1.觀察(1)如圖,在陽光下觀察直立于地面的旗桿AB及它在地面影子BC,旗桿所在直線與影子所在直線的位置關系是什么?(2)隨著時間的變化,影子BC的位置在不斷的變化,旗桿所在直線AB與其影子B’C’所在直線是否保持垂直?經(jīng)觀察我們知道AB與BC永遠垂直,也就是AB垂直于地面上所有過點B的直線。而不過點B的直線在地面內總是能找到過點B的直線與之平行。因此AB與地面上所有直線均垂直。一般地,如果一條直線與一個平面α內所有直線均垂直,我們就說l垂直α,記作l⊥α。2.定義:①文字敘述:如果直線l與平面α內的所有 直線都垂直,就說直線l與平面α互相垂直,記作l⊥α.直線l叫做平面α的垂線,平面α叫做直線l的垂面.直線與平面垂直時,它們唯一的公共點P叫做交點.②圖形語言:如圖.畫直線l與平面α垂直時,通常把直線畫成與表示平面的平行四邊形的一邊垂直.③符號語言:任意a?α,都有l(wèi)⊥a?l⊥α.
1.觀察(1)如圖,在陽光下觀察直立于地面的旗桿AB及它在地面影子BC,旗桿所在直線與影子所在直線的位置關系是什么?(2)隨著時間的變化,影子BC的位置在不斷的變化,旗桿所在直線AB與其影子B’C’所在直線是否保持垂直?經(jīng)觀察我們知道AB與BC永遠垂直,也就是AB垂直于地面上所有過點B的直線。而不過點B的直線在地面內總是能找到過點B的直線與之平行。因此AB與地面上所有直線均垂直。一般地,如果一條直線與一個平面α內所有直線均垂直,我們就說l垂直α,記作l⊥α。2.定義:①文字敘述:如果直線l與平面α內的所有 直線都垂直,就說直線l與平面α互相垂直,記作l⊥α.直線l叫做平面α的垂線,平面α叫做直線l的垂面.直線與平面垂直時,它們唯一的公共點P叫做交點.②圖形語言:如圖.畫直線l與平面α垂直時,通常把直線畫成與表示平面的平行四邊形的一邊垂直.
6.例二:如圖在正方體ABCD-A’B’C’D’中,O’為底面A’B’C’D’的中心,求證:AO’⊥BD 證明:如圖,連接B’D’,∵ABCD-A’B’C’D’是正方體∴BB’//DD’,BB’=DD’∴四邊形BB’DD’是平行四邊形∴B’D’//BD∴直線AO’與B’D’所成角即為直線AO’與BD所成角連接AB’,AD’易證AB’=AD’又O’為底面A’B’C’D’的中心∴O’為B’D’的中點∴AO’⊥B’D’,AO’⊥BD7.例三如圖所示,四面體A-BCD中,E,F(xiàn)分別是AB,CD的中點.若BD,AC所成的角為60°,且BD=AC=2.求EF的長度.解:取BC中點O,連接OE,OF,如圖?!逧,F分別是AB,CD的中點,∴OE//AC且OE=1/2AC,OF//AC且OF=1/2BD,∴OE與OF所成的銳角就是AC與BD所成的角∵BD,AC所成角為60°,∴∠EOF=60°或120°∵BD=AC=2,∴OE=OF=1當∠EOF=60°時,EF=OE=OF=1,當∠EOF=120°時,取EF的中點M,連接OM,則OM⊥EF,且∠EOM=60°∴EM= ,∴EF=2EM=
2.某小組有20名射手,其中1,2,3,4級射手分別為2,6,9,3名.又若選1,2,3,4級射手參加比賽,則在比賽中射中目標的概率分別為0.85,0.64,0.45,0.32,今隨機選一人參加比賽,則該小組比賽中射中目標的概率為________. 【解析】設B表示“該小組比賽中射中目標”,Ai(i=1,2,3,4)表示“選i級射手參加比賽”,則P(B)= P(Ai)P(B|Ai)= 2/20×0.85+ 6/20 ×0.64+ 9/20×0.45+ 3/20×0.32=0.527 5.答案:0.527 53.兩批相同的產品各有12件和10件,每批產品中各有1件廢品,現(xiàn)在先從第1批產品中任取1件放入第2批中,然后從第2批中任取1件,則取到廢品的概率為________. 【解析】設A表示“取到廢品”,B表示“從第1批中取到廢品”,有P(B)= 112,P(A|B)= 2/11 ,P(A| )= 1/11所以P(A)=P(B)P(A|B)+P( )P(A| )4.有一批同一型號的產品,已知其中由一廠生產的占 30%, 二廠生產的占 50% , 三廠生產的占 20%, 又知這三個廠的產品次品率分別為2% , 1%, 1%,問從這批產品中任取一件是次品的概率是多少?
