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Utah Valley University - PMGT - 321
Exam#2Conceptscheck#6Managing Volunteers1. Benefits of VolunteersoAdditional laboroSpecialized skills: Administrative, Crafts, ProfessionaloEnhanced support and advocacyoBetter understanding of needsoHumanize the experience for program partici
Utah Valley University - PMGT - 321
Exam #2 Concepts Check #5Excise Taxes (aka selective sales taxes)1. Benefits based- often % of price (hotel/motel, motor fuels)oRecover part of cost of a public service from people who benefit2. Sumptuary- generally per unit (sin tax: alcohol, cigare
Utah Valley University - PMGT - 321
Exam#2ConceptsCheck#41. WhatisIncome?oAdjustedGrossIncome(AGI)equalsGrossreceiptsfromtaxablesources(wages,salaries,interest,cashprizes)MinusAdjustment(healthinsurance,someretirementsavings,etc.)oTaxableincomeequalsAGIminusPersonalexemptionsDeduct
Utah Valley University - PMGT - 321
Exam #2 Concepts check #31. Base ValuationoWilling Buyer and Willing Seller Test- not the original purchase priceoHighest and best use of property- key issue2. Valuing the BaseoAppraised value: the amount the property would sell for in a normal tr
Utah Valley University - PMGT - 321
Market Economy Two competing economic systemso Command: directed by a centralized governmento Market: private enterprise. Flexible, practical, and changeable. No centerMay be practical, but they also rest upon the fundamental principle of individualf
Utah Valley University - PMGT - 321
Exam #2 Review Day 21. Incidence/Burden- who actually pays the tax2. Economic Incidence- group that has to pay the tax.oAre the taxed goods necessary?oEx: tax on apple farmers might actually be paid by owners of agricultural land orconsumers of app
S.F. State - ISYS - 565
ISYS 464, Section 1, Spring, 2012 (Beckman)Homework #2: DescriptionDue: Tuesday, February 21, 2012, start of classChapter 3, textbook page 144, Problems and Exercises, #9.(10 points)3.A bank has three types of accounts: checking, savings, and loan.
University of Minnesota - ECON - 3102
University of MinnesotaDepartment of EconomicsEcon 3102: Intermediate MacroeconomicsMidterm 2 - Answer KeyShort Answer (25 points)(a) (5 points) Brie explain in terms of income and substitution eects why an increase in the real wage mayycause a wor
University of Minnesota - ECON - 3102
Practice Midterm 1Econ 3102ATTENTION: The following questions DO NOT cover all the material that will be on the midterm. Iput questions about some of the material that was not in the problems sets, so that you can have an ideaof how they may show up i
University of Minnesota - ECON - 3102
University of MinnesotaDepartment of EconomicsEcon 3102: Intermediate MacroeconomicsProblem Set 1 Answer KeyExercise 1 (25 points) Download the Excel le "GDP and Ination.xls" from the course websites(https:/sites.google.com/site/marcelozouainpedron
University of Minnesota - ECON - 3102
University of MinnesotaDepartment of EconomicsEcon 3102: Intermediate MacroeconomicsProblem Set 2 Answer keyExercise 1 (20 points)Consider problem 7 in chapter 4, page 144, of the book assuming that the utility is Cobb-Douglas (U (C; `) =C ` ). Firs
University of Minnesota - ECON - 3102
University of MinnesotaDepartment of EconomicsEcon 3102: Intermediate MacroeconomicsProblem Set 3 Answer keyExercise 1 (30 points)Solve problem 1 in chapter 8 page 310 of the book.Answer:(a) The consumer lifetime wealth is given by:swe = yt+y 0
University of Minnesota - ECON - 3102
University of MinnesotaDepartment of EconomicsEcon 3102: Intermediate MacroeconomicsProblem Set 4 Answer keyExercise 1 (40 points)Consider the social security model developed in chapter 9 of the book. There are innite periods, and ineach period two
University of Minnesota - MATH - 2373
