Exercise 05-solution
PRML第三章习题答案
PRML第三章习题答案Chapter 3. Linear Models for Regression⽬录更新⽇志(截⾄20210710)习题简述线性回归3.1:\text{tanh}可由\sigma线性变换得到3.2:最⼩⼆乘解为正交投影:加权最⼩⼆乘对应数据依赖的噪声或重复数据:带噪声的输⼊相当于权重正则,类别神经⽹络的数据增强:带权重正则的最⼩⼆乘等同于带约束优化:多元线性回归的极⼤似然估计3.7:权重的最⼤后验估计:线性回归的序列学习:利⽤⾼斯线性模型证明3.8:计算预测分布:证明后验⽅差随数据量增⼤⽽减⼩3.12:权重分布的共轭先验:证明基于 Gaussian-gamma 先验的权重分布的预测分布是 Student's t 分布:正交基函数诱导的等价核的性质经验贝叶斯-3.19:evidence 的对数似然:推导\alpha的优化过程:利⽤⾏列式对数的导数优化\alpha:推导\beta的优化过程:计算 evidence 的边际分布:⽤贝叶斯公式重新推导 3.23习题详解Exercise 3.3Hint.\begin{aligned} E_D(\mathbf{w}) &=\frac{1}{2}\sum^N_{n=1}r_n\|\mathbf{t}_n-W^T\phi(\mathbf{x}_n)\|^2_2\\ &=-\sum^N_{n=1}\ln\mathcal{N} (\mathbf{t}_n|W^T\phi(\mathbf{x}_n),r_n^{-1}I)+\text{const.} \end{aligned}第⼀个等号可以把r_n理解为样本权重,第⼆个等号可以把r_n理解为样本噪声的精度(precision)。
Comment.Exercise 3.4Solution.由于样本独⽴,噪声与输⼊独⽴,只需考虑⼀个样本的情形。
\begin{aligned} \underset{\epsilon}{\mathbb{E}}\left[(w^T(x+\epsilon)-t)^2\right] &= \underset{\epsilon}{\mathbb{E}}\left[((w^Tx-t)+w^T\epsilon)^2\right]\\ &=\underset{\epsilon}{\mathbb{E}}\left[(w^Tx-t)^2\right] + \underset{\epsilon}{\mathbb{E}}\left[(w^T\epsilon)(\epsilon^Tw)\right] + \underset{\epsilon}{\mathbb{E}}\left[ (w^Tx-t)w^T\epsilon\right]\\ &=(w^Tx-t)^2 + w^T\underset{\epsilon}{\mathbb{E}}\left[\epsilon\epsilon^T\right]w + (w^Tx-t)w^T\underset{\epsilon}{\mathbb{E}}\left[ \epsilon\right]\\ &=(w^Tx-t)^2 + \sigma^2 w^T w \end{aligned}Comment.该结论对多元输出的情形亦成⽴,只需⽤迹技巧展开 F-范数。
大学英语(三)Unit 6 Passage A Exercise
proud
She refused to accept help in changing a tire.
2. Find out more supporting details in the passage for the following statements, especially the underlinedlowing questions.
1) When did the author arrive in Cairo? At midnight.
2) Who was the author traveling with? ___________________
Answer:Men would smile and respectfully and kindly chat with her.
Answer:She invites goodwill from the people she meets.
3 . Fill each of the blanks with an appropriate word.
approach core counterpart economy efficient
exception invite relief punctually vague
1. The mayor is scheduled to meet his American (您未填写此题) during his 7-day visit to San Francisco.
1. We stop at a worn car.(Para. 4)
It has seen a better day.
Answer:There are quite a few scrapes on its body.
StudentSolutionManual
follows as:
r(t, T ) = − 1 ln P .
(3)
τ 100
Finally, to obtain the semi-annually compounded yield for the 1-year T-bill, use:
r2(0, 1) = 2 ×
1 (P /100)1/2
−
1
Yield Duration Modified Macaulay
a. 6.95%
3
3
2.8993
b. 6.28% 2.9542 2.9974 2.9061
c. 6.66% 0.9850 0.9850 0.9689
d. 0.00% 0.5
0.5
0.5
e. 6.82% 0.5111 0.5111 0.4943
Exercise 7.
Price Duration $ Duration
a. $89.56 4.55
$407.88
b. $67.63 -7.00 ($473.39)
c. $79.46 3.50
$277.74
d. $100.00 0.5
$50.00
e. $100.00 -$102.70 -0.2763 ($28.38)
a. $10 mn
b. Compute the dollar duration of the cash flows in each bond, and then the dollar duration of the portfolio:
Security 6yr IF @ 20% - fl quart
4yr fl 45bps semi 5yr zero Portfolio
ch06partSolution
Selected Solutions for Exercises inNumerical Methods with Matlab:Implementations and ApplicationsGerald W.RecktenwaldChapter6Finding the Roots of f(x)=0The following pages contain solutions to selected end-of-chapter Exercisesfrom the book Numerical Methods with Matlab:Implementations andApplications,by Gerald W.Recktenwald,c 2000,Prentice-Hall,Upper Saddle River,NJ.The solutions are c 2000Gerald W.Recktenwald.ThePDF version of the solutions may be downloaded or stored or printed onlyfor noncommercial,educational use.Repackaging and sale of these solutionsin any form,without the written consent of the author,is prohibited.The latest version of this PDFfile,along with other supplemental material for the book,can be found at /recktenwald.2Finding the Roots of f(x)=0 6–2The function f(x)=sin(x2)+x2−2x−0.09has four roots in the interval−1≤x≤3.Given the m-file fx.m,which containsfunction f=fx(x)f=sin(x.^2)+x.^2-2*x-0.09;the statement>>brackPlot(’fx’,-1,3)produces only two brackets.Is this result due to a bug in brackPlot or fx?What needs to be changed so that all four roots are found?Demonstrate that your solution works.Partial Solution:The statement>>Xb=brackPlot(’fx’,-1,3)Xb=-0.15790.05262.1579 2.3684returns two brackets.A close inspection of the plot of f(x)reveals that f(x)crosses the x-axis twice near x=1.3.These two roots are missed by brackPlot because there default search interval is too coarse.There is no bug in brackPlot.Implementing a solution using afiner search interval is left as an exercise.6–11Use the bisect function to evaluate the root of the Colebrook equation(see Exercise8) for /D=0.02and Re=105.Do not modify bisect.m.This requires that you write an appropriate function m-file to evaluate the Colebrook equation.Partial Solution:Using bisect requires writing an auxiliary function to evaluate the Cole-brook equation in the form F(f)=0,where f is the friction factor.The following form of F(f)is used in the colebrkz function listed below.F(f)=1√f+2log10/D3.7+2.51Re D√fMany other forms of F(f)will work.function ff=colebrkz(f)%COLEBRKZ Evaluates the Colebrook equation in the form F(f)=0%for use with root-finding routines.%%Input:f=the current guess at the friction factor%%Global Variables:%EPSDIA=ratio of relative roughness to pipe diameter%REYNOLDS=Reynolds number based on pipe diameter%%Output:ff=the"value"of the Colebrook function written y=F(f)%Global variables allow EPSDIA and REYNOLDS to be passed into%colebrkz while bypassing the bisect.m or fzero functionglobal EPSDIA REYNOLDSff=1.0/sqrt(f)+2.0*log10(EPSDIA/3.7+2.51/(REYNOLDS*sqrt(f)));Because the bisect function(unlike fzero)does not allow additional parameters to be passed through to the F(f)function,the values of /D and Re are passed to colebrkz via global variables.Running bisect with colebrkz is left to the reader.For Re=1×105and /D=0.02the solution is f=0.0490.Copyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.Chapter6:Finding the Roots of f(x)=03 6–13Derive the g3(x)functions in Example6.4and Example6.5.(Hint:What is thefixed-pointformula for Newton’s method?)Partial Solution:Thefixed point iteration formulas designated as g3(x)in Example6.4 and Example6.5are obtained by applying Newton’s method.The general form of Newton’smethod for a scalar variable isx k+1=x k−f(x k) f (x k)Example6.4:The f(x)function and its derivative aref(x)=x−x1/3−2f (x)=1−13x−2/3Substituting these expressions into the formula for Newton’s method and simplifying givesx k+1=x k−x k−x1/3k−21−(1/3)x−2/3k=x k(1−(1/3)x−2/3k)−(x k−x1/3k−2)1−(1/3)x−2/3k=x k−(1/3)x1/3k−x k+x1/3k+21−(1/3)x−2/3k=(2/3)x1/3k+21−(1/3)x k=2x1/3k+63−x−2/3kRepeating this analysis for Example6.5is left as an exercise.Copyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.4Finding the Roots of f(x)=0 6–17K.Wark and D.E.Richards(Thermodynamics,6th ed.,1999,McGraw-Hill,Boston,Example 14-2,pp.768–769)compute the equilibrium composition of a mixture of carbon monoxide and oxygen gas at one atmosphere.Determining thefinal composition requires solving3.06=(1−x)(3+x)1/2 x(1+x)1/2for x.Obtain afixed-point iteration formula forfinding the roots of this equation.Implement your formula in a Matlab function and use your function tofind x.If your formula does not converge,develop one that does.Partial Solution:Onefixed point iteration formula is obtained by isolating the factor of (3+x)in the numerator.3.06x(1+x)1/21−x =(3+x)1/2=⇒x=3.06x(1+x)1/21−x2−3=⇒g1(x)=3.06x(1+x)1/21−x2−3Anotherfixed point iteration formula is obtained by solving for the isolated x in the denomi-nator to getx=(1−x)(3+x)1/23.06(1+x)=⇒g2(x)=(1−x)(3+x)1/23.06(1+x)Performing10fixed point iterations with g1(x)givesit xnew1-7.6420163e-012-2.5857113e+003-1.0721050e+014-7.9154865e+015-7.1666488e+026-6.6855377e+037-6.2575617e+048-5.8590795e+059-5.4861826e+0610-5.1370394e+07Thus,g1(x)does not converge.The g2(x)function does converge to the true root of x= 0.340327....Matlab implementations of thefixed point iterations are left as an Exercise. Copyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.Chapter6:Finding the Roots of f(x)=05 6–24Create a modified newton function(say,newtonb)that takes a bracket interval as input instead of a single initial guess.From the bracket limits take one bisection step to determine x0,the initial guess for Newton e the bracket limits to develop relative tolerances on x and f(x)as in the bisect function in Listing6.4.Solution:The newtonb function is listed below.The demoNewtonb function,also listed below, repeats the calculations in Example6.8with the original newton function and with the new newtonb function.Running demoNewtonb gives>>demoNewtonbOriginal newton function:Newton iterations for fx3n.mk f(x)dfdx x(k+1)1-4.422e-018.398e-01 3.526644293139032 4.507e-038.561e-01 3.521380147397333 3.771e-078.560e-01 3.521379706804574 2.665e-158.560e-01 3.5213797068045750.000e+008.560e-01 3.52137970680457newtonb function:Newton iterations for fx3n.mk f(x)dfdx x(k+1)1-4.422e-018.398e-01 3.526644293139032 4.507e-038.561e-01 3.521380147397333 3.771e-078.560e-01 3.521379706804574 2.665e-158.560e-01 3.5213797068045750.000e+008.560e-01 3.52137970680457The two implementations of Newton’s method give identical results because the input to newtonb is the bracket[2,4].This causes the initial bisection step to produce the same initial guess for the Newton iterations that is used in the call to newton.function demoNewtonb%demoNewtonb Use newton and newtonb to find the root of f(x)=x-x^(1/3)-2%%Synopsis:demoNewton%%Input:none%%Output print out of convergence history,and comparison of methodsfprintf(’\nOriginal newton function:\n’);r=newton(’fx3n’,3,5e-16,5e-16,1);fprintf(’\nnewtonb function:\n’);rb=newtonb(’fx3n’,[24],5e-16,5e-16,1);Copyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.6Finding the Roots of f(x)=0 function r=newtonb(fun,x0,xtol,ftol,verbose)%newtonb Newton’s method to find a root of the scalar equation f(x)=0%Initial guess is a bracket interval%%Synopsis:r=newtonb(fun,x0)%r=newtonb(fun,x0,xtol)%r=newtonb(fun,x0,xtol,ftol)%r=newtonb(fun,x0,xtol,ftol,verbose)%%Input:fun=(string)name of mfile that returns f(x)and f’(x).%x0=2-element vector providing an initial bracket for the root%xtol=(optional)absolute tolerance on x.Default:xtol=5*eps%ftol=(optional)absolute tolerance on f(x).Default:ftol=5*eps%verbose=(optional)flag.Default:verbose=0,no printing.%%Output:r=the root of the functionif nargin<3,xtol=5*eps;endif nargin<4,ftol=5*eps;endif nargin<5,verbose=0;endxeps=max(xtol,5*eps);feps=max(ftol,5*eps);%Smallest tols are5*epsif verbosefprintf(’\nNewton iterations for%s.m\n’,fun);fprintf(’k f(x)dfdx x(k+1)\n’);endxref=abs(x0(2)-x0(1));%Use initial bracket in convergence testfa=feval(fun,x0(1));fb=feval(fun,x0(2));fref=max([abs(fa)abs(fb)]);%Use max f in convergence testx=x0(1)+0.5*(x0(2)-x0(1));%One bisection step for initial guessk=0;maxit=15;%Current and max iterationswhile k<=maxitk=k+1;[f,dfdx]=feval(fun,x);%Returns f(x(k-1))and f’(x(k-1))dx=f/dfdx;x=x-dx;if verbose,fprintf(’%3d%12.3e%12.3e%18.14f\n’,k,f,dfdx,x);endif(abs(f/fref)<feps)|(abs(dx/xref)<xeps),r=x;return;endendwarning(sprintf(’root not found within tolerance after%d iterations\n’,k));Copyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.Chapter6:Finding the Roots of f(x)=07 6–27Implement the secant method using Algorithm6.5and Equation(6.13).Test your program by re-creating the results in Example6.10.What happens if10iterations are performed?Replace the formula in Equation(6.13)withx k+1=x k−f(x k)(x k−x k−1)f(x k k−1,whereεis a small number on the order ofεm.How and why does this change the results? Partial Solution:The demoSecant function listed below implements Algorithm(6.5)using Equation(6.13).The f(x)function,Equation6.3,is hard-coded into demoSecant.Note also that demoSecant performs ten iterations without checking for convergence.function demoSecant(a,b);%demoSecant Secant method for finding the root of f(x)=x-x^(1/3)-2=0%Implement Algorithm6.5,using Equation(6.13)%%Synopsis:demoSecant(a,b)%%Input:a,b=initial guesses for the iterations%%Output:print out of iterations;no return values.%copy initial guesses to local variablesxk=b;%x(k)xkm1=a;%x(k-1)fk=fx3(b);%f(x(k))fkm1=fx3(a);%f(x(k-1))fprintf(’\nSecant method:Algorithm6.5,Equation(6.13)\n’);fprintf(’n x(k-1)x(k)f(x(k))\n’);fprintf(’%3d%12.8f%12.8f%12.5e\n’,0,xkm1,xk,fk);for n=1:10x=xk-fk*(xk-xkm1)/(fk-fkm1);%secant formula for updating the rootf=fx3(x);fprintf(’%3d%12.8f%12.8f%12.5e\n’,n,xk,x,f);xkm1=xk;xk=x;%set-up for next iterationfkm1=fk;fk=f;endCopyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.8Finding the Roots of f(x)=0 Running demoSecant with an initial bracket of[3,4](the same bracket used in Example6.10) gives>>demoSecant(3,4)Secant method:Algorithm6.5,Equation(6.13)n x(k-1)x(k)f(x(k))0 3.00000000 4.00000000 4.12599e-011 4.00000000 3.51734262-3.45547e-032 3.51734262 3.52135125-2.43598e-053 3.52135125 3.52137971 1.56730e-094 3.52137971 3.52137971-8.88178e-165 3.52137971 3.52137971-2.22045e-166 3.52137971 3.521379710.00000e+007 3.52137971 3.521379710.00000e+00Warning:Divide by zero.