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旅游合同
根據(jù)國家有關(guān)旅游事業(yè)管理的規(guī)定，甲乙雙方經(jīng)協(xié)商一致，簽訂本合同，共同信守執(zhí)行。第一條 旅游的時間安排：由甲方在 年 月 日至 年 月 日為乙方提供旅游服務(wù)。第二條 旅游的地點及每個旅游景點的時間安排：甲方為乙方提供的旅游景點為 個。分別是 、 、 。每天的時間安排為上午 時至 時，下午 時至 時。第三條 旅游的生活安排：甲方為乙方提供食宿，每天伙食在 元至 元標(biāo)準(zhǔn)內(nèi)。第四條 導(dǎo)游服務(wù)：甲方為乙方提供導(dǎo)游服務(wù)，服務(wù)內(nèi)容： .第五條 旅游的費用：本次旅游的費用總計 元（包括食宿在內(nèi)）。在旅游出發(fā)前 日交清。第六條 旅游的交通工具：甲方為乙方提供交通工具。 、 、 等交通工具都由甲方聯(lián)系、提供，并保證乙方的旅游安全。第七條 甲方的權(quán)利義務(wù)及違約金：1.甲方應(yīng)按本合同的規(guī)定按時為乙方安排本次旅游。
新人教版高中英語選修2Unit 3 Reading for writing教學(xué)設(shè)計
The theme of this part is to write an article about healthy diet. Through reading and writing activities, students can accumulate knowledge about healthy diet, deepen their understanding of the theme of healthy diet, and reflect on their own eating habits. This text describes the basic principles of healthy diet. The author uses data analysis, definition, comparison, examples and other methods. It also provides a demonstration of the use of conjunctions, which provides important information reference for students to complete the next collaborative task, writing skills, vivid language materials and expressions.1. Teach Ss to learn and skillfully use the new words learned from the text.2. Develop students’ ability to understand, extract and summarize information.3. Guide students to understand the theme of healthy diet and reflect on their own eating habits.4. To guide students to analyze and understand the reading discourse from the aspects of theme content, writing structure, language expression, etc., 5. Enable Ss to write in combination with relevant topics and opinions, and to talk about their eating habits.1. Guide students to analyze and understand the reading discourse from the aspects of theme content, writing structure, language expression, etc.2. Enable them to write in combination with relevant topics and opinions, and to talk about their eating habits.3. Guide the students to use the cohesive words correctly, strengthen the textual cohesion, and make the expression fluent and the thinking clear.Step1: Warming upbrainstorm some healthy eating habits.1.Eat slowly.2.Don’t eat too much fat or sugar.3.Eat healthy food.4.Have a balanced diet.Step2: Read the passage and then sum up the main idea of each paragraph.
人教版高中數(shù)學(xué)選修3組合與組合數(shù)教學(xué)設(shè)計
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(jié)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(nèi)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(biāo)準(zhǔn):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).
新人教版高中英語選修2Unit 3 Using langauge-Listening教學(xué)設(shè)計
1． How is Hunan cuisine somewhat different from Sichuan cuisine?The heat in Sichuan cuisine comes from chilies and Sichuan peppercorns. Human cuisine is often hotter and the heat comes from just chilies.2.What are the reasons why Hunan people like spicy food?Because they are a bold people. But many Chinese people think that hot food helps them overcome the effects of rainy or wet weather.3.Why do so many people love steamed fish head covered with chilies?People love it because the meat is quite tender and there are very few small bones.4.Why does Tingting recommend bridge tofu instead of dry pot duck with golden buns?Because bridge tofu has a lighter taste.5 .Why is red braised pork the most famous dish?Because Chairman Mao was from Hunan, and this was his favorite food.Step 5: Instruct students to make a short presentation to the class about your choice. Use the example and useful phrases below to help them.? In groups of three, discuss what types of restaurant you would like to take a foreign visitor to, and why. Then take turns role-playing taking your foreign guest to the restaurant you have chosen. One of you should act as the foreign guest, one as the Chinese host, and one as the waiter or waitress. You may start like this:? EXAMPLE? A: I really love spicy food, so what dish would you recommend?? B: I suggest Mapo tofu.? A: Really ? what's that?
