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    programming Solution

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    Chapter 13 Answers to Exercises 1. Identify three objects that might belong to each of the following classes: a. Automobile b. NovelAuthor c. CollegeCourse The students will have a variety of answers for these questions. Some examples might be: a. myRedChevroletCamaro‚ theBlackfordMustangWithTheDentThatBobDrives‚ thePorschee911ThatDonaldTrumpOwns b. Terry Brooks‚ Steven King‚ Ray Bradbury c. English Composition‚ Calculus‚ Physics 2. Identify three different classes that might contain

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    SMBD Solutions

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    Smbd – Assignment 4 Submitted to Prof. Ishwar Murthy 1 Introduction Mr. Debashish Chatterjee‚ owner of hotel Aria‚ is working on room booking policy to maximize the revenues. And specifically‚ he is working on the weekend operations in which the number of bookings are maximum. There are two types of bookings‚ Class I – One day bookings which start from SAT noon to SUN noon or SUN noon to MON noon and Class II – Two day bookings which start from SAT noon to SUN noon. The tariff of Class I booking

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    MidtermSpring2011 Solution

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    UNIVERSITY OF TORONTO RSM 222 H1S MIDTERM EXAM‚ Winter 2011 Instructions: DURATION: AID ALLOWED: 1 hour 50 minutes Non-programmable calculator. Programmable calculators have to be reset before the exam. Total marks on the exam are 100 allocated as follows: Section I Marks Achieved 9 Multiple Choice Questions 3 marks each for a total of 27 marks II Short Answer Questions 18 marks III Problem 1: Job Costing 13 Marks Problem 2: Process Costing 12 marks Problem 3: CVP Problem 15 marks Problem

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    NIke solutions

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    Appendix I C1: Equity = Stock Price x Number of Shares Outstanding = $42.09 X 271.5 = $11‚427.435 million C2: Using Adjusted Beta formula: Adjusted Beta = 0.67* historical Beta + 0.33 = 0.67* 0.69 +0.33=0.79 C3: Using CAPM formula: KE = Krf + ß (Km-Krf) = 3.59%+0.79*6.7%=8.89% C4: Using rearranged DGM formula: KE =D1/P0 +g= 0.48(1+5.5%)/42.09 +5.5%=6.7% C5: Using redeemable bond formula: KD:  95.6= 100/ (1+KD/2)40 + 3.375(1-0.38)/(1+KD/2)n KD=4.52% C6: Using WACC formula: Rwacc =4.52*10.19% + 8

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    Wilkinson Solution

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    International Financial Management WILKINSON SWORD TRIALS AND TRIBULATIONS IN TURKEY CASE STUDY NUMBER 2 Performed by: Problematique In 2000‚ Wilkinson Sword-Turkey SA‚ (hereinafter “WST”) won the approval for a $12 million capital expenditure to finance the launch of a new product line‚ the Quattro shaving system‚ from its US-based parent company. Mrs. Ozcan‚ President and GM of the Turkish subsidiary‚ had to chose between two financing options: (1) Extension of the USD denominated intercompany

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    CHAPTER 1 Understanding the issues 1. (a) horizontal combination—both are marine engine manufacturers (b) vertical combination—manufacturer buys distribution outlets (c) conglomerate—unrelated businesses 2. By accepting cash in exchange for the net assets of the company‚ the seller would have to recognize an immediate taxable gain. However‚ if the seller were to accept common stock of another corporation instead‚ the seller could construct the transaction as a tax-free reorganization

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    !1 ! ! Summary of David Suzuki’s “Food Connections” In his essay “Food Connections‚” David Suzuki states that food is not just something we eat‚ but something that connects us to our earth and different places all around it. Fruits and vegetables come into our earth‚ straight from the soil‚ already fresh and healthy and we poison it with pesticides‚ antibiotics‚ etc. The way we live now‚ how we eat and treat our earth is unnatural. Our markets which are now turned into large supermarkets

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    Effects of the Mind-Body Connection on Learning Jason Grant COLL100 B161 American Military University Professor Allison Knox Effects of the Mind-Body Connection on Learning What is the “Mind-Body Connection” and does it have a profound effect on an individual’s learning? The mind-body connection can be explained as the physical and mental connection that our existence has on itself and the world around it. Some researchers have noted that this connection can be stronger in certain situations

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    econometric solution

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    Hieu Nguyen – FIN 5309 Section 1 Assignment 1 2.3 Table 2.2 X=0 X=1 Total Y=0 0.15 0.07 0.22 Y=1 0.15 0.63 0.78 Total 0.30 0.70 1.00 With W = 3+6X and V = 20-7Y‚ we have: (W|X=0)=3 (W|X=1)=9 Total (V|Y=0)=20 0.15 0.07 0.22 (V|Y=1)=13 0.15 0.63 0.78 Total 0.30 0.70 1.00 a. E(W) = 3 x 0.3 + 9 x 0.7 = 7.2 E(V) = 20 x 0.22 + 13 x 0.78 = 14.54 b. = (3 – 7.2)2 x 0.3 + (9 - 7.2)2 x 0.7 = 7.56 = (20 – 14.54)2 x 0.22 + (13 – 14.54)2 x 0.78 = 8.4084 c. cov

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    Eoq Solution

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    Q.1a) The following graph is EOQ model with planned shortages. Let the parameters from the basic EOQ model. d = constant demand rate K = setup cost for placing one order Q = order quantity h = inventory holding cost per unit of product per unit of time p = shortage cost per unit of product per unit of time S = inventory level just after an order of size Q arrives So‚ Q– S = Shortage in inventory just before an order of Q units is added Production or ordering cost per

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