(2)方法一:第一次取到一件不合格品,還剩下99件產品,其中有4件不合格品,95件合格品,于是第二次又取到不合格品的概率為4/99,由于這是一個條件概率,所以P(B|A)=4/99.方法二:根據(jù)條件概率的定義,先求出事件A,B同時發(fā)生的概率P(AB)=(C_5^2)/(C_100^2 )=1/495,所以P(B|A)=(P"(" AB")" )/(P"(" A")" )=(1/495)/(5/100)=4/99.6.在某次考試中,要從20道題中隨機地抽出6道題,若考生至少答對其中的4道題即可通過;若至少答對其中5道題就獲得優(yōu)秀.已知某考生能答對其中10道題,并且知道他在這次考試中已經(jīng)通過,求他獲得優(yōu)秀成績的概率.解:設事件A為“該考生6道題全答對”,事件B為“該考生答對了其中5道題而另一道答錯”,事件C為“該考生答對了其中4道題而另2道題答錯”,事件D為“該考生在這次考試中通過”,事件E為“該考生在這次考試中獲得優(yōu)秀”,則A,B,C兩兩互斥,且D=A∪B∪C,E=A∪B,由古典概型的概率公式及加法公式可知P(D)=P(A∪B∪C)=P(A)+P(B)+P(C)=(C_10^6)/(C_20^6 )+(C_10^5 C_10^1)/(C_20^6 )+(C_10^4 C_10^2)/(C_20^6 )=(12" " 180)/(C_20^6 ),P(E|D)=P(A∪B|D)=P(A|D)+P(B|D)=(P"(" A")" )/(P"(" D")" )+(P"(" B")" )/(P"(" D")" )=(210/(C_20^6 ))/((12" " 180)/(C_20^6 ))+((2" " 520)/(C_20^6 ))/((12" " 180)/(C_20^6 ))=13/58,即所求概率為13/58.
3.某縣農民月均收入服從N(500,202)的正態(tài)分布,則此縣農民月均收入在500元到520元間人數(shù)的百分比約為 . 解析:因為月收入服從正態(tài)分布N(500,202),所以μ=500,σ=20,μ-σ=480,μ+σ=520.所以月均收入在[480,520]范圍內的概率為0.683.由圖像的對稱性可知,此縣農民月均收入在500到520元間人數(shù)的百分比約為34.15%.答案:34.15%4.某種零件的尺寸ξ(單位:cm)服從正態(tài)分布N(3,12),則不屬于區(qū)間[1,5]這個尺寸范圍的零件數(shù)約占總數(shù)的 . 解析:零件尺寸屬于區(qū)間[μ-2σ,μ+2σ],即零件尺寸在[1,5]內取值的概率約為95.4%,故零件尺寸不屬于區(qū)間[1,5]內的概率為1-95.4%=4.6%.答案:4.6%5. 設在一次數(shù)學考試中,某班學生的分數(shù)X~N(110,202),且知試卷滿分150分,這個班的學生共54人,求這個班在這次數(shù)學考試中及格(即90分及90分以上)的人數(shù)和130分以上的人數(shù).解:μ=110,σ=20,P(X≥90)=P(X-110≥-20)=P(X-μ≥-σ),∵P(X-μσ)≈2P(X-μ130)=P(X-110>20)=P(X-μ>σ),∴P(X-μσ)≈0.683+2P(X-μ>σ)=1,∴P(X-μ>σ)=0.158 5,即P(X>130)=0.158 5.∴54×0.158 5≈9(人),即130分以上的人數(shù)約為9人.
