Tutorial 5 solution

Tutorial 5 solution - Name: _ ID: _ Hand in this sheet...

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Name: _______________________ ID: __________________________ Page 1 of 2 Hand in this sheet before leaving class. 1. Convert to base 10. a) (1101010.11) 2 1 × 2 6 + 1 × 2 5 + 1 × 2 3 + 1 × 2 1 + 1 × 2 –1 + 1 × 2 –2 = 64 + 32 + 8 + 2 + 0.5 + 0.25 = 106.75 b) (675.32) 8 6 × 8 2 + 7 × 8 1 + 5 × 8 0 + 3 × 8 -1 + 2 × 8 –2 = 384 + 56 + 5 + 0.375 + 0.03125 = 445.40625 c) (AF9.C) 16 = A × 16 2 + F × 16 1 + 9 × 16 0 + C × 16 -1 = 10 × 16 2 + 15 × 16 1 + 9 × 16 0 + 12 × 16 -1 = 2560 + 240 + 9 + 0.75 = 2809.75 2. Convert to binary: (17.54) 8 17.54 <write each digit in binary> = (001)(111).(101)(100) = 1111.1011 2 <chuck away extra zeros> 3. Convert to hexidecimal: (10110.101) 2 10110.101 <group into 4-digit chunks> = 00010110.1010 <convert each chunk to hex> = 1 6 10 = 16.A 16 <use single characters> 4. Convert to octal: (675.2) 10 675 ÷ 8 = 84 R 3 84 ÷ 8 = 10 R 4 10 ÷ 8 = 1 R 2 1 ÷ 8 = 0 R 1 <stop when you get 0> 0.2 × 8 = 1 . 6 0. 6 × 8 = 4 . 8 0. 8 × 8 = 6 . 4 0. 4 × 8 =
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This note was uploaded on 02/23/2010 for the course MATH 2070 taught by Professor Aruliahdhavidhe during the Spring '10 term at UOIT.

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Tutorial 5 solution - Name: _ ID: _ Hand in this sheet...

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