>In/werk/MATLAB_Book/SolutionManual/roots/mfiles/demoSecant.m at line228 3.52137971NaN NaN9NaN NaN NaN10NaN NaN NaNThe secant method has fully converged in6iterations.Continuing the calculations beyond convergence gives afloating point exception because f(x k)−f(x k−1)=0in the denominator of Equation(6.13).In general,it is possible to have f(x k)−f(x k−1)=0before the secant iterations reach convergence.Thus,thefloating point exception exposed by demoSecant should be guarded against in any implementation of the secant method.Implementing thefix suggested in the problem statement is left as an exercise for the reader.Copyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.Chapter6:Finding the Roots of f(x)=09 6–33Write an m-file function to compute h,the depth to which a sphere of radius r,and specific gravity s,floats.(See Example6.12on page281.)The inputs are r and s,and the output ish.Only compute h when s<0.5.The s≥0.5case is dealt with in the following Exercise.If s≥0.5is input,have your function print an error message and stop.(The built-in error function will be useful.)Your function needs to include logic to select the correct root from the list of values returned by the built-in roots function.Partial Solution:The floata function listed below performs the desired computations.We briefly discuss three of the key statements in floata The coefficients of the polynomial are stored in the p vector.Thenc=getreal(roots(p));finds the real roots of the polynomial.The getreal subfunction returns only the real elements of a ing getreal is a defensive programming strategy.The sample calculation in Example6.12obtained only real roots of the polynomial,so getreal would not be necessary in that case.Thek=find(c>0&c<r);statement extracts the indices in the c vector satisfying the criteria0≤c k≤r.Then h=c(k);copies those roots satisfying the criteria to the h vector.No assumption is made that only one root meets the criteria.If more than one root is found a warning message is issued before leaving floata.Testing of floata is left to the reader.Copyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.10Finding the Roots of f(x)=0 function h=floata(r,s)%float Find water depth on a floating,solid sphere with specific gravity<0.5%%Synopsis:h=floata(r,s)%%Input:r=radius of the sphere%s=specific gravity of the sphere(0<s<1)%%Output:h=depth of the sphereif s>=0.5error(’s<0.5required in this version’)elsep=[1-3*r04*s*r^3];%h^3-3*r*h+4*s*r^3=0c=getreal(roots(p));k=find(c>0&c<r);%indices of elements in c such that0<c(k)<rh=c(k);%value of elements in c satisfying above criterionendif length(h)>1,warning(’More than one root found’);end%==============================function cr=getreal(c)%getreal Copy all real elements of input vector to output vector%%Synopsis:cr=getreal(c)%%Input:c=vector of numerical values%%Output cr=vector of only the real elements of c%cr=[]if c has only imaginary elementsn=0;for k=1:length(c)if isreal(c(k))n=n+1;cr(n)=c(k);endendif n==0,cr=[];warning(’No real elements in the input vector’);endCopyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.。
多层螺旋CT三维重建技术与X线影像学检查在骨盆骨折的诊断应用比较
2017年GUANGDONG WEILIANG YUANSU KEXUE 第24卷第1期文章编号:1006-446X (2017)1-0060-04多层螺旋CT 三维重建技术与X 线影像学检查在骨盆骨折的诊断应用比较谢伟玲(佛山市高明区中医院,广东佛山528500)摘要:目的比较多层螺旋CT 三维重建技术与X 线影像学两种检查方式在骨盆骨折诊断中的应用效果。
方法选取2013年8月—2015年2月佛山市高明区中医院收治的70例骨盆骨折患者,所有患者均进行多层螺旋CT 三维重建和X 线影像学检查,比较两种检查方法的结果。
结果采用X 线平片扫描共检出43例骨折,检出率为61.43%;采用多层螺旋CT 三维重建技术共检出70例骨折,检出率为100.00%;多层螺旋CT 三维重建技术骨盆骨折检出率高于X 线影像学检查,差异显著(P <0.05)。
结论多层螺旋CT 三维重建技术可更好的对骨盆骨折进行诊断,诊断率较高,可准确反映患者骨折具体情况,对提高手术成功率和改善患者预后具有重要的意义,在临床应用中值得推广。
关键词:多层螺旋CT 三维重建;X 线影像学;骨盆骨折中图分类号:R445文献标识码:A 收稿日期:2016-12-07作者简介:谢伟玲(1973—),女,汉族,本科,放射主治医师,专业:医学影像学。
E -mail :vvvyyy1@Comparison of MSCT Three -Dimensional Reconstructionand X -Ray Radiology in Diagnosis of Pelvic FractureXIE Weiling(Brillant District ,Foshan City Hospital ,Guangdong Foshan 528500,China )Abstract :Objective To compare effect of multi -slice spiral CT (MSCT )three -dimensionalreconstruction and X -ray radiology in diagnosis of pelvic fracture (PF ).Methods70PF patients admitted in hospital from August 2013to February 2015received MSCT three -dimensional reconstruction and X -ray radiology detection ,and their results were compared.Results There was 43fracture in X -ray scanning and detection rate was 61.43%;There was 70fracture in MSCT three -dimensional reconstruction and detection rate was 100.00%;The detection rate of MSCT three -dimensional reconstruction was higher than X -ray (P <0.05).Conclusion MSCT three -dimensional reconstruction can better diagnose PF and diagnosis rate is higher ,it can accurately reflect detailed fracture situation ,increase success rate and improve prognosis.It is worthy of clinical application.Key words :multi -slice spiral CT three -dimensional reconstruction ;X -ray radiology ;pelvic fracture·06·DOI:10.16755/ki.issn.1006-446x.2017.01.0152017年GUANGDONG WEILIANG YUANSU KEXUE第24卷第1期骨盆由左右髂骨、坐骨、耻骨及骶尾椎骨组成。
IMA案例分析题样题答案
Solution or Key Points:1.1.Solution•Descriptions of the five responsibilities of the management accountant,as well as examples of practices and techniques of each,include the following:o Planning.The manager gains an understanding of the impact on the organization of planned transactions(i.e.,analyzing strengths and weaknesses)and economic events,both strategic and tactical,and sets obtainable goals for the organization.Thedevelopment of budgets in a bottom-up or top-down approach for both the short and longterm is an example of planning.o Evaluating.The manager judges and analyzes the implication of various past and expected events,and then chooses the optimum course of action.The manager alsotranslates data and communicates the conclusions.Graphical analysis(such as trend,barcharts,or regression)and net present value analysis are examples of evaluating.o Controlling.The manager ensures the integrity of financial information,monitors performance against budgets and goals,and provides information internally for decisionmaking.The successful implementation of internal control procedures is an example ofcontrolling.o Accountability.The manager implements a reporting system closely aligned to organizational goals that contributes to the measurement of the effective use of resourcesand safeguarding of assets.Internal reporting,such as comparisons of actuals to budget,is an example of accountability.o External reporting.The manager prepares reports in accordance with generally accepted accounting principles and then disseminates this information to shareholders(both current and potential),creditors,and regulatory and tax agencies.An annual reportor a credit application are examples of external reporting.Solution or Key Points:2.2.SolutionIn referring to the general roles of managers,a.Two interpersonal roles include acting as a1.Leader who hires,trains,motivates employees by integrating individual andorganization goals by using effective communication2.Liaison who deals with people outside the organizationb.Two informational roles include acting as a1.Monitor who seeks information about other internal departments,the externalenvironment,and other organizations2.Disseminator who communicates outside information internally,and internalinformation to the subordinates,while exercising judgment so there is minimalinformation overloadc.Three decisional roles include acting as a1.Negotiator between subordinates,suppliers,contracts,and union2.Resource allocator who distributes funds,equipment,time,etc.,in accordancewith organizational objectives3.Disturbance handler who solves issues of conflict within and without theorganizationSolution or Key Points:3.3.SolutionFor the original organizational structure proposed by Donald Jenkins,a.Advantages include1.Centralized decision making2.Close supervision of subordinates3.Increased opportunities for promotionDisadvantages includeck of flexibility/creativity5.More difficult vertical communication6.Increased costs for additional management personnelb.The span of control will1.Allow for tighter control over employees2.Afford timelier task completion and problem resolution,as the supervisors arecloser to everyday operationsc.As a result of the organizational structure proposed by Donald Jenkins,the employees arelikely to have a clear understanding of what is expected of them.However,because of thetighter span of control,the employees may avoid making decisions and taking risks. Solution or Key Points:4.4.SolutionAn informal group is a number of individuals attracted to one another by common geography,identity, values,and/or goals.The group identifies,then enforces,a set of behavior standards via sanctions and rewards.The informal group exists supplementary to the formal management system within which it is normally found.Solution or Key Points:5.5.SolutionThe working environment where it would be most appropriate to use the task-oriented and relationship-oriented leadership styles,respectively,are as follows:a.The task-oriented leadership style,to some extent,is necessary to make all work groupsfunction effectively.It is usually used where the job is repetitive,under time pressure,andrequires accuracy.b.The relationship-oriented leadership style is used when the work environment is open orunstructured,or the methodology for accomplishing a task is loosely defined and theexercise of judgment is important.Solution or Key Points:6.6.SolutionAccording to Maslow’s theory of motivation,human needs exist in a hierarchy of importance,namely, physiological,security,social,esteem,and self-actualization.Once the first level of physiologicalneeds is satisfied,the individual is no longer motivated by them and moves to the next level.The process continues until the individual reaches self-actualization.a.Unions are often associated with the ability to satisfy the lower level needs by providingcompetitive salaries,safe working conditions,job security,and social belonging.Itappears that these needs were already being satisfied at Spokane Computer Associates,as salaries were relatively high,there was little turnover,and the plant was considered asafe and pleasant place to work.b.The products at Pullman were successful and sold without much effort on the part of thesalespeople.As a result,the salespeople were doing well financially,and their lowerneeds were satisfied.By offering the top performers recognition,the firm was appealing tothem through a higher level motivator,esteem.The salespeople were motivated toincrease sales by the need for recognition and respect from their peers and others.c.Zach Chaudry is leaving Midwest Electronics to accept a teaching and research positionat the university to satisfy the highest level need,self-actualization.In this new position,Chaudry will have the opportunity to control his own research.Solution or Key Points:7.7.Solutiona.The need categories most often associated with professional employees are esteem and self-actualization,since the lower level needs are usually satisfied by professional employment and are no longer motivators(satisficers).b.The need categories most often associated with assembly-line workers(skilled labor)are thephysiological,safety,and social needs.The physiological and safety needs are usually satisfied by employment and are no longer motivators(satisficers).The social needs such as acceptance and affiliation must continue to be fulfilled.Solution or Key Points:8.8.SolutionAt least three general purposes of organizational communication are to1.Integrate efforts of employees2.Build a community of employees with high morale and mutual trust3.Acquire and disseminate informationSolution or Key Points:9.9.Solutiona.Four personal characteristics,attributable to both senders and receivers,that can causecommunications problems include1.Listening problems resulting from the receiver misunderstanding the intended message2.Poor channel or media selection,e.g.,sender has chosen a communication carrier that does notget to the receivers,or faulty encoding,e.g.,the use of jargon3.The