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Unformatted text preview: Fill those bits with 1 and the rest with 0’s 1 0 1 0 0 0 1 0 1 = 101000101 2 2 8 2 7 2 6 2 5 2 4 2 3 2 2 2 1 2 b) 111001001 2 to decimal As mentioned before, each bit represents a power of 2. Multiply these powers of 2 by the number in the slot then add the results together. 111001001 = 1 * 2 8 +1 * 2 7 +1 * 2 6 +0 * 2 5 +0 * 2 4 +1 * 2 3 +0 * 2 2 +0 * 2 1 +1 * 2 0 = 256 + 128 + 64 + 8 + 1 = 457 10 c) 0x4C3 to octal Binary can easily be converted between other powers of base 2 by grouping. First, we convert to binary x4 = 0100, xC = 1100, x3 = 0011 Thus the binary is: 0100 1100 0011 Next, we regroup the bits into 3’s, as 2 3 = 8 010 011 000 011 2 3 0 3 = 2303 8 d) 1062 7 to base 13 First we must convert to base 10 1062 7 = 1 * 7 3 + 0 * 7 2 + 6 * 7 1 + 2 * 7 = 343 + 0 + 42 + 2 = 387 10 Next we convert to base 13 387 / 13 = 29 r10 (any digit larger than 9 we use letters) 29 / 13 = 2 r3 2 / 13 = 0 r2 2 3 A = 23A 13...
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This note was uploaded on 01/06/2009 for the course CMPE 12 taught by Professor Diblas during the Fall '08 term at UCSC.
 Fall '08
 DIBLAS

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