人教版高中數(shù)學(xué)選修3超幾何分布教學(xué)設(shè)計
探究新知問題1：已知100件產(chǎn)品中有8件次品，現(xiàn)從中采用有放回方式隨機(jī)抽取4件．設(shè)抽取的4件產(chǎn)品中次品數(shù)為X，求隨機(jī)變量X的分布列.（1）：采用有放回抽樣，隨機(jī)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(jié)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(chǎn)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(chǎn)品中任取4件有 C_100^4 種不同的取法，從100件產(chǎn)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(shè)一批產(chǎn)品共有N件，其中有M件次品．從N件產(chǎn)品中隨機(jī)抽取n件(不放回），用X表示抽取的n件產(chǎn)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(jī)變量X服從超幾何分布．
人教版高中數(shù)學(xué)選修3全概率公式教學(xué)設(shè)計
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(biāo)的概率分別為0.85，0.64，0.45，0.32，今隨機(jī)選一人參加比賽，則該小組比賽中射中目標(biāo)的概率為________. 【解析】設(shè)B表示“該小組比賽中射中目標(biāo)”，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(chǎn)品各有12件和10件，每批產(chǎn)品中各有1件廢品，現(xiàn)在先從第1批產(chǎn)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(shè)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(chǎn)品，已知其中由一廠生產(chǎn)的占 30%, 二廠生產(chǎn)的占 50% , 三廠生產(chǎn)的占 20%, 又知這三個廠的產(chǎn)品次品率分別為2% , 1%, 1%，問從這批產(chǎn)品中任取一件是次品的概率是多少？
人教版高中數(shù)學(xué)選修3條件概率教學(xué)設(shè)計
(2)方法一:第一次取到一件不合格品,還剩下99件產(chǎn)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(jī)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(shè)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.
人教版高中數(shù)學(xué)選修3正態(tài)分布教學(xué)設(shè)計
3.某縣農(nóng)民月均收入服從N(500,202)的正態(tài)分布,則此縣農(nóng)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(nèi)的概率為0.683.由圖像的對稱性可知,此縣農(nóng)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(nèi)取值的概率約為95.4%,故零件尺寸不屬于區(qū)間[1,5]內(nèi)的概率為1-95.4%=4.6%.答案:4.6%5. 設(shè)在一次數(shù)學(xué)考試中,某班學(xué)生的分?jǐn)?shù)X~N(110,202),且知試卷滿分150分,這個班的學(xué)生共54人,求這個班在這次數(shù)學(xué)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.
人教版高中數(shù)學(xué)選修3排列與排列數(shù)教學(xué)設(shè)計
4.有8種不同的菜種,任選4種種在不同土質(zhì)的4塊地里,有 種不同的種法. 解析:將4塊不同土質(zhì)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(zhì)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(fù)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).
新人教版高中英語選修2Unit 3 Learning about Language教學(xué)設(shè)計
1. We'll need ten months at least to have the restaurant decorated.2.Some traditional Chinese dishes from before the Ming Dynasty are still popular today.3.My grandpa's breakfast mainly includes whole grain biscuits and a glass of milk.4.People in this area would eat nearly a kilo of cheese per week.5. We enjoyed a special dinner in a fancy restaurant where the waiters all wore attractive suits.6. He prefers this brand of coffee which, as he said, has an unusually good flavor.Key:1. at a minimum 2. prior to3. consist of4. consume5. elegant6. exceptionalStep 5：Familiarize yourself with some food idioms by matching the meaning on the right with the colored words on the left.1.Public