influence that knowledge level,attitude(lack of interest),and background(demographics)will play in sending and receiving an effective message4.Interpersonal dynamics and emotions,including trust in the senderb.Three organizational characteristics that may be barriers to effective communications includeanizational status and power differences,e.g.,lower levels reluctant to communicate withupper levels or upper level perception that lower levels have little to contributeck of formal channels,i.e.,organization must provide adequate upward,downward,andhorizontal communications in the form of surveys,open-door policies,newsletters,memos,task force,and liaison personnel3.Departmental needs and goals;each one views problems and needs in its own termsSolution or Key Points:10.Solution10.a.Responsibility accounting is a system that recognizes various decision or responsibility centersthroughout an organization,and reflects the plans and actions of each of these centers by assigning responsibility for particular revenues and costs(as well as assets and liabilities where pertinent)to the individual managers primarily responsible for making decisions about these revenues and costs.Costs are classified into controllable and non-controllable categories with the responsibility center only charged with costs under its control.b.The advantages a company attains through the use of responsibility accounting include thefollowing:1.Participation in the planning process enhances the acceptance and achievability of the guidelinesestablished by the company.2.Responsibility plans provide clear guidelines to managers for day-to-day decisions and freemanagement from daily operations.3.More involvement will lead to better decision making.c.The use of responsibility accounting is advantageous to the managers in the following ways:1.Responsibility accounting facilitates delegation of decision-making and,therefore,managers areafforded greater freedom of action without daily supervision.This results in greater job satisfaction and higher motivation to improve performance.2.Managers are aware of the basis of their performance evaluations.Solution or Key Points:11.Solution11.The major disadvantages in the procedures employed by Rouge Corporation in preparing and implementing the divisional budgets include the following:a.The procedure is top-down versus a bottom-up approach.The budget is prepared by topmanagement and imposed on the divisions.The divisions do not participate in the budgetpreparation.This lack of participation inhibits goal congruence.b.Corporate management says differences between divisions are factored into the budgetprocess.However,problems unique to each division are best recognized by divisionmanagers.These managers should be allowed to communicate problems to topmanagement.Additionally,division managers should be allowed to suggest modificationsto the budget during the budget period.c.The budget is not flexible.The budget is set for the entire year and is not modified ascircumstances change.Solution or Key Points:12.12.SolutionMultiple performance evaluation criteria would be appropriate for the evaluation for the Star Paper Division.The criteria suggested by George Harris would take into account more of the results of the key decisions being made by the manager,are not conflicting,and emphasize the balance of profits with the control of current assets.These three measures are controllable by the manager and,in conjunction with ROA,would be more representative of the success of the business.Solution or Key Points:13.Solution13.Corporate management has established an environment in which there is an incompatibility(lack of goal congruence)between the achievement of corporate objectives and personal ethics.Under the current situation,severe penalties have been imposed by top management whenever subordinates do not achieve the high levels of performance established by the predetermined objectives.This has caused lower level management to take unethical courses of action.Corporate management apparently utilizes an authoritarian,nonparticipative management style which does not consider contributions from lower level management.As a result of this type of management style,top management may have established unreasonable expectations and may not recognize the need to change the expectations in light of changing circumstances.These factors may result in subordinates choosing any means to reach the objectives.The penalty and reward system appears to be inappropriate.There is no positive feedback or encouragement for effective performance and the penalties are heavy for failure to meet objectives.No code of ethics exists,and penalties are apparently nonexistent or minor for violation of acceptable business practices that are compatible with personal ethical standards.Solution or Key Points:14.14.SolutionAdam Williams should follow the established policies of the organization bearing on the resolution of such conflict.If these policies do not resolve the ethical conflict,Williams should report the problem to successively higher levels of management up to the board of directors until it is satisfactorily resolved.There is no requirement for Williams to inform his immediate superior of this action because he isinvolved in the conflict.If the conflict is not resolved after exhausting all course of internal review, Williams may have no other recourse than to resign from the organization and submit an informative memorandum to an appropriate member of the organization.15.Solution or Key Points:15.SolutionActivities in which JLS could engage to meet a social responsibility with respect to the three areas identified include the following:a.Human resources1.Contribute to the educational development of people through improved trainingand career advancement programs for current employees2.Provide for the well-being of employees by maintaining safe working conditionsb.Physical resources and environmental contributions1.Contribute to improvement in the environment by reducing company pollution ofair and water2.Contribute to energy conservation by taking the lead in promoting community-wide energy conservationc.Product or service contributions1.Design and manufacture safe or better quality products2.Respond quickly to customer complaints3.Develop clear and concise labels and instructions for product usesSolution or Key Points:16.16.SolutionThe arrangements made by Farbell and Cross&Gill exceeded the activities normally performed by an internal auditor for the external auditor and thus should be considered unethical.The external auditor gave up too much responsibility,and Farbell should not have agreed to such a proposal as it diminishes the value of the audit.Internal auditors can assist in performing substantive tests or tests of compliance,but their competence and objectivity must be considered,and the work must be supervised and tested.。
电磁场与电磁波第二版 (周克定 翻译) 答案
1
0 (Eaρ
)
•( ρ dφ dzaρ
)
=
ρl
/
ε0
E = ρl 2πε 0 ρ
m ∴
E
=
ρl 2πε 0 ρ
aρ
w.co 选取点 A 作为电位参考点 (点 A 和点 P 的φ hda 和 z 坐标相同 ) ,点 A 的 ρ = a 。
www.k 自由空间中任意点 P 的电位为
PA
∫ ∫ 答案网 VP =
案
⎧0 答
r<a
E(r) = 课⎪⎨ a后rk(r2 − a2 ) /(2ε0r2 ) a ≤ r ≤ b
⎩⎪ark(b2 − a2 ) /(2ε0r2 求穿过 r = b 的球面的电通量
∫ Ψ的Qf=总是量半s,径D即为空•bd间的s中球=总面Q的所f 自(=包案由单2围π电位网的k荷是自(b。w库由2w仑−电w.)a荷k2h)
答 后 课
b a
17
exercise 3.11 Solution:
假设无限长均匀带电导线位于坐标系的 z 轴,
mP co 导线上电荷的线密度为 ρl 。 daw. 空间中的电场强度可表示为 E = E(ρ )aρ www.kh 在以 z 轴为中轴、半径为 ρ 、高度为 1 的圆柱面上可利用高斯定律的
应用(Gauss) Divergence Theorem:
∫ ∫ (
∇ • Fdv = F • ds
v
s
∴ ∫ r • ds = ∫v ∇ • rdv
v 是以坐标原点为球心、半径为
b
的
b
y
球的体积 )
网
r
=
xax
+
yay
Section_13_Exercises_Solutionspdf
Section 1.3 Exercises (Solutions)1.109, 1.110, 1.111, 1.114*, 1.115, 1.119*, 1.122, 1.125, 1.127*, 1.128*, 1.131*, 1.133*, 1.135*, 1.137*, 1.139*, 1.145*, 1.146 - 148.1.109Sketch some normal curves.(a)Sketch a normal curve that has mean 10 and standard deviation 3.(b)On the same x axis, sketch a normal curve that has mean 20 and standard deviation 3.(c)How does the normal curve change when the mean is varied but the standard deviationstays the same.Solution(a) It is easiest to draw the curve first, and then mark the scale on the axis. (b) Draw acopy of the first curve, with the peak over 20. (c) The curve has the same shape, but is translated left or right.The curve has the same shape, but is translated left or right.1.110The effect of changing the standard deviation.(a)Sketch a normal curve that has mean 10 and standard deviation 3.(b)On the same x axis, sketch a normal curve that has mean 10 and standard deviation 1.(c)How does the normal curve change when the standard deviation is varied but the meanstays the same.Solution(a) As in the previous exercise, draw the curve first, and then mark the scale on the axis.(b) In order to have a standard deviation of 1, the curve should be 1/3 as wide, and threetimes taller. (c) The curve is centered at the same place (the mean), but its height andwidth change. Specifically, increasing the standard deviation makes the curve wider and shorter; decreasing the standard deviation makes the curve narrower and taller.1.111Know your density. Sketch density curves that might describe distributions with the following shapes:(a)Symmetric, but with two peaks (that is, two strong clusters of observations).(b)Single peak and skewed to the left.Solution1.111. Sketches will vary.1.114Total scores. Here are the total scores of 10 students in an introductory statistics course. Previous experience with this course suggests that these scores should come from a distribution that is approximately normal with mean 70 and standard deviation 10.(a)Using these values for µ and σ, standardize the first exam scores of these 10 students.(b)If the grading policy is to give grades of A to the top 15% of scores based on the normaldistribution with mean 70 and standard deviation 10, what is the cut-off for an A in terms of a standardized score?(c)Which students earned a grade of A on the final exam for this course?Solution1.114. (a) For example, (68−70)/10 = −0.2. The complete list is given below.(b) The cut-off for an A is the 85th percentile for the N(0, 1) distribution. From Table A,this is approximately 1.04; software gives 1.0364. (c) The top two students (with scores of 92 and 98) received A’s.68 54 92 75 73 98 64 55 80 70-0.2 -1.6 2.2 0.5 0.3 2.8 -0.6 -1.5 1 01.115Assign more grades. Refer to the previous exercise. The grading policy says the cut-offs for the other grades correspond to the following: bottom 5% receive F, next 10% receive D, next 40% receive C, and next 30% receive B. These cut-offs are based on the N(70, 10) distribution.(a)Give the cut-offs for the grades in this course in terms of standardized scores.(b)Give the cut-offs in terms of actual total scores.(c)Do you think that this method of assigning grades is a good one? Give reasons for youranswer.Solution1.115. (a) We need the 5th, 15th, 55th, and 85th percentiles for a N(0, 1) distribution.These are given in the table below. (b) To convert to actual scores, take the standard-score cut-off z and compute 10z + 70. (c) Opinions will vary.Note: The cut-off for an A given in the previous solution is the lowest score that gets an A—that is, the point where one’s grade drops from an A to a B. These cut-offs are thepoints where one’s grade jumps up. In practice, this is only an issue for a score that falls exactly on the border between two grades.1.116A uniform distribution. If you ask a computer to generate “random numbers” between 0 and 1, you will get observations from a uniform distribution. Figure 1.34 graphs the density curve for a uniform distribution. Use areas under this density curve to answer the following questions.(a)Why is the total area under this curve equal to 1?(b)What proportion of the observations lie below 0.35?(c)What proportion of the observations lie between 0.35 and 0.65?Figure 1.34 The density curve of a uniform distribution, for Exercise 1.116.1.118Find the mean, the median, and the quartiles. What are the mean and the median of the uniform distribution in Figure 1.34? What are the quartiles?Solutions (1.116 - 1.118)1.119Three density curves.Figure 1.35 displays three density curves, each with three points marked on it. At which of these points on each curve do the mean and the median fall?Figure 1.35 Three density curves, for Exercise 1.119.Solution1.119. (a) Mean is C, median is B (the right skew pulls the mean to the right). (b) MeanA, median A. (c) Mean A, median B (the left skew pulls the mean to the left).1.120Length of pregnancies. The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. Draw a density curve for this distribution on which the mean and standard deviation are correctly located.Solution1.122Pregnancies and the 68–95–99.7 rule. The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. Use the 68–95–99.7 rule to answer the following questions.