...understand the conversion from binary to decimal numbers and decimal to binary numbers, quickly and easily…. Decimal to Binary Decimal numbers use Base 10 Binary numbers use Base 2 8 bits = 1 Byte Study the following chart and the comments by each line or section… | | 8 bits | 7 bits | 6 bits | 5 bits | 4 bits | 3 bits | 2 bits | 1bit | Base 2 | | 27 | 26 | 25 | 24 | 23 | 22 | 21 | 20 | DecimalWeight value of each bit | | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | Decimal to binary… | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | | 2 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | | 3 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | | 4 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | | 5 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | | 6 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | Do you see the pattern? 1 = 00000001 0+0+0+0+0+0+1 (the 1 bit is in the 20 position, the 20 position value="1). 2 = 00000010 0+0+0+0+0+0+1+0 = (the 1 is in the 21 position, the 21 position value="2) 3 = 00000011 0+0+0=0+0+0+1+1 = (the 1 is in the 21 and 20 positions, the 20 value="1" and the 21 value="2" therefore, 2 +1 = 3). 4 = 00000100 0+0+0+0+0+1+0+0 = (the 1 is in the 22 position. The 22 value="4) Always start from left to right on the chart (highest to lowest number). Try the following exercises… Change the following decimal numbers into binary and vice-versa. I have done the first one for you, you do the rest... Make sure you know where your digits are coming from… Decimal | Binary | “Reasoning” | Ex. 234 | 11101010...
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...Unit 1 Lab 1.2: Binary Math and Logic Exercise 1.2.1: Using Figure 1-9 as an example, determine the result of adding 1102 and 10012. | | | | | | 1 | 1 | 0 | | 1 | 0 | 0 | 1 | | 1 | 1 | 1 | 1 | = 15 | Exercise 1.2.2: Using Figure 1-9 as an example, determine the result of adding 1102 and 1012. 1 | | | | | | 1 | 1 | 0 | | | 1 | 0 | 1 | | 1 | 0 | 1 | 1 | = 11 | Exercise 1.2.3: Using Figure 1-9 as an example, determine the result of adding 1112 and 1112. 1 | 1 | 1 | | | | 1 | 1 | 1 | | | 1 | 1 | 1 | | 1 | 1 | 1 | 0 | = 14 | Exercise 1.2.4: Determine the result of 1002 OR 0112. 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | Exercise 1.2.5: Determine the result of 1112 AND 1002. 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | Exercise 1.2.6: Determine the result of NOT 10012. 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | Exercise 1.2.7: Using binary addition, what is the result of 10102 + 102? Using binary addition, how would you repeatedly increment a number by 2? Continue adding 102. | 1 | | | | 1 | 0 | 1 | 0 | | | | 1 | 0 | | 1 | 1 | 0 | 0 | = 12 | | 1 | 1 | 0 | 0 | = 12 | | | | 1 | 0 | | | 1 | 1 | 1 | 0 | = 14 | | | | | | | | 1 | 1 | | | | | 1 | 1 | 1 | 0 | = 14 | | | | 1 | 0 | | 1 | 0 | 0 | 0 | 0 | = 16 | Exercise 1.2.8: 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | Using the AND operator, what is the result of 11002 AND 11112? What can you conclude about using AND on any value with a string of 1s...
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...Binary Numbers Binary numbers are ”Base 2” numbers. Binary numbers follow all the same rules as Base 10 number (our decimal system) Binary numbers are made up of combinations of 0's and 1's Place Value – Base 10 Base 10 numbers have place value... … 1,000's 100's 10's 1's These are powers of 10.... 1 = 100 10 = 101 100 = 102 1000 = 103 etc... Each place value increases by a factor of 10... hence ”Base 10” Place Value – Base 2 Base 2 numbers have place value... … 8's 4's 2's 1's These are powers of 10.... 1 = 20 2 = 21 4 = 22 8 = 23 etc... Each place value increases by a factor of 2... hence ”Base 2” Binary to Decimal Take the binary number 10110011... To convert this to a decimal number, look at each digit and its place value. Start with the right most digit (1). It is in the 1's place so it has a value of 1 (1x1=1). The next right most digit is in the 2's place. It has a value of 2 (1x2=2). The next right most digit is in the 4's place. It has a vlaue of 0 (0x4=0). And so on.... 1x27 + 0x26 + 1x25 + 1x24 + 0x23 + 0x22 + 1x21 + 1x20 = 179 So... 101100112 = 17910 Place Value - again... So, to summarize... We are working with binary numbers as they pertain to IP addresses. Since an IP address contains 4 octets and each octed is a byte (or 8 bits), we need only remember the first 8 place values.... 128 64 32 16 8 4 2 1 Practice... Try convering the following to decimal... 11000011 11110000...