University of Minnesota - MATH - 2373
University of Minnesota - MATH - 2373
University of Minnesota - MATH - 2373
University of Toronto - ANT - ANT100Y1
ANT100LECTURE 123/09/2010Biological AnthropologyProfessor: Dr Shawn Lehman Office: AP404A (19 Russell Street) Office Hours: Tuesday 2:00-3:00 Section Goals: 1. Historical development of biological science 2. Diversity of life and natural processes pro
University of Toronto - ANT - ANT100Y1
ANT100LECTURE 230/09/2010Principles of EvolutionGoals: 1. Understand microevolution: Genetic basis of inheritance and biological evolution Population genetics Natural selection Adaptations 2. Understand macroevolution: Speciation How to read a cladogr
University of Toronto - ANT - ANT100Y1
ANT100LECTURE 307/10/2010Primate Behaviour and EcologyGoals: 1. What is a primate and how does it differ from other mammals 2. Basic taxonomic characters of living primates 3. Primate ecology and sociality 4. Main conservation issues for primatesPrim
University of Toronto - ANT - ANT100Y1
ANT100LECTURE 414/10/2010Primate and Early Human EvolutionGoals: 1. General patterns of morphology and phylogenetics for fossil primates 2. What a hominin is in terms of taxonomy 3. Morphological trends in hominin evolution: a. Bipedalism (locomotion
University of Toronto - ANT - ANT100Y1
ANT100LECTURE 521/10/2010Hominins, Human Variation, and Forensic AnthropologyGoals: 1. Australopithecus anamensis - Kanapoi, Kenya - 4.2-3.9 MYA - Tooth row is parallel (ape-like) Australopithecus afarensis - Ethiopia (Hadar, Omo, and Fejej) and Tanza
Purdue - MA - 224
Fall 2011MA 22400 Topics ListFunctions of One VariableI.Finding definite and indefinite integrals using integration by parts Both given the definite/indefinite integral and in word problemsII.Evaluating improper integrals Given the improper integr
Purdue - MA - 224
Exam I Study Guide1.AntiderivativeIf F (x) = f (x), in MA223, we call f (x) the derivative of F (x); in MA224, we also call F (x)an antiderivative of f (x). Just notice that all the antiderivatives of f (x) are in the formF (x) + C , where C is an arb
Purdue - MA - 224
Exam III Study Guide1.The Method of Least SquaresGiven two distinct points (x1 , y1 ) and (x2 , y2 ) on the x y plane, we can nd an equation ofa straight line going through them. But what if we are given n points (x1 , y1 ), , (xn , yn )?Well, in most
Purdue - MA - 224
MA 22400Midterm Exam 3Solution to Exam 31Spring 2012MA 22400Midterm Exam 3Spring 20121. (10 points) The equation of the least-squares line for the n points (x1 ,y1 ), (x2 ,y2 ), . ,(xn ,yn ) is y = mx + b, wheren xy x ym=n x2 ( x) 2x2nb=y
Purdue - MA - 224
Solution to Quiz 11Problem 1. (5 points) Find the indenite integral of f (x) = (x2 x)( x 1).x2Solution. The indenite integral of f (x) is found as follows:f (x)dx =====Problem 2. (5 points) Integrate11(x2 x)( 2 )dxxx1(x 1 1 + )dxx1(x
Purdue - MA - 224
Common Erros from Quiz 1ex+ C , but simply ex + C . Dont confuse it withx1.ex dx is not2.f (x)g (x)dx is NOTf (x)dxekx dx =ekx+ C.kg (x)dx. In general, we dont have a formula to dealwith the integral of the product of two functions. We only
Purdue - MA - 224
Solution to Quiz 2Problem 1. (5 points) Find the given denite integral10x2dx.(x3 + 1)2Solution. There are two ways to nd the denite integral if substitution is needed. We willfollow both methods.Method 1. This is the one getting the most votes in
Purdue - MA - 224
Common Erros from Quiz 21.11du is NOT ln |u2 | + C , it is + C . But we have2uu2.f (x)g (x)dx is NOTf (x)dx +1du = ln |u| + Cug (x)dx. In general, we dont have a formula todeal with the integral of the product of two functions. But we do ha
Purdue - MA - 224
Solution to Quiz 3Problem 1. (4 points) Circle the average value of f (x) = ex (1 ex ) over the interval1 x 1:e + e1A.2B. e e1 + 1e e1 2C.2e11D.+e .22Solution. The average value V of f (x) over the given interval is found as follows11f