(a)Between what values do the lengths of the middle 95% of all pregnancies fall?(b)How short are the shortest 2.5% of all pregnancies? How long do the longest 2.5% last?Solution1.122. See the sketch of the curve in the solution to Exercise 1.120. (a) The middle 95%fall within two standard deviations of the mean: 266 ± 2(16), or 234 to 298 days. (b) The shortest 2.5% of pregnancies are shorter than 234 days (more than two standarddeviations below the mean).1.123Horse pregnancies are longer. Bigger animals tend to carry their young longer before birth. The length of horse pregnancies from conception to birth varies according to a roughly Normal distribution with mean 336 days and standard deviation 3 days. Use the 68–95–99.7 rule to answer the following questions.(a)Almost all (99.7%) horse pregnancies fall in what range of lengths?(b)What percent of horse pregnancies are longer than 339 days?Solution1.123. (a) 99.7% of horse pregnancies fall within three standard deviations of the mean:336 ± 3(3), or 327 to 325 days. (b) About 16% are longer than 339 days since 339days or more corresponds to at least one standard deviation above the mean.1.125Acidity of rainwater. The Normal quantile plot in Figure 1.32 (page 66) shows that the acidity (pH) measurements for rainwater samples in Exercise 1.36 are approximately Normal. How well do these scores satisfy the 68–95–99.7 rule? To find out, calculate the mean andstandard deviation s of the observations. Then calculate the percent of the 105 measurements that fall between − s and + s and compare your result with 68%. Do the same for the intervalscovering two and three standard deviations on either side of the mean. (The 68–95–99.7 rule is exact for any theoretical Normal distribution. It will hold only approximately for actual data.) Solution1.125. The mean and standard deviation are x = 5.4256 and s = 0.5379. About 67.62%(71/105 = 0.6476) of the pH measurements are in the range x ± s = 4.89 to 5.96. About95.24% (100/105) are in the range x ± 2s = 4.35 to 6.50. All (100%) are in the rangex ± 3s = 3.81 to 7.04.1.126Find some proportions. Using either Table A or your calculator or software, find the proportion of observations from a standard Normal distribution that satisfies each of the following statements. In each case, sketch a standard Normal curve and shade the area under the curve that is the answer to the question.(a)Z > 1.65(b)Z < 1.65(c)Z > −0.76(d)−0.76 < Z < 1.65Solution1.126. Using values from Table A: (a) Z > 1.65: 0.0495. (b) Z < 1.65: 0.9505.(c) Z > −0.76: 0.7764. (d) −0.76 < Z <1.65: 0.9505 − 0.2236 = 0.7269.1.127Find more proportions. Using either Table A or your calculator or software, find the proportion of observations from a standard Normal distribution for each of the following events. In each case, sketch a standard Normal curve and shade the area representing the proportion.(a)Z ≤ −1.8(b)Z ≥ −1.8(c)Z > 1.6(d)−1.8 < Z < 1.6Solution1.127. Using values from Table A: (a) Z ≤ −1.8: 0.0359. (b) Z ≥ −1.8: 0.9641.(c) Z > 1.6: 0.0548. (d) −1.8 < Z < 1.6: 0.9452 − 0.0359 = 0.9093.1.128Find some values of z. Find the value z of a standard Normal variable Z that satisfies each of the following conditions. (If you use Table A, report the value of z that comes closest to satisfying the condition.) In each case, sketch a standard Normal curve with your value of z marked on the axis.(a)22% of the observations fall below z.(b)40% of the observations fall above z.Solution1.128. (a) 22% of the observations fall below −0.7722. (This is the 22nd percentile of thestandard Normal distribution.) (b) 40% of the observations fall above 0.2533 (the 60thpercentile of the standard Normal distribution).1.129Find more values of z. The variable Z has a standard Normal distribution.(a)Find the number z that has cumulative proportion 0.65.(b)Find the number z such that the event Z > z has proportion 0.45.Solution1.129. (a) z = 0.3853 has cumulative proportion 0.65 (that is, 0.3853 is the 65thpercentile of the standard Normal distribution). (b) If z = 0.1257, then Z > z hasproportion 0.45 (0.1257 is the 55th percentile).1.130Find some values of z. The Wechsler Adult Intelligence Scale (WAIS) is the most common IQ test. The scale of scores is set separately for each age group and is approximately Normal with mean 100 and standard deviation 15. People with WAIS scores below 70 are considered mentally retarded when, for example, applying for Social Security disability benefits. What percent of adults are retarded by this criterion?Solution1.130. 70 is two standard deviations below the mean (that is, it has standard score z =−2), so about 2.5% (half of the outer 5%) of adults would have WAIS scores below 70.1.131High IQ scores. The Wechsler Adult Intelligence Scale (WAIS) is the most common IQ test. The scale of scores is set separately for each age group and is approximately Normal with mean 100 and standard deviation 15. The organization MENSA, which calls itself “the high IQ society,” requires a WAIS score of 130 or higher for membership. What percent of adults would qualify for membership?There are two major tests of readiness for college, the ACT and the SAT. ACT scores are reported on a scale from 1 to 36. The distribution of ACT scores are approximately Normal with mean µ = 21.5 and standard deviation σ = 5.4. SAT scores are reported on a scale from 600 to 2400. The SAT scores are approximately Normal with mean µ = 1509 and standard deviation σ= 321. Exercises 1.132 to 1.141 are based on this information.Solution1.131. 130 is two standard deviations above the mean (that is, it has standardscore z = 2), so about 2.5% of adults would score at least 130.1.132Compare an SAT score with an ACT score. Tonya scores 1820 on the SAT. Jermaine scores 29 on the ACT. Assuming that both tests measure the same thing, who has the higher score? Report the z-scores for both students.Solution1.132. Tonya’s score standardizes to z = (1820−1509)/321 = 0.9688, while Jermaine’sscore corresponds to z = (29−21.5)/5.4 = 1.3889. Jermaine’s score is higher.1.133Make another comparison. Jacob scores 16 on the ACT. Emily scores 1020 on the SAT. Assuming that both tests measure the same thing, who has the higher score? Report the z-scores for both students.Solution1.133. Jacob’s score standardizes to z = (16−21.5)/5.4 = −1.0185, while Emily’s scorecorresponds to z = (1020−1509)/321 = −1.5234. Jacob’s score is higher.1.134Find the ACT equivalent. Jose scores 2080 on the SAT. Assuming that both tests measure the same thing, what score on the ACT is equivalent to Jose’s SAT score?Solution1.134. Jose’s score standardizes to z = (2080−1509)/321 = 1.7788, so an equivalentACT score is 21.5 + 1.7788 × 5.4 = 31.1. (Of course, ACT scores are reported as whole numbers, so this would presumably be a score of 31.)1.135Find the SAT equivalent. Maria scores 30 on the ACT. Assuming that both tests measure the same thing, what score on the SAT is equivalent to Maria’s ACT score?Solution1.135. Maria’s score standardizes to z = (30−21.5)/5.4 =1.5741, so an equivalent SATscore is 1509 + 1.5741 × 321 = 2014.1.136Find an SAT percentile. Reports on a student’s ACT or SAT usually give the percentile as well as the actual score. The percentile is just the cumulative proportion stated as a percent: the percent of all scores that were lower than this one. Maria scores 2090 on the SAT. What is her percentile?Solution1.136. Maria’s score standardizes to z = (2090−1509)/321 = 1.81, for which Table Agives 0.9649. Her score is the 96.5 percentile.1.137Find an ACT percentile. Reports on a student’s ACT or SAT usually give the percentile as well as the actual score. The percentile is just the cumulative proportion stated as a percent: the percent of all scores that were lower than this one. Jacob scores 19 on the ACT. What is his percentile?Solution1.137. Jacob’s score standardizes to z = (19−21.5)/5.4 = −0.4630, for which Table Agives 0.3228. His score is the 32.3 percentile.1.138How high is the top 10%? What SAT scores make up the top 10% of all scores?Solution1.138. 1920 and above: The top 10% corresponds to a standard score of z = 1.2816,which in turn corresponds to a score of 1509 + 1.2816 × 321 = 1920 on the SAT..1.139How low is the bottom 20%? What SAT scores make up the bottom 20% of all scores?Solution1.139. 1239 and below: The bottom 20% corresponds to a standard score ofz = −0.8416, which in turn corresponds to a score of 1509 − 0.8416 × 321 =1239 on the SAT.1.140Find the ACT quartiles. The quartiles of any distribution are the values with cumulative proportions 0.25 and 0.75. What are the quartiles of the distribution of ACT scores?Solution1.140. The quartiles of a Normal distribution are ±0.6745 standard deviations from themean, so for ACT scores, they are 21.5 ± 0.6745 × 5.4 = 17.9 to 25.1.1.141Find the SAT quintiles. The quintiles of any distribution are the values with cumulative proportions 0.20, 0.40, 0.60, and 0.80. What are the quintiles of the distribution of SAT scores?Solution1.141. The quintiles of the SAT score distribution are 1509 − 0.8416 × 321 = 1239,1509−0.2533×321 = 1428, 1509+0.2533×321 = 1590, and 1509+0.8416×321 = 1779. 1.142Do you have enough “good cholesterol?” High-density lipoprotein (HDL) is sometimes called the “good cholesterol” because low values are associated with a higher risk of heart disease. According to the American Heart Association, people over the age of 20 years should have at least 40 mg/dL of HDL cholesterol.36 U.S. women aged 20 and over have a mean HDL of 55 mg/dL with a standard deviation of 15.5 mg/dL. Assume that the distribution is Normal.(a)What percent of women have low values of HDL (40 mg/dL or less)?(b)HDL levels of 60 mg/dL are believed to protect people from heart disease. What percentof women have protective levels of HDL?(c)Women with more than 40 mg/dL but less than 60 mg/dL of HDL are in the intermediaterange, neither very good or very bad. What proportion are in this category?Solution1.142. For a Normal distribution with mean 55 mg/dl and standard deviation 15.5 mg/dl:(a) 40 mg/dl standardizes to z = (40−55)/15.5 = −0.9677. Using Table A, 16.60% ofwomen fall below this level (software: 16.66%).(b) 60 mg/dl standardizes to z = (60−55)/15.5 = 0.3226. Using Table A, 37.45(c) Subtract the answers from (a) and (b) from 100%: Table A gives 45.95% (software:45.99%), so about 46% of women fall in the intermediate range.1.143Men and HDL cholesterol. HDL cholesterol levels for men have a mean of 46 mg/dL with a standard deviation of 13.6. Answer the questions given in the previous exercise for the population of men.Solution1.143. For a Normal distribution with mean 46 mg/dl and standard deviation 13.6 mg/dl:(a) 40 mg/dl standardizes to z = (40−46)/13.6 = −0.4412. Using Table A, 33% of men fallbelow this level (software: 32.95%).(b) 60 mg/dl standardizes to z = (60−46)/13.6 = 1.0294. Using Table A, 15.15(c) Subtract the answers from (a) and (b) from 100%: Table A gives 51.85% (software:51.88%), so about 52% of men fall in the intermediate range.1.145Length of pregnancies. The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days.(a)What percent of pregnancies last less than 240 days (that’s about 8 months)?(b)What percent of pregnancies last between 240 and 270 days (roughly between 8 monthsand 9 months)?(c)How long do the longest 20% of pregnancies last?Solution1.145. (a) About 5.2%: x < 240 corresponds to z < −1.625. Table A gives 5.16% for−1.63 and 5.26% for −1.62. Software (or averaging the two table values) gives 5.21%.(b) About 54.7%: 240 < x < 270 corresponds to −1.625 < z < 0.25. The area to theleft of 0.25 is 0.5987; subtracting the answer from part (a) leaves about 54.7%.(c) About 279 days or longer: Searching Table A for 0.80 leads to z > 0.84, whichcorresponds to x > 266 + 0.84(16) = 279.44. (Using the software value z > 0.8416 gives x > 279.47.)1.146Quartiles for Normal distributions. The quartiles of any distribution are the values with cumulative proportions 0.25 and 0.75.(a)What are the quartiles of the standard Normal distribution?(b)Using your numerical values from (a), write an equation that gives the quartiles of theN(µ, σ) distribution in terms of µ and σ.