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Answer to exercises(2)

Answer to exercises(2). P74, 2.1 (a) 1101011 2 =6B 16. (b) 174003 8 =1 111 100 000 000 011 2. (d) 67.24 8 =110 111.010 1 2. (e) 10100.1101 2 =14.D 16. (f) F3A5 8 =1111 0011 1010 0101 2. (i) 101111.0111 2 =57.34 8. (j) 15C.38 16 =1 0101 1100.0011 1 2. Answer to exercises(2).

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Answer to exercises(2)

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  1. Answer to exercises(2) P74, 2.1 (a) 11010112=6B16 (b) 1740038=1 111 100 000 000 0112 (d) 67.248=110 111.010 12 (e) 10100.11012=14.D16 (f) F3A58=1111 0011 1010 01012 (i) 101111.01112=57.348 (j) 15C.3816=1 0101 1100.0011 12

  2. Answer to exercises(2) P74, 2.2 (e) 5436.158=101 100 011 110.001 1012 =B1E.3416 (f) 13705.2078=1 011 111 000 101.010 000 1112 =17C5.43816 P74, 2.3 (d) C35016=1100 0011 0101 00002 =1415208 (e) 9E36.7A16=1001 1110 0011 0110 .0111 1012 =117066.3648

  3. Answer to exercises(2) P75, 2.4 123456701238 = 1 010 011 100 101 110 111 000 001 010 0112 = 01 010 01110 010 11101 110 00001 010 0112  The octal values of the four 8-bit bytes are: 1238 2278 1608 1238 P75, 2.5 (d) 67.248=681+780+28-1+48 –2=55.312510 (e) 10100.11012= 124+122.12-1+ 12-2+ 12-4 = 20.812510 (j) 15C.3816= 1162+5161+C160.316-1+816-2 = 348.2187510

  4. (1 2 727 125 3489 62 (2 (1 (0 (4 (0 2 5 8 8 5 (1 29 54 31 2 436 145 (1 15 2 (1 2 7 (6 (0 (4 (6 8 5 8 5 3 5 (1 (1 2 1 0 6 5 0 (1 1 2 0 Answer to exercises(2) P75, 2.6 (a) 12510=?2 (g) 72710=?5 (b) 348910=?8 348910=66418 72710=104025 12510=11111012

  5. (D 61453 16 3840 (0 16 (0 240 16 15 16 (F 0 Answer to exercises(2) P73, 2.31 P76, 2.15 There are 28 different 3-bit binary encodings are possible for the traffic-light controller of Table 2-12. 6145310 =F00D16 So the result “FOOD” can whet your appetite.

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