4.有8種不同的菜種,任選4種種在不同土質的4塊地里,有 種不同的種法. 解析:將4塊不同土質的地看作4個不同的位置,從8種不同的菜種中任選4種種在4塊不同土質的地里,則本題即為從8個不同元素中任選4個元素的排列問題,所以不同的種法共有A_8^4 =8×7×6×5=1 680(種).答案:1 6805.用1、2、3、4、5、6、7這7個數(shù)字組成沒有重復數(shù)字的四位數(shù).(1)這些四位數(shù)中偶數(shù)有多少個?能被5整除的有多少個?(2)這些四位數(shù)中大于6 500的有多少個?解:(1)偶數(shù)的個位數(shù)只能是2、4、6,有A_3^1種排法,其他位上有A_6^3種排法,由分步乘法計數(shù)原理,知共有四位偶數(shù)A_3^1·A_6^3=360(個);能被5整除的數(shù)個位必須是5,故有A_6^3=120(個).(2)最高位上是7時大于6 500,有A_6^3種,最高位上是6時,百位上只能是7或5,故有2×A_5^2種.由分類加法計數(shù)原理知,這些四位數(shù)中大于6 500的共有A_6^3+2×A_5^2=160(個).
探究新知問題1:已知100件產品中有8件次品,現(xiàn)從中采用有放回方式隨機抽取4件.設抽取的4件產品中次品數(shù)為X,求隨機變量X的分布列.(1):采用有放回抽樣,隨機變量X服從二項分布嗎?采用有放回抽樣,則每次抽到次品的概率為0.08,且各次抽樣的結果相互獨立,此時X服從二項分布,即X~B(4,0.08).(2):如果采用不放回抽樣,抽取的4件產品中次品數(shù)X服從二項分布嗎?若不服從,那么X的分布列是什么?不服從,根據(jù)古典概型求X的分布列.解:從100件產品中任取4件有 C_100^4 種不同的取法,從100件產品中任取4件,次品數(shù)X可能取0,1,2,3,4.恰有k件次品的取法有C_8^k C_92^(4-k)種.一般地,假設一批產品共有N件,其中有M件次品.從N件產品中隨機抽取n件(不放回),用X表示抽取的n件產品中的次品數(shù),則X的分布列為P(X=k)=CkM Cn-kN-M CnN ,k=m,m+1,m+2,…,r.其中n,N,M∈N*,M≤N,n≤N,m=max{0,n-N+M},r=min{n,M},則稱隨機變量X服從超幾何分布.
二項式定理形式上的特點(1)二項展開式有n+1項,而不是n項.(2)二項式系數(shù)都是C_n^k(k=0,1,2,…,n),它與二項展開式中某一項的系數(shù)不一定相等.(3)二項展開式中的二項式系數(shù)的和等于2n,即C_n^0+C_n^1+C_n^2+…+C_n^n=2n.(4)在排列方式上,按照字母a的降冪排列,從第一項起,次數(shù)由n次逐項減少1次直到0次,同時字母b按升冪排列,次數(shù)由0次逐項增加1次直到n次.1.判斷(正確的打“√”,錯誤的打“×”)(1)(a+b)n展開式中共有n項. ( )(2)在公式中,交換a,b的順序對各項沒有影響. ( )(3)Cknan-kbk是(a+b)n展開式中的第k項. ( )(4)(a-b)n與(a+b)n的二項式展開式的二項式系數(shù)相同. ( )[解析] (1)× 因為(a+b)n展開式中共有n+1項.(2)× 因為二項式的第k+1項Cknan-kbk和(b+a)n的展開式的第k+1項Cknbn-kak是不同的,其中的a,b是不能隨便交換的.(3)× 因為Cknan-kbk是(a+b)n展開式中的第k+1項.(4)√ 因為(a-b)n與(a+b)n的二項式展開式的二項式系數(shù)都是Crn.[答案] (1)× (2)× (3)× (4)√
解析:因為減法和除法運算中交換兩個數(shù)的位置對計算結果有影響,所以屬于組合的有2個.答案:B2.若A_n^2=3C_(n"-" 1)^2,則n的值為( )A.4 B.5 C.6 D.7 解析:因為A_n^2=3C_(n"-" 1)^2,所以n(n-1)=(3"(" n"-" 1")(" n"-" 2")" )/2,解得n=6.故選C.答案:C 3.若集合A={a1,a2,a3,a4,a5},則集合A的子集中含有4個元素的子集共有 個. 解析:滿足要求的子集中含有4個元素,由集合中元素的無序性,知其子集個數(shù)為C_5^4=5.答案:54.平面內有12個點,其中有4個點共線,此外再無任何3點共線,以這些點為頂點,可得多少個不同的三角形?解:(方法一)我們把從共線的4個點中取點的多少作為分類的標準:第1類,共線的4個點中有2個點作為三角形的頂點,共有C_4^2·C_8^1=48(個)不同的三角形;第2類,共線的4個點中有1個點作為三角形的頂點,共有C_4^1·C_8^2=112(個)不同的三角形;第3類,共線的4個點中沒有點作為三角形的頂點,共有C_8^3=56(個)不同的三角形.由分類加法計數(shù)原理,不同的三角形共有48+112+56=216(個).(方法二 間接法)C_12^3-C_4^3=220-4=216(個).