(c)The length of human pregnancies from conception to birth varies according to adistribution that is approximately Normal with mean 266 days and standard deviation 16 days. Apply your result from (b): what are the quartiles of the distribution of lengths of human pregnancies?Solution1.146. (a) The quartiles for a standard Normal distribution are ±0.6745.(b) For a N(µ, σ) distribution, Q1 = µ − 0.6745σ and Q3 = µ + 0.6745σ.(c) For human pregnancies,Q1 = 266 − 0.6745 × 16 .=255.2 and Q3 = 266 + 0.67455 ×16 =276.8 days.1.147IQR for Normal distributions. Continue your work from the previous exercise. The interquartile range IQR is the distance between the first and third quartiles of a distribution.(a)What is the value of the IQR for the standard Normal distribution?(b)There is a constant c such that IQR = cσ for any Normal distribution N(µ, σ). What is thevalue of c?Solution1.147. (a) As the quartiles for a standard Normal distribution are ±0.6745,we have IQR = 1.3490.(b) c = 1.3490: For a N(µ, σ) distribution, the quartiles are Q1 = µ − 0.6745σ andQ3 = µ + 0.6745σ.1.148Outliers for Normal distributions. Continue your work from the previous two exercises. The percent of the observations that are suspected outliers according to the 1.5 × IQR rule is the same for any Normal distribution. What is this percent?Solution1.148. In the previous two exercises, we found that for a N(µ, σ) distribution,Q1 = µ − 0.6745σ, Q3 = µ + 0.6745σ, and IQR = 1.3490σ.Therefore, 1.5 × IQR = 2.0235σ, and the suspected outliers are belowQ1 − 1.5 × IQR = µ − 2.698σ, and above Q3 + 1.5 × IQR = µ + 2.698σ.The percentage outside of this range is 2 × 0.0035 = 0.70%.1.149Deciles of Normal distributions. The deciles of any distribution are the 10th, 20th,…, 90th percentiles. The first and last deciles are the 10th and 90th percentiles, respectively.(a)What are the first and last deciles of the standard Normal distribution?(b)The weights of 9-ounce potato chip bags are approximately Normal with mean 9.12ounces and standard deviation 0.15 ounce. What are the first and last deciles of thisdistribution?Solution1.149. (a) The first and last deciles for a standard Normal distribution are ±1.2816.(b) For a N(9.12, 0.15) distribution, the first and last deciles are µ − 1.2816σ= 8.93 andµ + 1.2816σ = 9.31 ounces.。
人教版新目标八年级英语上册各单元知识详解
Detailed Solution for Unit 2 What’s the matter.【重要词汇】◆ matter/5mAt[/n.事情,问题,过失◆ arm/B:m/n.臂,胳膊◆ back/bAk/n.后背,背脊◆ ear/I[/n.耳朵◆ eye/aI/n.眼睛◆ foot/fJt/n.脚,足〔pl. feet/fi:t/〕◆ leg/leg/n.腿,腿部◆ neck/nek/n.脖子,颈部◆ nose/n[Jz/n.鼻子◆ stomach/5stQm[k/n.胃,胃部◆ tooth/tU:W/n.牙齿(pl. teeth/ti:W/)◆ sore/sC:,sC[/adj.疼痛的◆ stomachache/5stQm[keIk/n.胃痛,腹痛,肚子痛◆ throat/Wr[Jt/n.喉头,喉咙◆ toothache/5tU:WeIk/n.牙痛◆ fever/5fi:v[/n.发烧◆ lie/laI/v.躺,卧,平放在某处(lay, lain)◆ honey/5hQnI/n.蜂蜜◆ dentist/5dentIst/n.牙医◆ illness/5IlnIs/n.疾病,生病◆ advice/[d5vaIs/n.劝告,忠告,建议◆ thirsty/5W\:stI/adj.渴的,口渴的◆ heal/hi:l/v. 医治,治愈◆ balance/5bAl[ns/n. 平衡,平衡状态,协调◆ beef/bi:f/n. 牛肉◆ lamb/lAm/n. 小羊,羊羔,羊羔肉◆ energy/5en[dVI/n.精力,活力,能力◆ life/laIf/n. 生命,人生◆ bean/bi:n/n.豆,豆科植物◆ stray/streI/v. 远离,走失,离题◆ balanced/5bAl[nst/adj.平衡的,协调的◆ diet/5daI[t/n. 饮食,节食◆ backache/5bAkeIk/n.背痛◆ improve/Im5prU:v/v. 改善,改进,提高◆ speaking/5spi:kIN/ n. 口语◆ headache/5hedeIk/ n. 头痛◆ homestay/5h[JmsteI/ n. 待在家里◆ important//Im5pC:t[nt/adj. 重要的,重大的,严重的【重要词组】◆ have a cold患感冒◆ stressed out 紧张的,有压力的◆ bean sprout豆芽◆ get tired感觉疲惫◆ stay healthy保持健康◆ at the moment此刻,现在◆ get/ have a cold患感冒◆ see a doctor /dentist看医生/牙医◆ go to the party去参加聚会◆ make sb sick使某人不舒服(患锁病)◆ have a sore throat嗓子痛◆ have a fever发烧,发热◆ have a toothache牙痛◆ have a backache背痛◆ lie down and rest 躺下休息◆ drink lots of water 喝大量水◆ drink hot tea with honey 喝热蜂蜜茶◆ have a headache头痛◆ get tired 累了【重要词汇扩展】身体部位:head头neck颈部shoulder肩back后背leg腿thigh大腿arm手臂hand手nose 鼻子eye眼睛ear耳朵mouth嘴tooth牙齿foot脚〔注意foot的复数形式为feet,tooth的复数形式为teeth〕stomach胃throat喉咙,咽喉chest胸部pit胸口abdomen lung肺heart心脏liver 肝脏腹部waist腰backbone脊骨,脊柱hip臂部joint关节muscle肌肉nerve神经brain 脑疾病:cancer癌 influenza , flu流感 cough咳嗽sore eyes眼睛疼stuffy nose鼻子不通气 insomnia失眠医院工作人员:doctor in charge主治医生 head nurse护士长internist , physician内科医生surgeon外科医生dentist牙科医生 ear-nose-throat doctor 耳鼻喉医生医院名称:children’s hospital儿童医院 clinic 诊疗所first - aid station 急救站 ward 病房medical department 内科 surgical department 外科registration office 挂号处 out - patient department(OPD) 门诊部in - patient department 住院部 nursing department 护理部waiting room 候诊室 emergency room 急诊室operation room 手术室 laboratory 化验室blood bank 血库 pharmacy , dispensary 药房表示感觉的形容词有:tired 累的 thirsty口渴的 hungry饿的 stressed out紧张的,有压力的【重要词句详解】1. How to talk about our health①问某人哪儿不舒服:When we are not feeling well , we often go to see the doctor . The doctor will ask : What’s wrong (with you) ?What’s the matter (with you) ?What’s your trouble ?What happens to you ?Is there anything wrong with you ?都表示“你怎么了?”。
线性控制系统工程课后习题答案Solution of module (2)
线性控制系统工程课后习题答案Module 1: Introduction to Linear Control SystemsExercise 1QuestionConsider a control system with transfer function:$$ G(s) = \\frac{10}{s+2} $$Find the steady-state value of the output when the input is a unit step function.SolutionThe Laplace transform of a unit step function is given by:$$ U(s) = \\frac{1}{s} $$The steady-state value of the output can be obtained by evaluating the transfer function at s=0:$$ G(s=0) = \\frac{10}{s+2} = \\frac{10}{2} = 5 $$Therefore, the steady-state value of the output is 5 when the input is a unit step function.Exercise 2QuestionConsider a control system with transfer function:$$ G(s) = \\frac{4}{s^2 + 4s + 3} $$Find the poles and zeros of the transfer function.SolutionThe poles are the roots of the denominator polynomial. Therefore, we need to find the roots of the equation:s2+4s+3=0Using the quadratic formula, we can find the roots:$$ s = \\frac{-b \\pm \\sqrt{b^2-4ac}}{2a} $$In this case, a=1, b=4, and c=3. Substituting these values into the formula, we get:$$ s_{1,2} = \\frac{-4 \\pm \\sqrt{4^2-4(1)(3)}}{2(1)} $$ Simplifying this expression, we get:s1=−1s2=−3Therefore, the poles of the transfer function are -1 and -3.To find the zeros, we need to find the roots of the numerator polynomial, which in this case is a constant. Since there are no variables in the numerator, there are no zeros. Therefore, the transfer function has no zeros.Module 2: System ModelingExercise 1QuestionConsider a control system described by the following differential equation:$$ 4\\frac{d^2y(t)}{dt^2} + 8\\frac{dy(t)}{dt} + 6y(t) =2u(t) $$Determine the transfer function of the system.SolutionTo find the transfer function, we need to take the Laplace transform of both sides of the differential equation. The Laplace transform of the left side can be written as: $$ \\mathcal{L}\\left(4\\frac{d^2y(t)}{dt^2} +8\\frac{dy(t)}{dt} + 6y(t)\\right) $$Using the linearity property of the Laplace transform, we can split this expression as:$$ 4\\mathcal{L}\\left(\\frac{d^2y(t)}{dt^2}\\right) +8\\mathcal{L}\\left(\\frac{dy(t)}{dt}\\right) +6\\mathcal{L}\\left(y(t)\\right) $$Taking the Laplace transform of the right side, we get:$$ \\mathcal{L}(2u(t)) = 2\\mathcal{L}(u(t)) $$The Laplace transform of the input function u(t) is denoted as U(s).Therefore, the differential equation in the Laplace domain can be expressed as:4s2s(s)−4ss(0)−4s′(0)+8ss(s)−8s(0)+6s(s)=2s(s)Rearranging this equation, we get:(4s2+8s+6)s(s)=4ss(0)+4s′(0)+8s(0)+2s(s) Dividing both sides of the equation by (4s^2 + 8s + 6), we can isolate Y(s):$$ Y(s) = \\frac{4sy(0) + 4y'(0) + 8y(0) + 2U(s)}{4s^2 + 8s + 6} $$Therefore, the transfer function of the system is given by: $$ G(s) = \\frac{Y(s)}{U(s)} = \\frac{4s + 4}{4s^2 + 8s + 6} $$Exercise 2QuestionConsider a control system described by the following difference equation:s(s+2)+s(s+1)+s(s)=s(s)+s(s−1)Determine the transfer function of the system.SolutionTo find the transfer function, we need to take the Z-transform of both sides of the difference equation. The Z-transform of the left side can be written as:$$ Z\\{y(k+2) + y(k+1) + y(k)\\} $$Using the linearity property of the Z-transform, we can split this expression as:$$ Z\\{y(k+2)\\} + Z\\{y(k+1)\\} + Z\\{y(k)\\} $$Taking the Z-transform of the right side, we get:$$ Z\\{u(k) + u(k-1)\\} = Z\\{u(k)\\} + Z\\{u(k-1)\\} $$ The Z-transform of the input function u(k) is denoted asU(z).Therefore, the difference equation in the Z-domain can be expressed as:s2s(s)−s2s(0)−ss(0)+ss(s)−s(0)+s(s)=s(s)+ss(s)Rearranging this equation, we get:(s2+s+1)s(s)=(s2+s)s(0)+s(0)+s(s)Dividing both sides of the equation by (z^2 + z + 1), we can isolate Y(z):$$ Y(z) = \\frac{(z^2 + z)y(0) + y(0) + U(z)}{z^2 + z + 1} $$Therefore, the transfer function of the system is given by:$$ G(z) = \\frac{Y(z)}{U(z)} = \\frac{(z^2 + z)y(0) +y(0)}{z^2 + z + 1} $$Module 3: Feedback Control SystemsExercise 1QuestionConsider a feedback control system with a transfer function G(s) = 10/(s+5).Determine the closed-loop transfer function for the system when the controller is a proportional controller with gain K.In a feedback control system, the closed-loop transfer function can be computed by combining the transfer function of the controller (C(s)) and the transfer function of the plant (G(s)).For a proportional controller, the transfer function is given by:s(s)=sTherefore, the closed-loop transfer function (T(s)) can be expressed as:$$ T(s) = \\frac{C(s) \\cdot G(s)}{1 + C(s) \\cdot G(s)} $$ Substituting the values of C(s) and G(s), we get:$$ T(s) = \\frac{K \\cdot \\frac{10}{s+5}}{1 + K \\cdot \\frac{10}{s+5}} $$Simplifying this expression, we get:$$ T(s) = \\frac{10K}{s+5+10K} $$Therefore, the closed-loop transfer function for the system with a proportional controller is T(s) = 10K / (s+5+10K).QuestionConsider a feedback control system with a transfer function G(s) = 1/(s+2).Determine the closed-loop transfer function for the system when the controller is a proportional-integral (PI) controller with gains Kp and Ki.SolutionIn a feedback control system, the closed-loop transfer function can be computed by combining the transfer function of the controller (C(s)) and the transfer function of the plant (G(s)).For a proportional-integral (PI) controller, the transfer function is given by:$$ C(s) = Kp + \\frac{Ki}{s} $$Therefore, the closed-loop transfer function (T(s)) can be expressed as:$$ T(s) = \\frac{C(s) \\cdot G(s)}{1 + C(s) \\cdot G(s)} $$ Substituting the values of C(s) and G(s), we get:$$ T(s) = \\frac{(Kp + \\frac{Ki}{s}) \\cdot\\frac{1}{s+2}}{1 + (Kp + \\frac{Ki}{s}) \\cdot \\frac{1}{s+2}} $$Simplifying this expression, we get:$$ T(s) = \\frac{Kp + \\frac{Ki}{s}}{s^2 + (2Kp+1)s +(2Ki+2)} $$Therefore, the closed-loop transfer function for the system with a proportional-integral (PI) controller is T(s) = (Kp + Ki/s) / (s^2 + (2Kp+1)s + (2Ki+2)).ConclusionThese are the solutions to the selected exercises in the module on linear control system engineering. These solutions cover topics such as transfer functions, poles and zeros, system modeling, and feedback control systems. By understanding these concepts and practicing the exercises, you can enhance your knowledge and skills in the field of linear control systems.。
通过运动改善睡眠英语作文
运动与睡眠:改善睡眠质量的秘密武器 In today's fast-paced world, where stress and anxiety are constant companions, it's becoming increasingly difficult to maintain a healthy sleep schedule. However, there's one simple yet effective solution that can help us improve our sleep quality: regular physical activity. Exercise not only benefits our physical health but also plays a crucial role in promoting better sleep.The connection between exercise and sleep is bidirectional. On one hand, regular exercise can help improve sleep quality by reducing stress, anxiety, and depression. It also raises the body's temperature, which then drops after exercise, triggering a feeling of sleepiness. On the other hand, poor sleep can negatively impact our ability to exercise, as lack of sleep can lead to decreased motivation, energy, and performance.There are several types of exercise that can be beneficial for improving sleep. Aerobic exercise, such as walking, running, or cycling, can help raise the body's temperature and promote deep sleep. Strength training, on the other hand, can increase muscle mass and improve sleepquality by reducing stress and anxiety. Yoga and tai chi are also excellent choices for improving sleep, as they focus on breathing and relaxation techniques that can help calm the mind and promote better sleep.The timing of exercise is also crucial for improving sleep. While exercising too close to bedtime may stimulate the body and make it difficult to fall asleep, exercising earlier in the day can help reduce stress and anxiety, thus promoting better sleep at night. Additionally, engaging in light exercise before bedtime, such as stretching or walking, can help relax the body and mind, making it easier to fall asleep.However, it's important to note that not all exerciseis suitable for everyone. Some people may find that certain types of exercise, such as intense cardio workouts or strength training, may actually interfere with their sleep. Therefore, it's essential to find the right type and intensity of exercise that works best for your body and sleep schedule.In conclusion, regular physical activity is a powerful tool for improving sleep quality. By reducing stress,anxiety, and depression, raising body temperature, and promoting relaxation, exercise can help us achieve better sleep and overall health. However, it's crucial to find the right type and timing of exercise that suits our individual needs and sleep preferences. With the right approach, wecan harness the power of exercise to enjoy deeper, more restful sleep and improved overall well-being.**运动与睡眠:改善睡眠质量的秘密武器**在如今快节奏的社会中,压力和焦虑成为了我们形影不离的伴侣,维持健康的睡眠习惯变得越来越困难。