concern for the health of farm animals has mushroomed in the UK2.Anderson may be young but he's certainly rolling to doing dough!3.George is a popular lecturer. He often peppers his speech with jokes.4.As the person to bring home the bacon, he needs to find a stable job.5 He is often regarded as a ham actor for his over emphasized facial expressions. The media reported that these companies had treated pollution as a hot potato. 6.The media reported that these companies had treated pollution as a hot potato.7.Don't worry about the test tomorrow. It's going to be a piece of cake!8. It's best to fold the swimming ring when it is as flat as a pancake.A. completely flatB. something that is very easy to do C.an issue that is hard to deal withD.to include large numbers of somethingE.to earn on e's living to support a familyF. wealthyG.to rapidly increase in numberH. an actor who performs badly, especially by over emphasizing emotions
小學(xué)數(shù)學(xué)人教版四年級下冊《第三課三角形的分類》教案說課稿
1. 知識與技能 通過學(xué)生活動，幫助學(xué)生理解三角形按角分類的方法，掌握直角三角形、銳角三角形、鈍角三角形的概念；知道等腰三角形、等邊三角形。培養(yǎng)學(xué)生觀察，動手操作和抽象概括的能力；發(fā)展空間觀念。2.過程與方法 使學(xué)生經(jīng)歷觀察、操作、比較、概括等過程，在分類中體會每一類三角形角的特點；發(fā)現(xiàn)邊的特點。滲透集合思想。3.情感態(tài)度與價值觀 激發(fā)學(xué)生的主動參與意識，使學(xué)生感受到成功的喜悅，更增強(qiáng)學(xué)習(xí)興趣?！窘虒W(xué)重點】 直角三角形、銳角三角形、鈍角三角形的概念?！窘虒W(xué)難點】發(fā)現(xiàn)三角形角的特點?！窘虒W(xué)方法】啟發(fā)式教學(xué)、自主探索、合作交流、討論法、講解法。【課前準(zhǔn)備】多媒體【課時安排】 1課時【教學(xué)過程】（一）復(fù)習(xí)導(dǎo)入 師：說一說下面的角各是什么角。
小學(xué)數(shù)學(xué)人教版六年級下冊《第三課圓柱的體積》教案說課稿
（一）復(fù)習(xí)導(dǎo)入 師：什么是體積？生：物體所占空間的大小是物體的體積。師：怎樣求長方體和正方體的體積？生：長方體的體積=底面積×高 正方體的體積=底面積×高師：圓的面積計算公式是怎樣推導(dǎo)出來的？課件出示：生：把圓轉(zhuǎn)化成長方形，長方形的長等于圓柱底面周長的一半，寬等于半徑，所以圓的面積：S = πr2猜測：把圓柱轉(zhuǎn)化成什么立體圖形來推導(dǎo)圓柱的體積公式呢？呢？今天我們一起來探討這個問題。板書課題：圓柱的體積。
人教版高中數(shù)學(xué)選修3成對數(shù)據(jù)的相關(guān)關(guān)系教學(xué)設(shè)計
由樣本相關(guān)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(guān)，且相關(guān)程度很強(qiáng)。脂肪含量與年齡變化趨勢相同.歸納總結(jié)1．線性相關(guān)系數(shù)是從數(shù)值上來判斷變量間的線性相關(guān)程度，是定量的方法．與散點圖相比較，線性相關(guān)系數(shù)要精細(xì)得多，需要注意的是線性相關(guān)系數(shù)r的絕對值小，只是說明線性相關(guān)程度低，但不一定不相關(guān)，可能是非線性相關(guān)．2．利用相關(guān)系數(shù)r來檢驗線性相關(guān)顯著性水平時，通常與0.75作比較，若|r|>0.75，則線性相關(guān)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(nèi)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(guān)，并通過樣本相關(guān)系數(shù)推斷居民年收入與A商品銷售額的相關(guān)程度和變化趨勢的異同.
人教版高中數(shù)學(xué)選修3離散型隨機(jī)變量的方差教學(xué)設(shè)計
3.下結(jié)論.依據(jù)均值和方差做出結(jié)論.跟蹤訓(xùn)練2. A、B兩個投資項目的利潤率分別為隨機(jī)變量X1和X2，根據(jù)市場分析， X1和X2的分布列分別為X1 2% 8% 12% X2 5% 10%P 0.2 0.5 0.3 P 0.8 0.2求：(1)在A、B兩個項目上各投資100萬元， Y1和Y2分別表示投資項目A和B所獲得的利潤，求方差D(Y1)和D(Y2)；(2)根據(jù)得到的結(jié)論，對于投資者有什么建議？ 解：(1)題目可知，投資項目A和B所獲得的利潤Y1和Y2的分布列為：Y1 2 8 12 Y2 5 10P 0.2 0.5 0.3 P 0.8 0.2所以 ;; 解：(2) 由(1)可知 ，說明投資A項目比投資B項目期望收益要高；同時 ，說明投資A項目比投資B項目的實際收益相對于期望收益的平均波動要更大.因此，對于追求穩(wěn)定的投資者，投資B項目更合適;而對于更看重利潤并且愿意為了高利潤承擔(dān)風(fēng)險的投資者，投資A項目更合適.
