Why Not to Chew Gum in Formation

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  • Topic: Binary numeral system, Hexadecimal, Decimal
  • Pages : 2 (287 words )
  • Download(s) : 66
  • Published : April 25, 2013
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Thomas Turner
June 28,2011
Lab2 document(pg1) (Task 1) Convert decimal number (125) into binary.
125 /2 = 62 remainder5 1(lsd)
62 /2 = 31 remainder0 o
31 /2 = 15 remainder5 1
15 /2 = 7 remainder5 1
7 /2 = 3 remainder5 1
3 /2 = 1 remainder 5 1
1 /2 = .5 remainder 0 1
.5 /2 = 0 remainder 0 0
Convert your answer back to decimal to prove your answer.
0 1 1 1 1 1 0 1
0+64+32+16+8+4+2+1=125
(task 2)
Convert the binary number(10101101) into decimal.
Thomas Turner
June 28,2011
Lab2 document(pg2)
Bites - 1 0 1 0 1 1 0 1
Weights- 128+32+8+4+1=173(decimal)
(task 3)
(1) Convert the decimal number (210) into hexadecimal.
210/16 = 13 remainder 2 (lsd)
13/16 = 0 remainder 13 (msd) 132= D2 (hexadecimal)
(2) Convert your hexadecimal result back into decimal to prove your answer. Weights 16 1
Digits 13 2 (16 * 3) + ( 1 * 2) = 208 + 2 =210(decimal) (task 4)
(1)Convert the hexadecimal E7 into decimal.
E7 = 231
(14*16^1)+(7*16^0)
224+7=231

Thomas Turner
June 28, 2011
Lab2 document(pg3)

(2)Convert the hexadecimal number E7 into binary then convert the binary into a decimal to prove your answers. 231 /2 =115 remainder 5 1
115 /2 =57 remainder 5 1
57 /2 =28 remainder 5 1
28 /2 =14 remainder 0 0
14 /2 =7 remainder 0 0
7 /2 =3 remainder 5 1
3 /2 =1 remainder 5 1
1 /2 =.5 remainder 0 1
Bites- 1 1 1 0 0 1 1 1(binary)
Weights-128+64+32+4+2+1=231(decimal)
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