? H2 Hay so sanh tdng tren vdi kich thudc miu • GV neu khai niem tin suit: Tdn siidtfi ciia gid tri x^ Id ti so giifa tdn sduj vd kich thudc mdu N f i = ^ ' N Sau dd neu bang phan bd tan sd - tin suit nhu bang • Neu chii y: a) Tren hdng tdn sd ngudi ta thudng danh mot de ghi kich thudc mail N Kich thudc mdu N bang tdng cdc tdn sd b) Cd the vie't bdng tdn so' — tdn suat dang "ngang" (nhU bdng 2) thdnh bdng "dgc" (chuyen hdng thdnh cot nhu bdng 3) Thuc hien H1 Hoat ddng cua GV Cau hdi Hay neu kich thirdc miu Cau hdi Neu tan suat diem Cau hdi Hay tinh cac tan suat lai va dien vao chd trd'ng Hoat ddng cua HS Ggi y t r a idi cau hdi Kich thudc miu : 400 Goi y t r a Idi cau hdi Tin suat diem la / = — = 13,75% ^ 400 Ggi y tra Idi cau hdi GV cho HS tinh va chia HS td mdi td cir dai dien len bang dien, GV nhan xet va cho diem 189 Sd trung binh la (tinh den hang phin tram): (a) 81; (c) 82,08; (b) 82; (d)83 Cau Tren dudng trdn don \'i mdt cung trdn co sd la 22536° Hay chgn sd radian cua cung trdn dd cac cung trdn cd sd sau day (a) K (b)f ; K (d) 125;r + (c) ; r - - ; IL K^ CAU HOI Tir LUAN Can Xac dinh m de cac tam thiic sau ludn duong vdi mgi x: a) (w - \)x - 2mx + m-6\ b) (m + l)x -2mx + iv + Cau Tinh n 2n 9n a) cos —+ COS I- + COS — 5 , , 71 271 971 b) sin —+ Sin — + sin — 5 Cau Sd tien dien cua 30 hg dan d mdt khu phd cho bdi bang sau: Ldp Ldp Tan sd [60; 69] [70; 79] 10 [80; 89] [90; 99] N = 30 399 a) Hay tinh sd tien trung binh mdi hg tieu thu dien mdt thang b) Tinh tin suat ciia mdi ldp Hudng ddn vd ddp dn: I CAU HOI TRAC NGHIEM KHACH QUAN Cau (I d) Chgn (c) Cau (1 d) Chgn (d) Cau (1 d) Chgn (c) Cau (1 d) Chgn (d) IL CAU HOI T U LUAN Cau (2 d, mdi cau d) a) Tam thiic ludn ludn duang va chi : 777 > - /?? + > 2 m - 777 + > 2 777 -(777 +4/77 + ) < A 777 > Cau (2 d mdi cau I diem) r r n n 7t Sn „,v ^ a) la CO c o s — = c o s ( H — ) = - c o s — ; c o s — = cos(7rH COS 971 5 ^ 471^ 471 = C0S(7H ) = -cos 5 71 271 971 D o d o : cos —+ C0S h + c o s — = cos7r = - l 5 b) Lam tuong tu cau a) DS 400 371 371 ^,.^ ) = -cos- Cau (3 diem, mdi cau 1,5 d) a)x « 80,2 (don vi nghin ddng) b) Tin suit tuong ring la : 16,7 ; 33,3 ; 26,7 ; 23,3 Oe s5 L CAU HOI TRAC NGHIEM KHACH QUAN Can Cho bat phuo'ng trinh 3 x+-^^>2+—— x-2 x-2 (1) Tap nghiem ciia bat phuong trinh la : (a) S = [2; +00); (b) S = (2; +oo); (c) S = R\ {2}; (d) ca ba phuong an tren deu sai Hay chgn ket qua diing Cau Thdng ke sd tien dien thoai 12 hg dan mot td dan sd ta dugc ket qua nhu sau : 55, 76, 85, 75, 53, 63, 72, 64, 82, 96, 112, 210 sd trung vi la (tinh de'n hang phin tram) (a) 63; (b) 72; (c) ; (d) 67,5 5K / , 19 Gdc cd so — cd so dd la 12 (a) 75; (b) 80; (c)85 ; (d)60 Cau Khdng giai bat phuang trinh x +7x - 12 < Hay dien diing-sai vao d trd'ng Bat phuang