人教版高中數(shù)學(xué)選修3離散型隨機(jī)變量的均值教學(xué)設(shè)計
對于離散型隨機(jī)變量，可以由它的概率分布列確定與該隨機(jī)變量相關(guān)事件的概率。但在實際問題中，有時我們更感興趣的是隨機(jī)變量的某些數(shù)字特征。例如，要了解某班同學(xué)在一次數(shù)學(xué)測驗中的總體水平，很重要的是看平均分；要了解某班同學(xué)數(shù)學(xué)成績是否“兩極分化”則需要考察這個班數(shù)學(xué)成績的方差。我們還常常希望直接通過數(shù)字來反映隨機(jī)變量的某個方面的特征，最常用的有期望與方差.二、 探究新知探究1.甲乙兩名射箭運動員射中目標(biāo)靶的環(huán)數(shù)的分布列如下表所示：如何比較他們射箭水平的高低呢？環(huán)數(shù)X 7 8 9 10甲射中的概率 0.1 0.2 0.3 0.4乙射中的概率 0.15 0.25 0.4 0.2類似兩組數(shù)據(jù)的比較,首先比較擊中的平均環(huán)數(shù),如果平均環(huán)數(shù)相等，再看穩(wěn)定性.假設(shè)甲射箭n次，射中7環(huán)、8環(huán)、9環(huán)和10環(huán)的頻率分別為：甲n次射箭射中的平均環(huán)數(shù)當(dāng)n足夠大時，頻率穩(wěn)定于概率，所以x穩(wěn)定于7×0.1+8×0.2+9×0.3+10×0.4=9.即甲射中平均環(huán)數(shù)的穩(wěn)定值(理論平均值)為9,這個平均值的大小可以反映甲運動員的射箭水平.同理，乙射中環(huán)數(shù)的平均值為7×0.15+8×0.25+9×0.4+10×0.2=8.65.
人教版高中數(shù)學(xué)選修3二項式系數(shù)的性質(zhì)教學(xué)設(shè)計
1.對稱性與首末兩端“等距離”的兩個二項式系數(shù)相等,即C_n^m=C_n^(n"-" m).2.增減性與最大值 當(dāng)k(n+1)/2時,C_n^k隨k的增加而減小.當(dāng)n是偶數(shù)時,中間的一項C_n^(n/2)取得最大值;當(dāng)n是奇數(shù)時,中間的兩項C_n^((n"-" 1)/2) 與C_n^((n+1)/2)相等,且同時取得最大值.探究2.已知(1+x)^n =C_n^0+C_n^1 x+...〖+C〗_n^k x^k+...+C_n^n x^n 3.各二項式系數(shù)的和C_n^0+C_n^1+C_n^2+…+C_n^n=2n.令x=1 得(1+1)^n=C_n^0+C_n^1 +...+C_n^n=2^n所以，(a+b)^n 的展開式的各二項式系數(shù)之和為2^n1. 在(a+b)8的展開式中,二項式系數(shù)最大的項為 ,在(a+b)9的展開式中,二項式系數(shù)最大的項為 . 解析:因為(a+b)8的展開式中有9項,所以中間一項的二項式系數(shù)最大,該項為C_8^4a4b4=70a4b4.因為(a+b)9的展開式中有10項,所以中間兩項的二項式系數(shù)最大,這兩項分別為C_9^4a5b4=126a5b4,C_9^5a4b5=126a4b5.答案:1.70a4b4 126a5b4與126a4b5 2. A=C_n^0+C_n^2+C_n^4+…與B=C_n^1+C_n^3+C_n^5+…的大小關(guān)系是( )A.A>B B.A=B C.A<B D.不確定 解析:∵(1+1)n=C_n^0+C_n^1+C_n^2+…+C_n^n=2n,(1-1)n=C_n^0-C_n^1+C_n^2-…+(-1)nC_n^n=0,∴C_n^0+C_n^2+C_n^4+…=C_n^1+C_n^3+C_n^5+…=2n-1,即A=B.答案:B
旅游策劃方案模板
酒店門口集中出發(fā)廣東奧林匹克體育中心9:30-11:20 嶺南印象園個色彩各異的看臺區(qū)就像一片花瓣,合起來一看,廣州市市花——木棉花赫然入目。
人教版高中數(shù)學(xué)選修3一元線性回歸模型及其應(yīng)用教學(xué)設(shè)計
1.確定研究對象，明確哪個是解釋變量，哪個是響應(yīng)變量；2.由經(jīng)驗確定非線性經(jīng)驗回歸方程的模型；3.通過變換，將非線性經(jīng)驗回歸模型轉(zhuǎn)化為線性經(jīng)驗回歸模型；4.按照公式計算經(jīng)驗回歸方程中的參數(shù)，得到經(jīng)驗回歸方程；5.消去新元，得到非線性經(jīng)驗回歸方程；6.得出結(jié)果后分析殘差圖是否有異常 ．跟蹤訓(xùn)練1.一只藥用昆蟲的產(chǎn)卵數(shù)y與一定范圍內(nèi)的溫度x有關(guān)，現(xiàn)收集了6組觀測數(shù)據(jù)列于表中： 經(jīng)計算得： 線性回歸殘差的平方和: ∑_(i=1)^6?〖(y_i-(y_i ) ?)〗^2=236,64,e^8.0605≈3167.其中 分別為觀測數(shù)據(jù)中的溫度和產(chǎn)卵數(shù)，i=1,2,3,4,5,6.(1)若用線性回歸模型擬合，求y關(guān)于x的回歸方程 (精確到0.1）；(2)若用非線性回歸模型擬合，求得y關(guān)于x回歸方程為 且相關(guān)指數(shù)R2＝0.9522． ①試與(1)中的線性回歸模型相比較，用R2說明哪種模型的擬合效果更好 ?②用擬合效果好的模型預(yù)測溫度為35℃時該種藥用昆蟲的產(chǎn)卵數(shù).(結(jié)果取整數(shù)).
新人教版高中英語選修2Unit 4 Reading for writing教學(xué)設(shè)計