4.They were going to find someone to take part in their bet when they saw Henry walking on the street outside.[歸納]1.過去將來時的基本構成和用法過去將來時由“would+動詞原形”構成,主要表示從過去某一時間來看將要發(fā)生的動作(尤其用于賓語從句中),還可以表示過去的動作習慣或傾向。Jeff knew he would be tired the next day.He promised that he would not open the letter until 2 o'clock.She said that she wouldn't do that again.2.表示過去將來時的其他表達法(1)was/were going to+動詞原形:該結構有兩個主要用法,一是表示過去的打算,二是表示在過去看來有跡象表明將要發(fā)生某事。I thought it was going to rain.(2)was/were to+動詞原形:主要表示過去按計劃或安排要做的事情。She said she was to get married next month.(3)was/were about to+動詞原形:表示在過去看來即將要發(fā)生的動作,由于本身已含有“即將”的意味,所以不再與表示具體的將來時間狀語連用。I was about to go to bed when the phone rang.(4)was/were+現(xiàn)在分詞:表示在過去看來即將發(fā)生的動作,通常可用于該結構中的動詞是come,go,leave,arrive,begin,start,stop,close,open,die,join,borrow,buy等瞬間動詞。Jack said he was leaving tomorrow.
常跟雙賓語的動詞有:(需借助to的)bring, ask, hand, offer, give, lend, send, show, teach, tell, write, pass, pay, promise, return等;基本句型 五S +V + O + OC(主+謂+賓+賓補)特點:動詞雖然是及物動詞,但是只跟一個賓語還不能表達完整的意思,必須加上一個補充成分來補足賓語,才能使意思完整。 判斷原則:能表達成—賓語 是…/做…注:此結構由“主語+及物的謂語動詞+賓語+賓語補足語”構成。賓語與賓語補足語之間有邏輯上的主謂關系或主表關系,若無賓語補足語,則句意不夠完整。可以用做賓補的有:名詞,形容詞,副詞,介詞短語,動詞不定式,分詞等。如:He considers himself an expert on the subject.他認為自己是這門學科的專家。We must keep our classroom clean.我們必須保持教室清潔。I had my bike stolen.我的自行車被偷了。We invited him to come to our school.我們邀請他來我們學校。I beg you to keep secret what we talked here.我求你對這里所談的話保密。用it做形式賓語,而將真正的賓語放到賓語補足語的后面,以使句子結構平衡,是英語常用的句型結構方式。即:主語+謂語+it+賓補+真正賓語。如:We think it a good idea to go climb the mountain this Sunday.