【北美精算师考试】lossmodels
Errata and clarifications for Loss Models: From Data to DecisionsPage 32, at the end of the line prior to the formula at the bottom of the page, add the footnote: The tilde on the indicates that this is an approximation due to grouping.I Page 44, line 6: Replace "valuet" with "value"Page 67, Theorem 2.3: The coveriance matrix in the third to last line of the statement of the Theorem must be divided by to read .n Ð`ÑÐ`ÑÎ8g g w D Page 68, Example 2.27: Only the first factor in the third term on line 2 of the string of equations has a zero expectation. This is sufficient for that term to be zero. The secondterm is only approximately zero because is only asymptotically unbiased./.5s s Î##Page 74, Corollary 2.6: Change part of it to read "The expected payment, per payment,provided 1, is"J Ò.ÎÐ" <ÑÓ Page 75, Definition 2.28: In the last line, replace "" with "";and exist exists IÐ\•.Ñand in the second to last paragraph: In the first line, replace "moments" with "firstmoment"; in the third line replace "could be either" with "is"; in the fourth and fifth lines,delete "or because does not converge. If the second limit exists,lim .Ä!.'B0ÐBÑ.B IÐ\•BÑ will still exist."; and in the seventh, eighth, and ninth lines, delete "If the first limit exists but the second does not, both and will fail to exist, but the IÐ\ÑIÐ\•BÑdifference will."Page 76, Definition 2.29: After "it is" add ", provided 1,"J Ð.Ñ \Page 76, line 5: Replace "due to having a heavy tail," with "(for example, if the \distribution is heavy-tailed)"Page 94, Definition 2.34: It should be noted that must be positive.!Page 137, second line of Example 2.74: Replace "This data are" with "This data set is"Page 165, Exercise 2.44: The last sentence should be: "Keep the two samples separate so that (2.14) can be used. Then construct a 95% confidence interval for the mean."Page 168, line 1 of Exercise 2.67: Replace "gamma" with "inverse Weibull"Page 190, Table 2.38: The fifth row of the table should be 4–13 and not 5–13.Page 199, Introduction: It should be made clear that throughout Chapter 3 all random variables have the non-negative integers as the set of possible values.Page 205, near the bottom: It should be noted that .8œ8D 5œ!_5Page 206, last line: Replace "numbers of claims per day" with "number of days with a given number of claims"Page 210, Definition 3.2: It should be noted that the definition that uses the gamma function applies only if 1.B 5 Page 241, last line: Replace "" with ""T ÐDÑT ÐDÑX X #Page 257, In Definition 3.6, it should be noted that refers to the square root of .3 "Page 258, In the last line of Equation (3.38), the exponent in the expectation should be an uppercase theta () rather than lower case.@Page 264, first paragraph after the end of Exmaple 3.33: Note that while the binomial distribution is not infinitely divisible,if is an integer, then the binomial 78Î8‡distribution is preserved.Page 265, just before Example 3.35: Note that the "number of data points" is the number of years, not the number of exposures.Page 267, ZM Poisson entry in Tble 3.21: The greek (nu) in the third term in the /numerator should be the letter (vee).@Page 271, Exercise 3.5: The information referred to is in the table at the top of the page.Page 279, Exercise 3.34: Add a condition of independence. That is, let have W 3independent compound Poisson frequency distributions...Page 296, At the end of equation (4.4) add for Also, on the following line 5œ"ß#ßáÞchange the text to "with probability zero on negative values." Finally, the equation after (4.4) holds only for For , this equation reduces to .5œ#ß$ßáÞ5œ"0ÐBÑœ0ÐBÑ\‡"\The same holds for the equation at the top of page 297. At the bottom of page 296,change "with probabilities at" to "with positive probabilities only at"Page 308, in the footnote, change the expression on the second to last line toT <Ð\œC Ñ !3.Page 316, add the following after line 6: "A reason to favor matching zero or onemoment is that the resulting probabilities will always be non-negative. When matching two or more moments, this cannot be guaranteed."Page 358, Exercise 4.5: Change the standard deviation of "Room" from 300 to 500.Page 361, Exercise 4.18: In the second table, the entries under should be 1 and 10, notB0 and 10.Page 363, Exercise 4.24: There should not be an exercise number here. This paragraph is the conclusion to Exercise 4.23. To match the answers in the solutions manual, all subsequent exercises in this chapter should be numbered one less.policyholder,where n is the Page 413, Example 5.19: In the second line, after "" insert "number of exposure units,"Page 417, First paragraph of solution to Example 5.23: Replace from the sentence0œQœbeginning "Now 175" to the end of the solution with "Now 175 and we assume8œœthat will be 140 for this group. Thus, with 210 and 1,082.41 we use (5.38) 5-!0\œwith estimated by 150 to obtainÈ1501402101,082.410.472^œÐÎÑÎœ\(note that is the average of 210 claims, so approximate normality is assumed by the central limit theorem). Thus the net premium per life insured is0.4721500.528175163.2.TœÐÑÐÑ ÐÑÐÑœ-The net premium for the whole group is 125163.220,400.ÐÑœÐÑœÐÑœPage 449, after (5.68): The assumption must be changed to 0.1)1)!"Page 461, line three of third paragraph: Replace "by" with "be"Page 468, paragraph before Example 5.47: It reads better if the second and third to last sentences are switched.Page 488, Exercise 5.44: Add "expected" before "number of claims" in the second to last line., Exercise 5.47(c) and , Exercise 5.48(c): In both cases add "for each Page 489page 490)" at the end.Page 491, Exercise 5.50: In line 5 change "mean" to "means" and "variance" to "variances"1)1)ÐÑœÐÑœPage 493, Exercise 5.56(f): Change the requirement to 0.!"Page 493, Exercise 5.57(b): Change the left hand side to IÐ\l\ â \œ=Ñ8 ""8Page 495, Exercise 5.67: On the second line change the decimals to the exact fractions 1/6 and 5/36 and on the fourth line change the decimals to the exact fractions 5/6 and5/36.Page 496, Exercise 5.68: Change the Bühlmann numbers to 2.72, 7.71, and 10.57.Page 496, Exercise 5.67(b): Place a 0 before the decimal point in .25.Page 497, Exercise 5.73: There is an extra | in a subscript in the second bullet.page 499, Exercise 5.79: In the third-fifth bullets, and should be upper case to .-indicate that they are random variables.Page 510, first number in the column headed by "4": Change "1 ,441" to "1,441"Page 609, last line: Add an "" after ""2Ð4 "Î#ÑPage 610, last line: Add an "" after ""2Ð4 "Î#ÑPage 614, line 4 of section H.4: Replace "" with ""X +X +88Page 620, answer to Exercise 3.19(a), change the answer toFor 1,2,3,4,5,6,7the values are 0.276, 2.315, 2.432, 2.891, 4.394, 2.828, and 4.268,5œwhich, if anything, are increasing. The negative binomial or geometric model may work well.Page 622, answer to Exercise 4.19, the difference is 14.78.Page 626, answer to Exercise 5.68 should be expressed as 1210.6.IÐ\l\œÑœ#"Errata for the Solutions Manual to Loss Models: From Data to DecisionsExercise 2.3, page 2: The expression in parentheses in the equation for should be <Ð. -. -Ñ##.Exercise 2.6, page 3: The denominator should be 96,000 and not 60,000. The answer is correct.Exercise 2.39, page 10: All of the ratios of with a subscript over should be have the -)) in the numerator and the subscripted in the denominator. The answer is correct.-Exercise 2.45, page 14: The information matrix isMÐßÑœs s !"”•1.707900.01668610.01668610.000176536.Exercise2.67, page 20: Change the solution to read as follows:The parameter estimates are 0.963933 and 94.3501.7)s œœs IÐ\•ÑœÒ" Ð ÑÓ10,00094.35010.0374165; 0.011163> Ò" /B:Ð ÑÓœ10,0000.011163516.09.Similarly, 5,000440.28 and the answer is .8516.09440.2860.648.IÐ\•ÑœÐ ÑœThe per payment expected value is60.648500060.6480.021542,815.60.ÎÒ" J ÐÑӜΜExercise 2.115(b), page 37: In the expression 3510 in the final ratio, replace the exp Ð ÎÑ10 with 20. The answer is correct.Exercise 2.118, page 37: In the last line, change the "" to a comma.Q Exercise 3.6(c), page 44: The test statistic is 0.23 2.73 2.60 5.56.œExercise 3.10, page 48: The solution is 0.471494 and 0.352072.!)"œ<œœœExercise 3.12, page 48: Both (b) parts have the confidence interval missing. The first one is 0.1001 1.960.000011012 or 0.10010.0065 and the second one is …ÐÑ…"Î#0.166 1.960.000193556 or 0.1660.027.…ÐÑ…"Î#Exercise 3.19(a), page 53: Replace the solution with the following:For 1,2,3,4,5,6,7the values are 0.276, 2.315, 2.432, 2.891, 4.394, 2.828, and 4.268,5œwhich, if anything, are increasing. The negative binomial or geometric model may work well.Exercise 3.47, page 68: Replace the last sentence with the following:Because the variance is , the goodness-of-fit test statistic equals the Poisson /Ð" Ñ5""test statistic divided by or 6.191.09772 5.64. The geometric model is accepted." Îœ"Exercise 3.49, page 68: The first paragraph of the correct solution is:@œÒ" J ÐÑÓÎÒ" J ÐÑÓœÎÞ1000500916 The original frequency distribution is P-ETNB with 3, 0.5, and 2. The new frequency distribution is P-ZMNB-"œ<œ œwith 3, 0.5, 291698, and 0.37472.-"œ<œ œÐÎÑœÎ:œœQ !!œ$ Ð"(Î)Ñ !Ð"(Î)Ñ" $Þ&Þ&Þ&Þ&This is equivalent to a P-ETNB distribution with 310.37472 1.87584,-œÐ Ñœ<œ œÎœœÎ0.5, and 98, which is P-IG with 1.87584 and 98."-"In the first line of the second paragraph, replace the equals sign in the numerator with a minus sign.Exercise 4.14, page 75: All of the + signs should be subscripts, not superscripts. The letter should be used in place of for the random variable.\W Exercise 4.16, page 76: 505025 and the solution is 2/3.IÐEÑœ5 55œ "Exercise 4.19, page 77: The second line for the variance should be1061265,000. The premium is 100514.18 for a difference of 14.78.7;Ð ;Ñœ7; #Exercise 4.22, page 77: The third term in the sum should be 0.1Pr .Š‹^ 10030027 ÈExercise 4.24 and beyond in chapter 4: In the text the problem numbers are one higher than the number given here.Exercise 4.25, page 79: The final equation is0.9Pr 1.53431 1.534. The answer isœÒW Ÿ ÎÓœB .B œ Ò ÎÓ--ÈÈ'""Þ& $Î% %-Ècorrect.Exercise 4.62, page 90: The solution is correct, but the notation is misleading. ChangeJ ÐBÑJ ÐBÑ8J ‡8W 8W ‡Ð8 "Ñ to to reflect the distribution using all policies and change toJ 8 "8 " to reflect the distribution using only the first policies.Exercise 5.22, page 106: In answer (c) the integral is 13, which is 0.35. Therefore,Î only (a) is possible.Exercise 5.38, page 111: In Table 4.4, the first entry in the column headed "2" should be 0.0625.Exercise 5.38, page 112: In part (h) the last four probabilities should have the events separated by commas and not by bars. Also, two of the probabilities need to be changed:Pr 0.118329 and Pr 3 0.046367. In part (i), the Ð\œ#ß\œ#ÑœÐ\œß\œ#Ñœ#"#"probability to divide by is 0.343333.Exercise 5.50, page 114: 0.050.125,0000.11001,400.+œÒ ÐÑÓÐÑ ÐÑÐÑœ=###5œÎœ^œÎÐ Ñœ5,0001,400 3.5714, 33 3.57140.4565,T œÐÎÑ ÐÑœ-0.456520030.54351035.87.Exercise 5.56(d), page 119: Add horizontal space between (d) and (e).Exercise 5.58(b), page 122: Need a minus sign inZ ÐÑœÐ Ñ Ðœ2123456 3.5) 2.917.#######Exercise 5.59(b), page 122: Because the support of the prior distribution is 1, that is ) also the support of the posterior distribution. Therefore, it is not a gamma distribution.The constant to make it a density is'’“"_ #"' # #""'"'x #%)#"'Ð"&Ñ/.œ/ â œ/)))"(1,179,501,863. The posteriormean is then '’“"_ #" #"""x #%)#"Ð"Ñ/.Î/œ âÎ)))777761,179,501,8631,179,501,863 which"8is 10,025,765,8381,179,501,8638.5. (Actually, the ratio is slightly less than 8.5.)ÎœExercise 5.67, page 125: For part (a), because 56253, other changes .ÐFÑœÐÎÑÐÑœÎare 76, 1/4, 1859, and 36221. For part (b),.œÎ+œ5œÎ^œÎÐÎÑÐÑ ÐÎÑÐÎќΜ362210.25185221761,3491,326 1.01735.Exercise 5.68, page 125: Write 12 two times in place of .IÐ\l\œÑIÐ\l\œX Ñ#"#"Exercise 5.84, page 133: In the expression for the term in parentheses should be +s divided by 2 and the number 3263should not be squared. The answer is correct.)*Exercise 5.85, page 134: The formula for is correct, however the value is 22,401.00.@s Substituting this correct value into the rest of the development produces 617.54,+œs 5œ^œ^œ^œs 36.27, 0.8585, 0.8663, 0.9330. The three estimates are then"#$203.61, 225.53, and 181.39. For the alternative method, the new values must be used ^and so 204.32 and the three estimates are 204.50, 226.37, and 181.81..s œExercise 5.86, page 135: Change the latter part to read 5,62513,7500.4091,5œÎœs ^œÎœ œ33.40910.8800. The premium is 0.8800(475)0.1200(600)490.。