Purdue - MA - 224
Common Error to Quiz 31dx = 1 + C1dx = x + C1
Purdue - MA - 224
Solution to Quiz 4xex dx.Problem 1. (5 points) FindSolution. A rule of thumb is that if you have ex in the integrand, let it be in dv instead ofletting u = ex . So we have the following as advertised in classu = x dv = ex dxdu = dx v =ex dx = ex (w
Purdue - MA - 224
Common Error to Quiz 4DONt confuse derivative with integral:ln xdx =1+CxIn fact,ln xdx = x ln x x + CThis is obtained by integration by parts.1
Purdue - MA - 224
Solution to Quiz 4Problem 1. (5 points) Find the improper integral+13x24dx.Solution. By the denition of improper integral,+N1=3 dx4N +x23x 2 dxlim4N2lim N +x422= lim ( + )N +4N=1=Problem 2. (5 points) Use the trapezoidal
Purdue - MA - 224
Common Error to Quiz 51. DONT confuse integral with derivative:325x 2 1x dx = 3+ C = x 2 + C.52 132Instead,3x 2 +12+ C = + C.x dx = 3x2 + 1322. In the trapezoidal rule, the coecient pattern is 1,2,2,2,.,2,2,1,not the others.1
Purdue - MA - 224
Solution to Quiz 62Problem 1. (5 points) Given f (x, y ) = ln yex ,(1) nd fx and fy .Solution. To nd fx , only x is treated as the variable while y is thought of as a constant.So2fx (x, y ) = ln y 2xex .Similarly, to nd fy , x is treated as a cons
Purdue - MA - 224
Common Error to Quiz 61. Given f (x, y ), to nd fx, we treat y as a constant. But this doesnt2mean that there is no y -part. For instance, given f (x, y ) = ln yex ,2fx = 2xex ,but2fx = ln y 2xex .2. Sometimes the usage of product rule is necessa
Purdue - MA - 224
dzProblem 1. (5 points) Given z = xy ; x = e2t , and y = et , use chain rule to ndwhendtt = 0.Solution.dxdydz= zx+ zydtdtdt= y 2e2t + x(et )Since t = 0, we have x = e20 = 1 and y = e0 = 1. Therefore,dz= 1(2 1) + 1(1)dt=1Problem 2. (5 p
Purdue - MA - 224
Solution to Quiz 8Problem 1. (5 points) Use the Lagrange multiplier method to nd the values of x, y wheref (x, y ) = x2 + y 2 gets the minimum under the constraint xy = 1.Solution. Letting g (x, y ) = xy and knowing f (x, y ) = x2 + y 2 , we nd the exp
Purdue - MA - 224
Solution to Quiz 9Problem 1. (5 points) Evaluate the given double integral2xx2 ydydx0Solution.2022x(x ydydx =02x00====1ydy )dx0x22=xy)dx2002xx2 ( 0)dx2023xdx022x424 02x2 (Problem 2. (5 points) Given the geom
Purdue - MA - 224
Solution to Quiz 10Problem 1. (5 points) Find the radius of convergence of the following power series:k=4xk2k1Solution.Lets nd the following L(x) which depends on x:xk+1 2k1|k 2(k+1)1 xkx= lim | |k 2x=| |2We know if L(x) < 1, i.e., |x/2| <
Purdue - MA - 224
Solution to Quiz 11Problem 1. (5 points) Write down the power series of1dx1xin summation notation.Solution.Let us rst write the integrand1as its power series and then integrate term by term:1x1dx =1xxn dxn=0xn dx=n=0=n=0xn+1n+1Probl
Purdue - MA - 224
Solution to Quiz 12Problem 1. (5 points) Solve the dierential equationdy= 2ydxSolution.Dividing both sides by y and multiplying by dx, we seperate the variables as follows:dy= 2dx.ydy= 2dx.yThen we get ln |y | = c + 2x, which yields |y | = ec
Purdue - MA - 224
Important limits(1) For k, p > 0,Nplim kN = lim N pekN = 0N eN This is because in the long run (N ), ekN grows much fasterthan N p.Examplex2x1/3lim= 0; lim x/4 = 0.N e3xN e(2) For p > 0,ln(N )= lim ln(N )N p = 0N N N plimThis is becau
Purdue - MA - 224
MA 22400Midterm Exam 1Solution to Exam 11Spring 2012MA 22400Midterm Exam 1Spring 20121. (14 points) Evaluate the following indenite integrals12 )dx2xxSolution Since none of the two multiplicative terms is part of the derivative of the other
Purdue - MA - 261
MA 261 - Spring 2011Study Guide # 3You also need Study Guides # 1 and # 2 for the Final Exam1. Line integral of a function f (x, y ) along C , parameterized by x = x(t), y = y (t) and a t b, isf (x, y ) ds =(bf (x(t), y (t)Cadxdt)2(+dydt)