【教材分析】This teaching period mainly deals with the grammar: the restrictive relative clauses.This period carries considerable significance to the cultivation of students’ writing competence and lays a solid foundation for the basic appreciation of language beauty. The teacher is expected to enable students to master this period thoroughly and consolidate the knowledge by doing some exercise of good quality.【教學目標與核心素養(yǎng)】1. Get students to have a good understanding of the basic usages of the restrictive relative clauses.2. Enable students to use the restrictive relative clauses flexibly.3. Develop students’ speaking and cooperating abilities.4. Strengthen students’ great interest in grammar learning.【教學重難點】How to enable students to have a good understanding of the restrictive relative clauses, especially the uses of the relative words such as which, that, who, whom.【教學過程】Step1: 語法知識呈現(xiàn)定語從句(一)—關系代詞的用法在復合句中, 修飾名詞或代詞的從句叫定語從句。定語從句通常由關系代詞或關系副詞引導,說明事物的具體信息,從句位于被修飾詞之后。被定語從句修飾的詞叫先行詞,引導定語從句的詞叫關系詞,關系詞指代先行詞,并在定語從句中充當成分。關系詞有兩種:關系代詞who, whom ,whose, that, which, as和關系副詞when, where, why。
II Learn the technical terms-2.1. What can be used as “Subject, Object, Predicative, Direct Object, Indirect Object and objective complement” in a sentence?2. What can be used as “adverbial” in a sentence?3. What can be used as “verb” in a sentence?Answers to questions 1-3:1. Nouns, pronouns and appellations can be used as “Subject, Object, Predicative, Direct Object, Indirect Object and Objective Complement”. Besides, adjectives can be used as “Predicative and Objective Complement” in a sentence.2. Adverbs and prepositional phrases can be used as “Adverbial”.3. Verbs with actual meaning can be used as “Verb” in a sentence. Auxiliary verbs alone cannot be used as “Verb” in a sentence.III Learn to recognize the sentence structures.1. SV structure. For Example:(1) A bird flies.S V(2) A monkey jumps.S V(3) A fish swims.S V√ In SV structures, verbs are “intransitive verbs”.2. SVO structure. For Example:(1) A sheep eats grass.S V O(2) They like bananas.S V O(3) He wants candy.S V O√ In SVO structures, verbs are “transitive verbs”.3. SP structure. For Example:(1) This is great.S P (2) He looks well.S P (3) She became a teacher.S P √ In SP structures, Predicatives are formed by “l(fā)ink verbs” and “adjectives or nouns”.√ link verbs: be, become, grow, look, feel, taste, etc.4. SV IO DO structure. For Example:(1) He asked me a question.S V IO DO(2) Danny wrote me a letter.S V IO DO(3) Billy brought Sam a kite.S V IO DO√ In SV IO DO structures, the verbs are transitive and are followed by two objectives – pronouns or nouns as Indirect Objective, and nouns as Direct Objectives.
4. When he got absorbed in his world of music, he felt as if he could “see” the beauty of the world around him, like he had in his previous life.P·P as adverbial: _________________________________________________________________.Function: _______________________________________________________________________.Step 5 Solid Complete the passage with the words in brackets in their correct forms.Well known as a successful band, the Impact members show quite a few striking qualities. They never ever give up. When _____________(question) by the media, they are not _____________(discourage) and practise even harder. They are improving themselves by attending several master training class. They are united. _____________(fill with) team spirit, they act as a whole, always aiming for glory. Step 6 Difference and similarity from -ingObserve the following examples.1. He went out, shutting the door behind him.=He went out, ________________________________________________________.2. Not knowing what to do, he went to his parents for help.