第一章概率论习题解答
教 案概率论与数理统计(Probability Theory and Mathematical Statistics )Exercise 1.1 向指定目标射三枪,观察射中目标的情况。
用1A 、2A 、3A 分别表示事件“第1、2、3枪击中目标”,试用1A 、2A 、3A 表示以下各事件:(1)只击中第一枪;(2)只击中一枪;(3)三枪都没击中;(4)至少击中一枪。
Solution (1)事件“只击中第一枪”,意味着第二枪不中,第三枪也不中。
所以,可以表示成 1A 32A A 。
(2)事件“只击中一枪”,并不指定哪一枪击中。
三个事件“只击中第一枪”、“只击中第二枪”、“只击中第三枪”中,任意一个发生,都意味着事件“只击中一枪”发生。
同时,因为上述三个事件互不相容,所以,可以表示成 123A A A +321A A A +321A A A .(3)事件“三枪都没击中”,就是事件“第一、二、三枪都未击中”,所以,可以表示成 123A A A .(4)事件“至少击中一枪”,就是事件“第一、二、三枪至少有一次击中”,所以,可以表示成 321A A A 或 123A A A +321A A A +321A A A +1A 32A A +321A A A +321A A A + 321A A A .Exercise 1.2 设事件B A ,的概率分别为21,31 .在下列三种情况下分别求)(A B P 的值:(1)A 与B 互斥;(2);B A ⊂ (3)81)(=AB P .Solution 由性质(5),)(A B P =)()(AB P B P -.(1) 因为A 与B 互斥,所以φ=AB ,)(A B P =)()(AB P B P -=P(B)=21 (2) 因为;B A ⊂所以)(A B P =)()(AB P B P -=)()(A P B P -=613121=-(3) )(A B P =)()(AB P B P -=838121=- Exercise 1.3 一袋中有8个大小形状相同的球,其中5个黑色球,三个白色球。
chap15-solutions
Solution to Exercise 15.2-5Each time the l -loop executes,the i -loop executes n l C 1times.Each time the i -loop executes,the k -loop executes j i D l 1times,each time referencing m twice.Thus the total number of times that an entry of m is referenced while computing other entries is P n l D 2.n l C 1/.l 1/2.Thus,n X i D 1n X j D i R.i;j /Dn X l D 2.n l C 1/.l 1/2D 2n 1X l D 1.n l/l D 2n 1X l D 1nl 2n 1X l D 1l 2D 2n.n 1/n 6D n 3 n 22n 3 3n 2C n 3:15-2Selected Solutions for Chapter 15:Dynamic Programmingsubproblem results is thus the product of the number of ways to do the left half and the number of ways to do the right half. For each possible place to split the matrix chain,R ECURSIVE -M ATRIX -C HAINfinds the best way to parenthesize the left half,finds the best way to parenthesize the right half,and combines just those two results.Thus the amount of work to combine the left-and right-half subproblem results is O.1/.Section 15.2argued that the running time for enumeration is .4n =n 3=2/.We will show that the running time for R ECURSIVE -M ATRIX -C HAIN is O.n3n 1/.To get an upper bound on the running time of R ECURSIVE -M ATRIX -C HAIN ,we’ll use the same approach used in Section 15.2to get a lower bound:Derive a recur-rence of the form T.n/ :::and solve it by substitution.For the lower-bound recurrence,the book assumed that the execution of lines 1–2and 6–7each take at least unit time.For the upper-bound recurrence,we’ll assume those pairs of lines each take at most constant time c .Thus,we have the recurrenceT.n/ c if n D 1;c C n 1X k D 1.T.k/C T.n k/C c/if n 2:This is just like the book’s recurrence except that it has c instead of 1,and so we can be rewrite it asT.n/ 2n 1X i D 1T.i/C cn :We shall prove that T.n/D O.n3n 1/using the substitution method.(Note:Any upper bound on T.n/that is o.4n =n 3=2/will suffice.You might prefer to prove one that is easier to think up,such as T.n/D O.3:5n /.)Specifically,we shall show that T.n/ cn3n 1for all n 1.The basis is easy,since T.1/ c D c 1 31 1.Inductively,for n 2we haveT.n/ 2n 1X i D 1T.i/C cn2n 1X i D 1ci3i 1C cnc 2n 1X i D 1i3i 1C n !D c 2 n3n 1.3 1/2 C n (see below)D cn3n 1C c1 3n 2.2n C 1 3n / cn3n 1for all c >0,n 1.Selected Solutions for Chapter15:Dynamic Programming15-3 Running R ECURSIVE-M ATRIX-C HAIN takes O.n3n 1/time,and enumerating all parenthesizations takes .4n=n3=2/time,and so R ECURSIVE-M ATRIX-C HAIN is more efficient than enumeration.Note:The above substitution uses the following fact:n 1X i D1ix i 1D nx n 1.x 1/2:This equation can be derived from equation(A.5)by taking the derivative.Letf.x/D n 1X i D1x i D x n 1x 1C1 x nSolution to Exercise15.4-4When computing a particular row of the c table,no rows before the previous roware needed.Thus only two rows—2 Y:length entries—need to be kept in memoryat a time.(Note:Each row of c actually has Y:length C1entries,but we don’t needto store the column of0’s—instead we can make the program“know”that thoseentries are0.)With this idea,we need only2 min.m;n/entries if we always callLCS-L ENGTH with the shorter sequence as the Y argument.We can thus do away with the c table as follows:Use two arrays of length min.m;n/,pre ious-row and current-row,to hold the appropriate rows of c.Initialize pre ious-row to all0and compute current-row from left to right.When current-row isfilled,if there are still more rows to compute,copy current-row into pre ious-row and compute the new current-row.Actually only a little more than one row’s worth of c entries—min.m;n/C1en-tries—are needed during the computation.The only entries needed in the tablewhen it is time to compute cŒi;j are cŒi;k for k j 1(i.e.,earlier entries inthe current row,which will be needed to compute the next row);and cŒi 1;k fork j 1(i.e.,entries in the previous row that are still needed to compute the restof the current row).This is one entry for each k from1to min.m;n/except thatthere are two entries with k D j 1,hence the additional entry needed besides theone row’s worth of entries.We can thus do away with the c table as follows:Use an array a of length min.m;n/C1to hold the appropriate entries of c.At the time cŒi;j is to be computed,a will hold the following entries:aŒk D cŒi;k for1 k<j 1(i.e.,earlier entries in the current“row”),aŒk D cŒi 1;k for k j 1(i.e.,entries in the previous“row”),15-4Selected Solutions for Chapter15:Dynamic ProgrammingaŒ0 D cŒi;j 1 (i.e.,the previous entry computed,which couldn’t be put into the“right”place in a without erasing the still-needed cŒi 1;j 1 ).Initialize a to all0and compute the entries from left to right.Note that the3values needed to compute cŒi;j for j>1are in aŒ0 D cŒi;j 1 ,aŒj 1 D cŒi 1;j 1 ,and aŒj D cŒi 1;j .When cŒi;j has been computed,move aŒ0 (cŒi;j 1 )to its“correct”place,aŒj 1 ,and put cŒi;j in aŒ0 .Selected Solutions for Chapter15:Dynamic Programming15-5 cŒj D(0if j D0;.cŒi 1 C lcŒi;j /if j>0:min1 i jNote that the way we defined lc ensures thatall choices made willfit on the line(since an arrangement with lc D1cannot be chosen as the minimum),andthe cost of putting words i;:::;j on the last line will not be0unless this really is the last line of the paragraph(j D n)or words i:::jfill the entire line.We can compute a table of c values from left to right,since each value depends only on earlier values.To keep track of what words go on what lines,we can keep a parallel p table that points to where each c value came from.When cŒj is computed,if cŒj is based on the value of cŒk 1 ,set pŒj D k.Then after cŒn is computed,we can trace the pointers to see where to break the lines.The last line starts at word pŒn and goes through word n.The previous line starts at word pŒpŒn and goes through word pŒn 1,etc.In pseudocode,here’s how we construct the tables:P RINT-N EATLY.l;n;M/let extrasŒ1::n;1::n ,lcŒ1::n;1::n ,and cŒ0::n be new arrays//Compute extrasŒi;j for1 i j n.for i D1to nextrasŒi;i D M l ifor j D i C1to nextrasŒi;j D extrasŒi;j 1 l j 1//Compute lcŒi;j for1 i j n.for i D1to nfor j D i to nif extrasŒi;j <0lcŒi;j D1elseif j==n and extrasŒi;j 0lcŒi;j D0else lcŒi;j D.extrasŒi;j /3//Compute cŒj and pŒj for1 j n.cŒ0 D0for j D1to ncŒj D1for i D1to jif cŒi 1 C lcŒi;j <cŒjcŒj D cŒi 1 C lcŒi;jpŒj D ireturn c and pQuite clearly,both the time and space are‚.n2/.In fact,we can do a bit better:we can get both the time and space down to‚.nM/. The key observation is that at most d M=2e words canfit on a line.(Each word is15-6Selected Solutions for Chapter15:Dynamic Programmingat least one character long,and there’s a space between words.)Since a line withwords i;:::;j contains j i C1words,if j i C1>d M=2e then we knowthat lcŒi;j D1.We need only compute and store extrasŒi;j and lcŒi;j forj i C1 d M=2e.And the inner for loop header in the computation of cŒjand pŒj can run from max.1;j d M=2e C1/to j.We can reduce the space even further to‚.n/.We do so by not storing the lcand extras tables,and instead computing the value of lcŒi;j as needed in the lastloop.The idea is that we could compute lcŒi;j in O.1/time if we knew thevalue of extrasŒi;j .And if we scan for the minimum value in descending orderof i,we can compute that as extrasŒi;j D extrasŒi C1;j l i 1.(Initially,extrasŒj;j D M l j.)This improvement reduces the space to‚.n/,since nowthe only tables we store are c and p.Here’s how we print which words are on which line.The printed output ofG IVE-L INES.p;j/is a sequence of triples.k;i;j/,indicating that words i;:::;jare printed on line k.The return value is the line number k.G IVE-L INES.p;j/i D pŒjif i==1k D1else k D G IVE-L INES.p;i 1/C1print.k;i;j/return kThe initial call is G IVE-L INES.p;n/.Since the value of j decreases in each recur-sive call,G IVE-L INES takes a total of O.n/time.。
chap5-solutions
Solution to Exercise5.2-1Since H IRE-A SSISTANT always hires candidate1,it hires exactly once if and onlyif no candidates other than candidate1are hired.This event occurs when candi-date1is the best candidate of the n,which occurs with probability1=n.H IRE-A SSISTANT hires n times if each candidate is better than all those who wereinterviewed(and hired)before.This event occurs precisely when the list of ranksgiven to the algorithm is h1;2;:::;n i,which occurs with probability1=nŠ.5-2Selected Solutions for Chapter 5:Probabilistic Analysis and Randomized AlgorithmsThus,E ŒX D E"n X i D 1X i #Dn X i D 1E ŒX i (linearity of expectation)Dn X i D 11=nD 1;and so we expect that exactly 1customer gets back his own hat.Note that this is a situation in which the indicator random variables are not inde-pendent.For example,if n D 2and X 1D 1,then X 2must also equal 1.Con-versely,if n D 2and X 1D 0,then X 2must also equal 0.Despite the dependence,Pr f X i D 1g D 1=n for all i ,and linearity of expectation holds.Thus,we can use the technique of indicator random variables even in the presence of dependence.Selected Solutions for Chapter5:Probabilistic Analysis and Randomized Algorithms5-3D n2!1 21 4:Thus the expected number of inverted pairs is n.n 1/=4.Solution to Exercise5.3-4P ERMUTE-B Y-C YCLIC chooses offset as a random integer in the range1offset n,and then it performs a cyclic rotation of the array.That is,BŒ..i C offset 1/mod n/C1 D AŒi for i D1;2;:::;n.(The subtractionand addition of1in the index calculation is due to the1-origin indexing.If wehad used0-origin indexing instead,the index calculation would have simplied toBŒ.i C offset/mod n D AŒi for i D0;1;:::;n 1.)Thus,once offset is determined,so is the entire permutation.Since each value ofoffset occurs with probability1=n,each element AŒi has a probability of endingup in position BŒj with probability1=n.This procedure does not produce a uniform random permutation,however,sinceit can produce only n different permutations.Thus,n permutations occur withprobability1=n,and the remaining nŠ n permutations occur with probability0.。
算法导论第三版答案
Procedure BINARY-SEARCH takes a sorted array A, a value , and a range Œlow : : high of the array, in which we search for the value . The procedure compares to the array entry at the midpoint of the range and decides to eliminate half the range from further consideration. We give both iterative and recursive versions, each of which returns either an index i such that AŒi D , or NIL if no utions for Chapter 2: Getting Started
AŒlow : : high contains the value . The initial call to either version should have the parameters A; ; 1; n.