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...Binary Numbers Overview Binary is a number system used by digital devices like computers, cd players, etc. Binary is Base 2, unlike our counting system decimal which is Base 10 (denary). In other words, Binary has only 2 different numerals (0 and 1) to denote a value, unlikeDecimal which has 10 numerals (0,1,2,3,4,5,6,7,8 and 9). Here is an example of a binary number: 10011100 As you can see it is simply a bunch of zeroes and ones, there are 8 numerals in all which make this an 8 bit binary number. Bit is short for Binary Digit, and each numeral is classed as a bit. The bit on the far right, in this case a 0, is known as the Least significant bit (LSB). The bit on the far left, in this case a 1, is known as the Most significant bit (MSB) notations used in digital systems: 4 bits = Nibble 8 bits = Byte 16 bits = Word 32 bits = Double word 64 bits = Quad Word (or paragraph) When writing binary numbers you will need to signify that the number is binary (base 2), for example, let's take the value 101. As it is written, it would be hard to work out whether it is a binary or decimal (denary) value. To get around this problem it is common to denote the base to which the number belongs, by writing the base value with the number, for example: 1012 is a binary number and 10110 is a decimal (denary) value. Once we know the base then it is easy to work out the value, for example: 1012 = 1*22 + 0*21 + 1*20 = 5 (five) 10110 = 1*102 + 0*101 +...
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... public class Binary_Decimal { Scanner scan; int num; void getVal() { System.out.println("Binary to Decimal"); scan = new Scanner(System.in); System.out.println("\nEnter the number :"); num = Integer.parseInt(scan.nextLine(), 2); } void convert() { String decimal = Integer.toString(num); System.out.println("Decimal Value is : " + decimal); } } class MainClass { public static void main(String args[]) { Binary_Decimal obj = new Binary_Decimal(); obj.getVal(); obj.convert(); } } ------------------------------------------- import java.util.Scanner; public class Binary_Octal { Scanner scan; int num; void getVal() { System.out.println("Binary to Octal"); scan = new Scanner(System.in); System.out.println("\nEnter the number :"); num = Integer.parseInt(scan.nextLine(), 2); } void convert() { String octal = Integer.toOctalString(num); System...
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...1. Decimal to binary: 47.375= ( 101101.011 )2 2. Binary to decimal (unsigned): 1011.0102= ( 11.25 ) 3. Base 6 to base 5 2016= ( 243 )5 4. Base 3 to base 7 1023= ( 14 )7 5. Octal to binary 6278= ( 110 010 111 )2 6. Hex to binary OXA3E9= ( 1010 0011 1110 1001 )2 7. Octal to hex 7268=0x( 1D6 ) 8. Base 9 to base 3 8479= ( 22 11 21 )3 9. Base 6 to base 3 1356= ( 2012 )3 10. Decimal to hex 86= 0x( 56 ) 11. Convert the following decimal number to sign and magnitude format (8 bits) -17=10010001 12. Convert the following decimal number to 1’s complement (8 bits) -17=11101110 13. Convert the following decimal number to 2’s complement (8 bits) -17=11101111 14. Convert the following decimal number to IEEE 754 ( 32 bits) 47.37510=0 1000 0100 0111 1011 0000 0000 0000 000 15. Convert the following IEEE 754 to decimal number 0 10000011 010001100000000 00000000= ( 20.75 ) 16. 4 bits unsigned number addition (Overflows or not?) 0101+1100=(1)0001 there is an over flow 17. 4 bits 1’s complement addition (Overflows or not?) 1101+1010= (1)0111 there is an overflow 18. 4 bits 1’s complement addition (Overflows or not?) 1100+1000 (1) 0100 there overflow 19. 4 bits 2’s complement addition (Overflows or not?) 1101+1011=(1)1000 not 20. 4 bits 2’s complement addition (Overflows or not?) 1100+1010(1)0110...
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...Read Only & Binary Files In order to write data to a binary file within a program, you would use the command ‘writeallbytes’. For example: Dim CustomerData As Byte() = (From c In customerQuery).ToArray() My.Computer.FileSystem.WriteAllBytes("C:\MyDocuments\CustomerData", CustomerData, True) The commands displayed above will save the data from an array into the file CustomerData. With using the switch True at the end, tells VB to append the file. If the switch were set to False, then VB would create the file if it did not exist, if it did then it would overwrite everything currently in the file with the data from the array. Const ReadOnly = 1 Set objFSO = CreateObject("Scripting.FileSystemObject") Set objFolder = objFSO.GetFolder("C:\Scripts") Set colFiles = objFolder.Files For Each objFile in colFiles If objFile.Attributes AND ReadOnly Then objFile.Attributes = objFile.Attributes XOR ReadOnly End If Next In order to write data to a “read-only” file, you will start off by declaring the variable that you will use inside of the program for the file. After that you will open the file and run a test on the file using the command line: If obj.File.Attributes AND ReadOnly Then. If the both of the statements return with a true then the file is “read-only,” and will need to be switched before any changes can be made to the file. If the Attributes and the ReadOnly comes back False then the file is not “read-only,” and changes can be made freely to...