假定你是英國的Jack，打算來中國旅行，請你給你的中國筆友李華寫一封信，要點如下：1．你的旅行計劃：北京→泰山→杭州；2．征求建議并詢問他是否愿意充當(dāng)你的導(dǎo)游。注意：1.詞數(shù)80左右(開頭和結(jié)尾已給出，不計入總詞數(shù))；2．可以適當(dāng)增加細(xì)節(jié)，以使行文連貫。參考詞匯：故宮 the Forbidden City；泰山 Mount TaiDear Li Hua，I'm glad to tell you that 'm going to visit China.First，I am planning to visit Beijing，the capitalof China，where I am looking forward to enjoying the Great Wall，the Forbidden City and somebeautiful parks.Then I intend to go to visit Mount Tai in Shandong Province.I've heard that it is one ofthe most famous mountains in China and I can't wait to enjoy the amazing sunrise there.After that，I amalso going to Hangzhou.It is said that it is a beautiful modern city with breathtaking natural sights，among which the West Lake is a well- known tourist attraction.What do you think of my travel plan? Will you act as my guide? Hope to hear from you soon.
新人教版高中英語選修2Unit 2 Reading and thinking教學(xué)設(shè)計
Her tutor told her to acknowledge __________ other people had said if she cited their ideas, and advised her _______（read） lots of information in order to form __________wise opinion of her own.Now halfway __________ her exchange year, Xie Lei felt much more at home in the UK. She said __________ (engage) in British culture had helped and that she had been__________ (involve) in social activities. She also said while learning about business, she was acting as a cultural messenger __________（build） a bridge between the two countries. keys:Xie Lei, a 19yearold Chinese student, said goodbye to her family and friends in China and boarded (board) a plane for London six months ago in order to get a business qualification. She was ambitious(ambition) to set up a business after graduation. It was the first time that she had left (leave) home.At first, Xie Lei had to adapt to life in a different country. She chose to live with a host family, who can help with her adaptation (adapt) to the new culture. When she missed home, she felt comforted (comfort) to have a second family. Also Xie Lei had to satisfy academic requirements. Her tutor told her to acknowledge what other people had said if she cited their ideas, and advised her to read lots of information in order to form a wise opinion of her own.Now halfway through her exchange year, Xie Lei felt much more at home in the UK. She said engaging (engage) in British culture had helped and that she had been involved (involve) in social activities. She also said while learning about business, she was acting as a cultural messenger building a bridge between the two countries.