Selected Solutions for Chapter 2: Getting Started
2-3
d. We follow the hint and modify merge sort to count the number of inversions in ‚.n lg n/ time.
To start, let us define a merge-inversion as a situation within the execution of merge sort in which the MERGE procedure, after copying AŒp : : q to L and AŒq C 1 : : r to R, has values x in L and y in R such that x > y. Consider an inversion .i; j /, and let x D AŒi and y D AŒj , so that i < j and x > y. We claim that if we were to run merge sort, there would be exactly one mergeinversion involving x and y. To see why, observe that the only way in which array elements change their positions is within the MERGE procedure. Moreover, since MERGE keeps elements within L in the same relative order to each other, and correspondingly for R, the only way in which two elements can change their ordering relative to each other is for the greater one to appear in L and the lesser one to appear in R. Thus, there is at least one merge-inversion involving x and y. To see that there is exactly one such merge-inversion, observe that after any call of MERGE that involves both x and y, they are in the same sorted subarray and will therefore both appear in L or both appear in R in any given call thereafter. Thus, we have proven the claim.
南航双语矩阵论matrix theory第3章部分习题参考答案
1
Exercise 8
Let S be the subspace of C[a, b] spanned by e x , xe x , and x 2 e x . Let D be the differentiation operation of S, i.e., D( f ) f ' . Find the matrix representing D with respect to [e x , xe x , x2 e x ] Solution
3
This will imply that k1 must be zero since n 1 ( v) 0 . By repeating the process above, we obtain that k0 , k1 , , kn 1 must be all zero. This proves that
Solution
(1) 1
1
1
( x)
1 / 2 0
( x) 1 0 1
Hence, A
1 1/ 2 1 0
1/ 2 1 1/ 2 0
4
If B is nonsingular, then AB B1 ( BA) B . AB and BA are similar.
A P diБайду номын сангаасg(1 , 2 , , n ) P1 then Ak P diag(1k , 2k ,
, n . Hence, diag(1 , 2 ,
, n k ) P1 .
Ak O implies that i 0 for i 1, 2,
运动对我们的健康有好处英文作文
运动对我们的健康有好处英文作文The Awesome Power of Exercise!Do you ever feel tired, sluggish, or grumpy? Well, I've got a super solution for you - exercise! Getting physical and moving your body is like giving it a magic energy boost. I used to think exercise was just for jocks and health nuts, but now I'm totally hooked on the amazing feeling it gives me.Let me break it down for you on why exercise is so amazingly awesome:Exercise Makes You StrongerWhen you exercise, you're actually making your muscles bigger and more powerful. It's like leveling up in a video game! The more you move around and get physical, the stronger your muscles become. That extra muscle power helps you do cool things like run faster, jump higher, and lift heavier objects. Who doesn't want to be a super strong kid?Exercise Boosts Your EnergyInstead of feeling tired and lazy after exercise, you actually get this crazy energy burst. Your heart starts pumping and pushing more oxygen through your body, almost like an energydrink but without the sugar crash after. I used to come home from school feeling like a zombie, but now a simple game of basketball or bike ride has me feeling recharged and ready to take on the world.Exercise Sharpens Your MindAlong with giving you more physical energy, exercise is like miracle grow for your brain. All that blood pumping and oxygen flowing helps your mind work better. You'll feel more alert, have an easier time focusing, and might even get better grades! Exercise literally makes you smarter. Who wouldn't want a strong body and a smart brain?Exercise Melts Away StressWe all get stressed sometimes, like when there's a big test coming up or we're going through a rough patch with friends. Instead of holding all that tension inside, exercise is an amazing way to release it. You get this awesome feeling of calm and relaxation after a good workout. It's like giving your body a refresh and recharge. No more feeling anxious or grumpy!Exercise Improves Your MoodHave you ever noticed how good you feel after playing outside or going for a run? Exercise releases all these coolchemicals in your brain called endorphins that make you happier and more positive. It's like an instant pick-me-up in a bottle, but even better because you get it naturally! Just a few jumping jacks or a quick dancearound your room is enough to boost your spirits.Exercise Helps You Sleep BetterIf you have trouble falling or staying asleep, exercise can be the key to sweet dream land. All that physical activity during the day helps tire out your body in the best way possible. You'll be able to fall asleep faster and get higher quality sleep. No more lying awake at night with a buzzing mind - exercise is like a natural sleep aid!Exercise Sets You Up for a Healthier FutureThe really exciting thing about exercise is that it pays off big time as you get older. All the healthy exercise habits you build now will stick with you and protect you from diseases and health issues down the road. You're actually giving your future self the gift of a stronger heart, better lungs, lower risk of diabetes and cancer, and much more!As you can see, exercise is pretty much a magical cure-all for feeling better physically and mentally. And the best part is, itdoesn't have to be super hard or complicated. Simple things like walking instead of riding, playing actively at recess, or joining a recreational sport all count. You can exercise pretty much anywhere - at home, at school, at the park. Just get that body moving!I get that exercise might not seem like fun at first, but trust me, once you start, you'll be hooked. That feeling of exhaustion but in a good way, the sweat, the heavy breathing - it's all worth it for the amazing boost it gives you. Don't be a couch potato, be a fitness superhero instead!So next time you're feeling blah, lethargic, or grumpy, bust out some jumping jacks, go for a bike ride, or play an active game. Your body and mind will thank you for giving it the gift of exercise. This is one habit that will seriously level up your life!。
英语排序专项练习题
英语排序专项练习题### English Sorting Practice QuestionsThis set of practice questions is designed to help you improve your English sorting skills, which is essential for understanding the structure and flow of written English. Below are various types of sorting exercises that will test your ability to arrange sentences in a logical order.#### Exercise 1: Narrative SortingRead the following sentences and arrange them in the correct order to form a coherent narrative.1. The sun was setting, casting a golden glow over the city.2. I had always dreamed of visiting Paris, and finally, my dream came true.3. I walked along the Seine, admiring the beautiful bridges.4. As I arrived, the Eiffel Tower loomed in the distance.5. The first thing I did was to fi nd a charming café for a cup of coffee.Correct Order: 2, 5, 4, 1, 3#### Exercise 2: Descriptive SortingSort the following sentences to create a descriptive passage about a garden.1. The roses were in full bloom, their fragrance filling the air.2. A small pond with koi fish added a touch of tranquility.3. The garden was a riot of colors, with flowers of every hue.4. A stone path meandered through the lush greenery.5. At the center was a beautiful marble statue.Correct Order: 3, 4, 1, 2, 5#### Exercise 3: Argumentative SortingArrange the sentences to present a logical argument for the benefits of reading.1. Reading can improve vocabulary and comprehension skills.2. It is a common belief that reading is a solitary activity.3. However, it can also foster empathy and understanding of different perspectives.4. Reading is a gateway to knowledge and personal growth.5. Many people find comfort in the company of a good book.Correct Order: 4, 1, 3, 5, 2#### Exercise 4: Process SortingSort the following steps to describe the process of making a cup of tea.1. Boil water in a kettle or pot.2. Place a teabag in a cup.3. Pour the hot water over the teabag.4. Allow the tea to steep for a few minutes.5. Add sugar or milk if desired.Correct Order: 1, 2, 3, 4, 5#### Exercise 5: Problem-Solution SortingArrange the sentences to outline a problem and its solution.1. Pollution is a major environmental concern.2. Many cities suffer from poor air quality.3. Implementing stricter emission standards can help reduce pollution.4. Encouraging the use of public transportation is another effective solution.5. It is important to find ways to combat this issue.Correct Order: 2, 1, 5, 3, 4#### Exercise 6: Chronological SortingSort the sentences to describe the chronological events of a day in the life of a student.1. The alarm clock rang, signaling the start of a new day.2. After a quick breakfast, I rushed to catch the school bus.3. I spent the day attending classes and taking notes.4. I returned home, eager to complete my homework.5. Finally, I went to bed, ready for another day.Correct Order: 1, 2, 3, 4, 5#### Exercise 7: Cause-Effect SortingArrange the sentences to show the cause and effect relationship between actions and outcomes.1. The athlete trained hard every day.2. As a result, she won the championship.3. She decided to enter a marathon.4. Her dedication paid off.5. She felt a sense of accomplishment.Correct Order: 1, 3, 4, 2, 5By practicing these exercises, you will enhance your ability to understand and create well-structured English sentences and paragraphs. Remember, the key to effective sorting is to identify the logical flow and coherence of the content. Happy practicing!。
- 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
Exercise Five -Solution1. Suppose you saw that the following exchange rates quoted in the foreign exchange market:1 US$1 US$ 1 US$ 1 US$1 € 1 £1 £= = = = = = = 7.7465 Chinese Renminbi0.5143 UK Pound Sterling ( £ )0.7599 Euros (€)11.1697 Mexican Pesos 14.7121 Mexican Pesos22.7166 Mexican pesos 15.0612 Chinese Renminbi a) The hypothetical exchange rates above contain an arbitrage opportunity. Explain how to exploit it, using nomore than three of the currencies listed above.Answer: Assume that you have 1 US$ to begin with. Pick currency pairs and try to exchange one currency foranother. If you end up with more than one then that is your arbitrage profits. Otherwise you could workbackwards.Let’s start with 1 US$, sell it for Chinese Renminbi, then sell Chinese Renminbi for UK Pounds and then sell itfor US dollar again. USD GBP USD CNY GBP USD CNY USDUSD == 00007.15143.010612.1517465.71=××× The difference between 1.00007 and 1 is so small that triangular arbitrage hardly exists here.Now try a different three currencies – US$, Peso, Pound. Sell 1 US$ for Peso, sell Peso for Pound and finallysell Pound for USD. USD GBP USD MXN GBP USD MXN USDUSD == 956052.05143.017166.2211697.111=××× The final amount you get is smaller than original one. So you must reverse your trading sequence for aprofitable arbitrage opportunity: Buy Pound with the USD, buy Pesos with the Pound and then buy USD with Pesos. USD MXNUSD GBP MXN USD GBP USD USD == 046.11697.1117166.225143.01=××× So 4.6% will be your arbitrage profit on each dollar. b) You can remove the arbitrage opportunity by changing just one of the exchange rates given. (“The market”should itself accomplish this, via arbitrage.) Which exchange rate is it, and what would the new correct value be?Answer: If you check other currency pairs in triangular arbitrage trading, you will see that it is the Peso/£ rate that is out of line, rather than one of the dollar exchange rates. For example, verify the followingarbitrage trading order:9992.0==EURUSD MXN EUR USD MXN GBP USD CNY GBP USD CNY USD USD Apart from the rounding error, all currencies along this order are priced correctly, which leaves only the the Peso/£ rate to be incorrect. The Pound is overvalued relative to the Peso. This is why buying Peso for Pound earned a profit.The correct Peso/£ rate is : 7183.225143.011697.11=×==GBP USD USD MXN GBP MXN [notes: If one wants to be precise to 4 decimal points, then 0.7599 should have been 0.7592. But these Calculations are not intended to apply to the fourth decimal point, which is the size of transaction costs And therefore not exploitable by arbitrage.]2. Suppose you are a manager of a book store. You can sell copies of World Trade and Payments by Caves, Frankel & Jones, at $135 per copy. The book’s U.S. publisher will sell it to you for $110 a copy. At the same time the book World Trade and Payments is also wholesaled by two foreign firms: A Chinese firm sells it for CNY 700, and a British firm for £ 50 (exchange rates are given above in Problem 1).a)Explain very briefly how you would maximize your profits from textbook purchases and sales.Answer: Given the following exchange rates,1 US$ = 7.7465 Chinese Renminbi1 US$=0.5143 UK Pound Sterling ( £ )The US$ price of buying from China would be $90.36, and the US$ price from the British firm is$97.22. Here is an apparent opportunity for arbitrage of goods – a mismatch of prices at the differentgeographical locations. The strategy “Buy low, sell high” again governs: buying books from theChinese Firm and selling them in the US will maximize profits. As long as the original price plustransportation is less than 110$, you do better buying from China.b)Your skeptical work-study assistant makes some international phone calls, and learns that the copies from the British publisher are of the previous edition, and the copies from the Chinese publisher are actually Xeroxes (影印版, it is illegal to sell it in USA). Now what will you do?Answer: You cannot sell Xerox copies in the US without proper publisher permission. Even if the store were willing to take chances by violating copyright laws, students will certainly not pay the same $135 for a Xerox one; thus, the Chinese alternative no longer offers a pure arbitrage opportunity. Similarly, the previous edition offered by the British firm is not a perfect substitute for the current edition. Consequently, there is no real arbitrage situation anymore, even if the difference in price exceeds transportation costs.c) Give the main reason(s) why it is harder to take advantage of an apparent arbitrage opportunity in a goods market than in a financial market.Answer: The immediate reason is the transportation cost. Arbitrage will start to operate only if the price in one country exceeds the price in the other by more than the shipping costs. Otherwise it is not worthboth to bear the shipment cost and wait for days or weeks to get the product. In this respect, geographicdistance and border effects play important roles. Tariffs and other trade barriers are the second reasonwhy it is harder to take advantage of an apparent arbitrage opportunity in a goods market. Thirdly, as aresult of imperfect information, consumers/retailers may not aware of the price difference in differentlocations, or of the quality of the product that they are importing. Currencies or securities traded infinancial markets, by contrast, have lower transactions costs than goods, and are often standardized.。