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...Decimal-Binary-Hexadecimal Conversion Chart This chart shows all of the combinations of decimal, binary and hexadecimal from 0 to 25 5 decimal. When m aking a change in a C V this chart will show the conversion for different nu mb ering system s. Som e deco ders sp lit the C V in to tw o pa rts. W hen y ou mo dify a CV you need to w rite back all 8 bits. T his cha rt will help deter min e the co rrect bit va lue a C V. Decimal Binary Hex Decimal Binary Hex Decimal Binary Hex Decimal Binary Hex Bit N o.> 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 76543210 00000000 00000001 00000010 00000011 00000100 00000101 00000110 00000111 00001000 00001001 00001010 00001011 00001100 00001101 00001110 00001111 00010000 00010001 00010010 00010011 00010100 00010101 00010110 00010111 00011000 00011001 00011010 00011011 00011100 00011101 00011110 00011111 00100000 00100001 00100010 00100011 00100100 00100101 00100110 00100111 00101000 00101001 00101010 00101011 00101100 00101101 00101110 00101111 00110000 00110001 00110010 00110011 00110100 00110101 00110110 00110111 00111000 00111001 00111010 00111011 00111100 00111101 00111110 00111111 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 20 21 22 23 24 25 26 27 28 29 2A 2B 2C 2D 2E 2F 30 31 32 33 34 35 36 37 38 39 3A 3B 3C 3D 3E 3F 64 65 66 67 68 69 70 71 72 73 74 75 76 77...
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...Ciannavei Lab 2: Number Conversion Lab Task 1: Below is an example that shows how to turn the decimal number ‘125’ into a binary number. 125/2=62 R1 62/2=31 R0 31/2=15 R1 15/2=7 R1 7/2=3 R1 3/2=1 R1 2/1=1 R1 Binary number = 1111101 Task 2: Add correlating weights together to gain decimal value from binary number. 1-2-4-8-16-32-64< Weights 1-1-1-1-1-0-1<Bits 64+32+16+8+4+1=125 Task 3 on next page Gian Ciannavei; Lab 2 Task 3: Below is an example on how to turn the decimal ‘210’ into a hexadecimal using the division by 16 methods. 210/16=13 R2 13 (lsd) 2 (msd) 16-1 <weights 13-2 <Digits=D2 Next is an example of how to turn a hexadecimal into a decimal, in this case, back to ‘210’. 16*13=208 1*2=2 2+208=210 Note: You can also convert the decimal number into binary and turn the binary number into a hexadecimal. 210/2=105 R0 105/2= 52 R1 52/2=26 R0 26/2=13 R0 13/2=6 R1 6/2=3 R0 3/2=1 R1 ½=1 R1 210=11010010 1101=13 0010=2 13(LSD) 2(MSD) =D2 Task 4: Convert hexadecimal number E7 into a decimal. 14(LSD) 7(MSD) =E7 14*16=224 7*1=7 224+7=231 Gian Ciannavei; Lab2 Convert hexadecimal E7 into binary, and then back to decimal to check answers. 14(LSD) 7(MSD) =E7 14*16=224 7*1=7 224+7=231 231/2=115 R1 115/2=57 R1 57/2=28 R1 28/2=14 R0 14/2=7 R0 7/2=3 R1 3/2=1 R1 ½=1 R1 Binary =...
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...2 OR 011 2 . 111 Exercise 1.2.5 1. Determine the result of 111 2 AND 100 2 . 100 Exercise 1.2.6 1. Determine the result of NOT 1001 2 . 0110 Exercise 1.2.7 1. Using binary addition, what is the result of 1010 2 + 10 2 ? Using binary addition, how would you repeatedly increment a number by 2? 1100 Exercise 1.2.8 1. Using the AND operator, what is the result of 1100 2 AND 1111 2 ? What can you conclude about using AND on any value with a string of 1s? 1100 When using AND on any value with a string of 1s, a 1 will be the result. Exercise 1.2.9 1. Using the OR operator, what is the result of 1100 2 OR 1111 2 ? What can you conclude about using OR on any value with a string of 1s? What value can you use with an OR operator to preserve the other input number in the logical equation? 1111 Lab 1.2 Review 1. Determine the result of 10010000 2 + 1101110 2 . Show the mapping that you created to solve this addition problem. 11111110 2. Determine the result of 11001100 2 AND 11111100 2 . Show the mapping or truth table that you used to solve this logic problem. 11001100 3. Determine the result of 11001100 2 OR 11111100 2 . Show the mapping or truth table that you used to solve this logic problem. 11111100 4. Using parentheses in binary works just like it does in decimal, where the operation inside the parentheses should be performed first. With this rule, determine the result of NOT( 11001100 2 AND 11111100 2 ). Show the...