trinh (a) cd hai nghiem; 1_| 401 D D D (b) ludn cd nghiem; (c) vd nghiem (d) cd nghiem la mdt khoang IL TULUAN Can L Cho ham sd tam thiic /(x) = x" - x + /7/-l a) Xac dinh m de tam thiic cd hai nghiem phan biet b) Xac dinh m de f(x) hay - m + I > hay m < b) f(x) < X V X G ( - ; 1) va chi : f/(l) simplify ( s i n ( x ) ' ' - c o s (x)'^4) ; - cos(x)" Vay ta thay sin x + cos x = l - c o s x > s i m p l i f y ( s i n (x) ^ + c o s (x) ^^6) ; - cos(x)'^ + cos(x)'* Vay ta thha'y sin x + cos x = l - S c o s " x + 3cos x Bing each tuong tu ta cd: > simplify (sin (x) ^10+cos (x) '^10) ; -5 cos(x)-+ lOcos(x)^- 10cos(x)^ + cos(x)^ > simplify(sin(x)^10-cos(x) ^10) ; -2cos(x)'"+ -5 cos(x)-+ lOcos(x)^- 10 cos(x)'V cos(x)^ > simplify (cos (x)''2-*sin (x)'"4-cos (x)'^4*sin (x) ^^2) ; - cos(x)'^ + cos(x)' + cos(x)^ Ta CO the riit ggn bieu thuc theo y mudn Vidu t h l : = ( ( s i n ( x ) ' " - c o s (x) ' " ) + s q r t ( l + * x + x ^ ) + e x p (x) ^2) ; //7l:=sin(x)'^-cos(x)'"+V(l + x)- +{e Ta riit ggn phan mu: > simplify(thi,power); s i n ( x ) ' " - c o s ( x ) ' " + csgn( + x ) ( l + x ) + exp-^ 404 Bing each lam nhu vay ta ludn riit gon dugc bieu thiic lugng giac ndi rieng va bieu thii'c ndi chung II NHlmG TINH TOAN THONG KE B A N G MAPLE De tim cac sd dac trung cua bang so lieu thdng ke, ta phai sip xep cac sd lieu thdng ke theo kieu tap hgp hay kieu mang Maple lam dieu that ki dieu Tuy nhien chiing ta cung phai dung nhung ngdn ngQ' lap trinh maple Sau day la mot sd vi du > ord > if :=proc(a,b) a end: > sort([4.7,8,9,22,45,3,6,5,4 9,23,34],ord); [3 4 6, 8, 9, 22 23 34 45] > taphopl:={4,7,8,9,22,45,3,6,5,4,9,23,34}; taphopl := { 5, 6, 7, 8, 22, 23 34, 45 i Tong cua tap hgp > nops(taphopl); 11 Ta chu y ring lenh sort la de sip xep ; lenh nops de dem sd phin tir ciia tap hgp hoac danh sach Tir ta tinh dugc cac sd dac trung: Tong sd danh sach la : > ds:=[3, 4, 4, 5, 6, 1, 8, 9, 22, 23, 34, 45]; ds := [3, 4, 6, 9, 9, 22, 23, 34 45 ] Tong cua danh sach > convert(ds,"+ ); 179 Sd phin tir cua danh sach: > convert(ds,table) ; table([\ = 3,2 = 4,3 = 4,4 = 5,5 = 6,6 = 7 = 8,8 = 9,9 = 9,10 = 22,11 = 23,12 = 34,13 = 45]) 405 > nops(ds); 13 So trung binh cdng la: > TB:=(179/13); 179 TB:= 13 Sd trung vi la sd thii danh sach: > TV:=op(7,ds); rF:=8 Mdt: Nhin vao danh sach da sip xep ta thay Mdt Id va Nhu vay dung Maple ta cd the tinh toan cac sd lieu thdng ke mdt each chinh xac III GIAI BAT PHl/ONG TRINH B A N G MAPLE Ciing nhu phuong trinh va he phuong trinh Maple giai bat phuang trinh mdt each hieu qua Ta cd the giai bait phirong trinh mot each true tiep b^ng lenh solve Vi du Giai bat phuong trinh x < > solve({x^2iniql:=(x+m+4/(x+m)solve(iniql,{x}); ' {X bpt2:=(4*(x+l/2)>sqrt(5*x^2+61*x)); bpt2:=^5x^ +6\x < x + > solve(bpt2,{X}) ; {0