1 Describe the graphical coordinate system in Java. Where is the origin? What units apply to the x‚y coordinates? Java AWT Graphic component is different with respect to x and y coordinate that we have learnt in math. It defines its own coordinate system to specify where and how drawing operation will take place. Here the origin (0‚0) is at the top left corner of the component. Here X coordinate increases as we move right and Y coordinate increases as we move down. . The units for X and Y is in
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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 – 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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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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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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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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451913 Marketing Plan LEOPARD CAFÉ Heather (Rowe) Holland March 4‚ 2015 Table of Contents Executive Summary 5 Company Introduction 6 Mission Statement 6 Vision Statement 7 Product and/or Services 7 Differentiation 9 Industry Profile 10 Situational Analysis 12 Growth in Silver Bow County 13 Growth of “The Leopard Café “ 14 Segmentation 14 Customer Profile 15 Market Demographics/Size 16 Customer Base Present and Future 18 Marketing Strategy/Market Share 21 Marketing
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PART A Courts all over the world have set precedence’s of treating directors as trustees which means in the performance of their assigned legal and corporate duties‚ they stand in a fiduciary relation to the shareholders of the company. A director as a trustee shall act in the best of his ability to benefit the company and not in furtherance of his own interest. Each of the four directors of the company stand in a fiduciary position to the company and thus liable for their acts of omission and
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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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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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