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...What is the difference between data and information: -Data is the lowest level of knowledge and information is the second level -Data by itself alone is not significant. Information is significant by itself. -Observations and recordings are done to obtain data while analysis is done to obtain information. 2. Discuss how computer represents data and basic units of measuring information (up to GB): Computer represents data using binary numeral system: - bit= binary unit (to 256) - bute- 8 bits - kb= 1024 byrtes - mb= 1024 Kb - gb= 1024 mb 3. Compare ASCII, EBCDIC and Unicode: ASCII- American Standard Code for Information Interchange ASCII is a 7 digit (7- bit) code EBCDIC-Extended Binary- Coded Decimal Interchange Code (INM mainframes) EBCDIC- represents each character as a unique combination of 8 bits 8 bites=256 unique combinations Unicode- Universal International coding standard design to represent text based data. Unicode identifies each character using, 0s and 1s. It’s a larger code, consisting of 1 to 4 bytes 4. Discuss the difference between computers: Mainframe-used by large organizations (hospital, university, bank, government offices) -manages a large amount of fertilized data. - located in climate-controlled data centers and connect to the rest of the company computers .via computer network Supercomputer: -$2 mil -\220 million - 50 – 50000 times faster than PC. - A large number of processors interconnected...
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...Gerald Hoover NT1110 1/24/2014 MOTHERBOARDS Motherboards and CPU sizes and shapes are determined by form factor. There are three types of form factors ATX, BTX, AND NLX. ATX FORM FACTOR 1. Replaces older AT’S 2. They are smaller 3. Better organization 4. Supports a wide variety of I/O devices 5. Power supply P1 (20)(24) pin 6. Mice and keyboard connectors consist of PS/2 and USB 7. Memory slots known as DIMM 8. Expansion slots are PCI and AGP 9. Usually has 2 IDE or 2 SATA connectors 10. Soft power button on front of computer A. Does not completely cut power B. Allows normal OS shutdown process 11. Power switch on back of computer A. Immediately kills all power 12. Wake on LAN A. Configures in CMOS B. Keyboard activity C. Network activity BTX FORM FACTOR 1. 20 or 24 pin P1 power connection 2. Better air flow for better cooling 3. Intake vent at front 4. Exhaust vent at back 5. CPU sink fins and memory modules installed parallel to air flow NLX FORM FACTOR 1. Supports riser cards 2. Fits into expansion slots 3. Provides connectors for additional expansion cards 4. Another name is also known as daughter boards ...
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...! 1. Gender Binary: Based off of prior knowledge and information from this class, gender binary is described as the two genders that identify with being either feminine or masculine. Though society often believes in this gender binary, we tend to put people into boxes by categorizing them and it isn’t always correct. In the Radical teacher on page 62 this quote sums up some truth and problems, “the mistaken belief that if your body has/does that thing it is a particular gender.” 2. Intersex: An individual who does not have the typical XY or XX body type. These individuals have a mixture of the two. As we saw in the film XXY, Alex is an example of an intersex person. Most people who are intersex often go through medical procedures to for “normalizing”. Society has this way of making these people seen as something that needs...
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...Deconstruction is a mode of interpretation which works by a careful and cautious entering of each layer within the story: “ The deconstructive critic seeks to find the thread in the text in question which will unravel it all or the loose stone which will pull down the whole building”. Deconstructive literary criticism uses binary oppositions. Binary oppositions can be defined as “ a pair of related terms or concepts that are opposite in meaning.” “Waiting for Godot”, a classic of modern theatre, is a tragicomedy in two acts which tells the story of two men, Vladimir and Estragon, who are waiting to meet a man named Godot. By using deconstructive literary criticism, the play can be analyzed threw the following binary oppositions: passive/active hopelessness/hope, forgetfulness/remembrance and staying/going. Vladimir and Estragon are in a constant state of waiting for Godot: “Nothing to be done. / I'm beginning to come round to that opinion."(Waiting for Godot). Although they are being passive they try to occupy themselves while waiting for Godot. Derrida states that in binary oppositions there is a unspoken hierarchy in which the first term functions as superior to the second term which is considered inferior: “ Derrida’s procedure is to invert the hierarchy in which the first term functions as privileged and superior and the second term as derivative and inferior. By showing that the primary term can be made out to be derivative from or a special case of the secondary term”...
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