Purdue - MA - 261
MA261 0021&0022 Quiz 1Spring 2011Problem 1. Find the angle between the vectors 1, 1, 1 and 1, 1, 2 . (Hint: Lookat the dot product of these two vectors.)Since1, 1, 1 1, 1, 2 = 1 + 1 2 = 0,we know these two vectors are perpendicular. Thus, the angle
Purdue - MA - 261
MA261 0021&0022 Quiz 2Spring 2011Problem 1. Given two planes x + y + z = 2 and x + y = 1,(a) Find the two normal vectors associated to each plane, respectively. (Notice thatx + y = 1 is the same thing as x + y + 0 z = 1)Solution. We read the normal v
Purdue - MA - 261
MA261 0021&0022 Quiz 3Spring 2011Problem. Consider the ellipsex2 y 2+= 1, z = 041in the three dimensional space. One system of parametric equations of it is r(t) =x(t), y (t), z (t) wherex(t) = 2 cos ty (t) = sin t , t [0, 2 ].z (t) = 0(a) Fi
Purdue - MA - 261
MA261 0021&0022 Quiz 3Spring 2011Problem. Consider the ellipsex2 y 2+= 1, z = 041in the three dimensional space. One system of parametric equations of it is r(t) =x(t), y (t), z (t) wherex(t) = 2 cos ty (t) = sin t , t [0, 2 ].z (t) = 0(a) Fi
Purdue - MA - 261
MA261 0021&0022 Quiz 4Spring 2011Problem 1. The position function of a particle is given byr(t) = 3 sin t, 3 cos t, 4t .(1) Find the velocity v(t).Solution. v(t) = r(t) = 3 cos t, 3 sin t, 4 .(2) Find the speed |(t)|.vSolution. |v(t)| = 32 + 42 =
Purdue - MA - 261
MA261 0021&0022 Quiz 5Spring 20112Problem 1 (Spring 2006). If u(x, y ) = yexy ,(a) nd ux .22Solution. ux = yy 2 exy = y 3 exy .(b) nd uxy .222Sollution. uxy = (ux )y = 3y 2 exy + y 3 2xyexy = y 2 exy (3 + 2xy 2 ).Problem 2 (Spring 2001). Give
Purdue - MA - 261
MA261 0021&0022 Quiz 6Spring 2011Problem 1. Suppose y is a function of x and they satisfy F (x, y ) = 0.Take the partial derivative with respect to x of both sides of the above equation (UseChain Rule ) to show thatFdyx= F .dxyProof. Given F (x
Purdue - MA - 261
MA261 0021&0022 Quiz 7Spring 2011Problem 1. Given f (x, y ) = x2 y 2 ,(a) Find the critical point of f (x, y ).Solution. Setting f (x, y ) to be zero, we have f (x, y ) = fx (x, y ), fy (x, y ) =2x, 2y = 0, 0 . So, (x, y ) = (0, 0) is the only critic
Purdue - MA - 261
MA261 0021&0022 Quiz 8Spring 2011Problem 1 (modied from Problem 2 in the second midterm of Spring22009). We will nd the maximum of f (x, y ) = xy on the ellipse x + y 2 = 1 using the4Lagrange multiplier method.2(a) Let g (x, y ) = x + y 2 . Write
Purdue - MA - 261
MA261 0021&0022 Quiz 9Spring 2011Problem 1 (Test II.8, Spring 2008). [Warning: No use of cell phonebrowsing the internet during the quiz!] Let D be the part of disk centered at 0with radius 2 that lies to the right of the line x = 1. Then which of the
Purdue - MA - 261
MA261 0021&0022 Quiz 10Spring 2011Problem 1. Find the length of a wire C in the shape of a helix described by theparametric equationC : x = cos t, y = sin t, z = t, 0 t 4Solution. To get the length of the wire, let us integrate the constant function
Purdue - MA - 261
MA261 0021&0022 Quiz 11Spring 2011Problem 1.(a) Given F(x, y ) = y exy , xexy , nd a function f (x, y ) such that f (x, y ) = F(x, y ).Solution. Since we are told f (x, y ) = F(x, y ), we have fx = yexy and fy = xexy .After integrating the rst equali
Purdue - MA - 261
MA261 0021&0022 Quiz 12Spring 2011Problem 1.(a) Given a function f (x, y, z ) = xyz , nd its gradient f (x, y, z ) and then the curlof the gradient, namely curl( f (x, y, z ).Solution. Given f (x, y, z ) = xyz , we know f (x, y, z ) = fx , fy , fz =
Purdue - MA - 303
MA 303: Dierential and Partial DierentialEquations for Engineering and SciencesFall 2011, Final Examination(Instructor: Aaron N. K. Yip) This test booklet has TWENTY FIVE questions totaling 200 points for the wholetest. You have 120 minutes to do thi