# Labs 1.1-1.4

**Topics:**Binary numeral system, Character encoding, Unicode

**Pages:**9 (586 words)

**Published:**April 24, 2014

Mapping for decimal number 2931

10^3 10^210^110^0

2931

2x1000=2000+9x100=900+3x10=30+1x1=1 = 2931

Exercise 1.1.2

Mapping for binary number 110 base 2

421

***

110

===

4 +2 +0= 6

Exercise 1.1.3

Mapping for binary number 11 base 2

21

**

11

==

2 +1= 3

Exercise 1.1.4

Mapping for binary number 10010 base 2

168421

*****

10010

=====

16 +0 +0 +2 +0= 18

Exercise 1.1.5

Mapping for binary number 11100010 base 2

1286432168421

********

11100010

========

128643200020= 226

Exercise 1.1.6

Binary conversion for decimal 156 base 10

156-128=28-16=12-8=4-4=0

128 64 32 168421

1 0011100

Exercise 1.1.7

Binary conversion for decimal 255

255-128=127-64=63-32=31-16=15-8=7-4=3-2=1-1=0

1286432168421

1111111`1

Exercise 1.1.8

Binary conversion for decimal 200

200-128=72-64=8-8=0

1286432168421

11001000

Lab 1.1 Reviews

1)Convert the decimal value 127 to binary.

127-64=63-32=31-16=15-8=7-4=3-2=1-1=0

6432168421

1111111

2)Explain why values 10 base 2 and 0010 base 2 are equivalent- Two digits start over, and next digit to convert from a base 10 integer numeral to its base 2 equivalent. 3)When you multiply 2 by the power of you get 1,2,4,8,16,32 etc. 4)It wouldn’t be that difficult you just have to ally the correct formulas.

Exercise 1.2.1

Binary addition for 110 base 2 + 1001 base 2

1001

+++

110

===

1111

Exercise 1.2.2

Adding binary 110 base 2 + 101 base 2

110

+++

101

===

1111 One plus one means you carry the one its one zero not 10

Exercise 1.2.3

Adding binary 111 base 2 and 111 base 2

Carry11

111

+++

111

===

1110 one plus one plus one is one one to be carried

Exercise 1.2.4

Determine the result of 100 base 2 OR 011 base 2

100

+++

011

===

111

Exercise 1.2.5

Determine the result of 111 base 2 AND 100 base 2

111

+++

100

===

100

Exercise 1.2.6

Determine the result of NOT 1001 base 2

0110

Exercise 1.2.71010 base 2 + 10 base 2

1 Carried

1010

++

10

====

1110

Exercise 1.2.8 1100 base 2 AND 1111 base 2

1100

1111

====

1100

I conclude that when using AND with a string of 1’s the answer is 1 Exercise 1.2.9

1100 base 2 OR 1111 base 2

1100

++++

1111

====

1111

If there is a 1 present the answer is 1.

Lab 1.2 Reviews

10010000 base 2 + 1101110 base 2

1)10010000

+++++++

1101110

========

11111110

11001100 base 2 AND 11111100 base 2

2)11001100

++++++++

11111100

========

11001100

11001100 base 2 OR 11111100 base 2

3)11001100

++++++++

11111100

========

11111100

NOT 11001100 base 2 AND 11111100 base 2

4)11001100

++++++++

11111100

========

11001100

NOT =

00110011

Exercise 1.3.1

Decimal value for Byte 1 = 65,208

Decimal value for Byte 2= 255

Exercise 1.3.2

The decimal value with them combined is 65,463

It is lager when combined.

Exercise 1.3.3

120MB x 1024 bytes = 122,880 bytes

Exercise 1.3.4

16 x 1024 = 16,384MB x 1024 = 16,777,216kb x 1024 = 1.717986918 x 10^10 bytes x 8 = 1.374389535 x 10^011 bits / 32 = 4,294,967,296

Exercise 1.3.5

Binary value of 110110 base 2 in hexadecimal

32168421

110110

_____________________________

6c base 16

Exercise 1.3.6

Hexadecimal f6 base 16 = binary 11110110

Decimal 246

Lab 1.3 Reviews

1)This way you will be familiar with sizing a hard drive for what it will be used for whether...

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