on the reactant (R) side of the paper from the last round should be moved to the product side. One-fourth of the candies that were on the product (P) side of the paper from the last round should be moved to the reactant side. (If you end up with a decimal for the number to exchange‚ round up.) 4. At the end of each round‚ count the number of candies on each side of the paper and keep track of the numbers in a data table. Round Candies on R side Candies on P side 0 40 0 1 20 20
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Digital Design With an Introduction to the Verilog HDL This page intentionally left blank Digital Design With an Introduction to the Verilog HDL FIFTH EDITION M. Morris Mano Emeritus Professor of Computer Engineering California State University‚ Los Angeles Michael D. Ciletti Emeritus Professor of Electrical and Computer Engineering University of Colorado at Colorado Springs Upper Saddle River Boston Columbus San Franciso New York Indianapolis London Toronto Sydney Singapore
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-12‚345 ’-12345’ ’ -12345’ ’-12345 ’ ’-0000000000000012345’ ’-12345’ ’ -12345’ ’-12345 ’ ’-0000000000000012345’ ’-12345’ Floating point numbers We consider the floating point numbers 12·34567 and -12·34. The %f formatting code presents data in decimal notation. The %e code does it in exponential form. Formatting code %f %20f %-20f %020f %+f %+20f %+-20f %+020f %.4f %20.4f %-20.4f %020.4f %+.4f %+20.4f %+-20.4f %+020.4f Formatting code %e %20e %-20e %020e %+e %+20e %+-20e %+020e %.4e %20.4e %-20
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Digital Digital Electronics (EE202) (EE202) NUMBER NUMBER SYSTEMS • Decimal 0~9 • Binary 0~1 • Octal 0~7 • Hexadecimal 0~F DECIMAL DECIMAL The decimal system is composed of 10 numerals or symbols. These 10 symbols are 0‚ 1‚ 2‚ 3‚ 4‚ 5‚ 6‚ 7‚ 8‚ 9; using these symbols as digits of a number‚ we can express any quantity. The decimal system‚ also called the base-10 system because it has 10 digits. EXAMPLE: 47 = (4 X 101)+(7 X 100) = (4 X 10) + (7 X 1) = 40+ 7 EXERCISE : 568.23 = BINARY
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by humans? Used in computers? Decimal 10 0‚ 1‚ … 9 Yes No Binary 2 0‚ 1 No Yes Octal 8 0‚ 1‚ … 7 No No Hexadecimal 16 0‚ 1‚ … 9‚ ‚ ‚ ‚ A‚ B‚ … F No No 1 6/28/2012 Quantities/Counting (1 of 3) Decimal 0 1 2 3 4 5 6 7 Binary 0 1 10 11 100 101 110 111 HexaOctal decimal 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 p. 33 Quantities/Counting (2 of 3) Decimal 8 9 10 11 12 13 14 15 Binary
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COURSE SYLLABUS SICS 1533: FOUNDATIONS OF COMPUTER SCIENCE "Whatever you vividly imagine‚ ardently desire‚ sincerely believe and enthusiastically act upon must inevitably come to pass!" Paul J. Meyer a "To be successful‚ you must decide exactly what you want to accomplish‚ then resolve to pay the price to get it." - Bunker Hunt b [Academic Year / Semester] 2013 / 2014‚ First Semester [Class Location] City Campus‚ Computer Lab [Class Meeting Time(s)] (Depending
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Lesson 1 TECHNOLOGY TROUGH THE YEARS In history‚ people have always sought out ways to make things easier to do and to understand. This is the reason why‚ after so many thousands of years‚ a simple tool like the wheel is now more than just a wooden disk that rolls on the ground. It is also the reason why a conversation is now possible between people who are on opposite sides of the world; and why information‚ which used to be found only in books‚ can now be accessed from the Internet with a computer
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DC = $ 2‚600‚000 AOC = $ 4‚727.27 AHC = $ 33‚000 TC = $ 2‚637‚727.27 Using the Economic Order Quantity (Q=EOQ) Calculate EOQ. (USE A WHOLE NUMBER‚ NO DECIMALS) EOQ = 416 Using the calculated EOQ‚ How often will orders be placed? Weeks. DO NOT ROUND (Show 2 decimal places) 1.67 Weeks According to the information supplied‚ what would be the annual inventory cost if they used the calculated EOQ? Also‚ separately
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Section 11: Data Manipulation Mark Nicholls – ICT Lounge IGCSE ICT – SECTION 11 DATA MANIPULATION MICROSOFT ACCESS STEP BY STEP GUIDE Mark Nicholls ICT Lounge Page |1 Section 11: Data Manipulation Mark Nicholls – ICT Lounge Contents Task 35 details………………………………………………………………………………………… Page 3 Opening a new Database…………………………………………………………………………. Page 4 Importing .csv file into the Database………………………………………………………. Page 5 - 9 Amending Field Properties……………………………………………………………………….. Page 10 - 11 Taking
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Exercise 1.1.5 – 111000102 = 22610 Exercise 1.1.6 – 15610 = 100111002 Exercise 1.1.7 – 25510 = 111111112 Exercise 1.1.8 – 20010 = 110010002 Exercise 1.1.9 – 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 128 64 32 16 8 4 2 0 binary 0 0 0 0 1 0 0 1 0 0 0 0 8 0 0 0 decimal 8 Exercise 1.1.10 – 101110012 = 18510 Exercise 1.1.11 – 1011112 = 4610 Exercise 1.1.12 – 100000012 = 12910 Lab 1.2: Binary Math and Logic Exercise 1.2.1 – 1111 Exercise 1.2.2 – 1011 Exercise 1.2.3 –1110 Exercise 1.2.4 – 111 Exercise 1.2
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