=__________________________________________, he went to his parents for help.Similarity: _______________________________________________________________________________________________________________________________________________________.Difference : _______________________________________________________________________________________________________________________________________________________.Step Practice1. ________ in a hurry, this article was not so good. 因為寫得匆忙, 這篇文章不是很好。2. ________ carefully, he found something he hadn’t known before. 他仔細讀書時, 發(fā)現(xiàn)了一些從前不知道的東西。3. ________ why he did it, the monitor said it was his duty. 當被問及他為什么要這么做時, 班長說這是他的職責
Step1:自主探究。1.(教材P52)Born(bear) in the USA on 2 January 1970, Whitacre began studying music at the University of Nevada in 1988.2.(教材P52) Moved(move) by this music, he said, “It was like seeing color for the first time.”3.(教材P56)I was very afraid and I felt so alone and discouraged(discourage).4.(教材P58)Encouraged(encourage) by this first performance and the positive reaction of the audience, I have continued to play the piano and enjoy it more every day.Step2:語法要點精析。用法1:過去分詞作表語1).過去分詞可放在連系動詞be, get, feel, remain, seem, look, become等之后作表語,表示主語所處的狀態(tài)Tom was astonished to see a snake moving across the floor.湯姆很驚訝地看到一條蛇正爬過地板。Finally the baby felt tired of playing with those toys.終于嬰兒厭倦了玩那些玩具。注意:1).過去分詞作表語時與被動語態(tài)的區(qū)別過去分詞作表語時,強調主語所處的狀態(tài);而動詞的被動語態(tài)表示主語是動作的承受者,強調動作。The library is now closed.(狀態(tài))圖書館現(xiàn)在關閉了。The cup was broken by my little sister yesterday.(動作)昨天我妹妹把杯子打碎了。2)感覺類及物動詞的現(xiàn)在分詞與過去分詞作表語的區(qū)別過去分詞作表語多表示人自身的感受或事物自身的狀態(tài),常譯作“感到……的”;現(xiàn)在分詞多表示事物具有的特性,常譯作“令人……的”。
【教材分析】This teaching period mainly deals with the grammar: tag questions.This period carries a considerable significance to the cultivation of students’ spoken English. The teacher is expected to enable students to master this period thoroughly and consolidate the knowledge by doing some exercise of good quality.【教學目標與核心素養(yǎng)】1. Get students to have a good understanding of the basic usages of tag questions.2. Enable students to use the basic phrases structures flexibly.3. Develop students’ speaking and cooperating abilities.4. Strengthen students’ great interest in grammar learning.【教學重難點】1. How to enable students to have a good understanding of the basic usages of tag questions.2. How to enable students to use the basic usages of tag questions flexibly.【教學過程】Step1: 語法自主探究一、基本組成方法1.肯定式陳述部分+否定附加疑問部分(前肯后否) You often play badminton, don’t you? 你經(jīng)常打羽毛球,是嗎?You are going to the gym with me, aren’t you?你要和我一起去健身房,是嗎?She’s been to shanghai before, hasn’t she? 她以前去過上海,是嗎?2.否定式陳述部分+肯定附加疑問部分(前否后肯) It isn't a beautiful flower, is it? 那不是美麗的花,是嗎?You didn't go skating yesterday, did you? 你昨天沒去滑冰,是嗎?They can’t finish it by Friday, can they?他們不能在星期五之前完成,是嗎?
【教學目標】知識目標:理解直線的點斜式方程、斜截式方程、橫截距、縱截距的概念;掌握直線的點斜式方程、斜截式方程的確定.能力目標:通過求解直線的點斜式方程和斜截式方程,培養(yǎng)學生的數(shù)學思維能力與數(shù)形結合的數(shù)學思想.情感目標:通過學習直線的點斜式方程和斜截式方程,體會數(shù)形結合的直觀感受.【教學重點】直線的點斜式方程、斜截式方程的確定.【教學難點】直線的點斜式方程、斜截式方程的確定.
【教學重點】直線的點斜式方程、斜截式方程的確定.【教學難點】直線的點斜式方程、斜截式方程的確定.【教學過程】1、對特殊三角函數(shù)進行鞏固復習;表1 內特殊三角函數(shù)值 不存在圖1 特殊三角形2、鞏固復習直線的傾斜角和斜率相關內容;直線的傾斜角:,;直線的斜率: , ;設點為直線l上的任意兩點,當時,
由樣本相關系數(shù)??≈0.97,可以推斷脂肪含量和年齡這兩個變量正線性相關,且相關程度很強。脂肪含量與年齡變化趨勢相同.歸納總結1.線性相關系數(shù)是從數(shù)值上來判斷變量間的線性相關程度,是定量的方法.與散點圖相比較,線性相關系數(shù)要精細得多,需要注意的是線性相關系數(shù)r的絕對值小,只是說明線性相關程度低,但不一定不相關,可能是非線性相關.2.利用相關系數(shù)r來檢驗線性相關顯著性水平時,通常與0.75作比較,若|r|>0.75,則線性相關較為顯著,否則不顯著.例2. 有人收集了某城市居民年收入(所有居民在一年內收入的總和)與A商品銷售額的10年數(shù)據(jù),如表所示.畫出散點圖,判斷成對樣本數(shù)據(jù)是否線性相關,并通過樣本相關系數(shù)推斷居民年收入與A商品銷售額的相關程度和變化趨勢的異同.
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