Answers are at the end of the chapter. 1. What weight does the digit 7 have in each of the following numbers? (a) 1370 (b) 6725 (c) 7051 (d) 58.72 2. Express each of the following decimal numbers as a sum of the products obtained by multiplying each digit by its appropriate weight: (a) 51 (b) 137 (c) 1492 (d) 106.58

Computer Networking: A Top-Down Approach (7th Edition)
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Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
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SECTION 2-1 CHECKUP
Answers are at the end of the chapter.
1. What weight does the digit 7 have in each of the following numbers?
(a) 1370 (b) 6725 (c) 7051 (d) 58.72
2. Express each of the following decimal numbers as a sum of the products obtained by
multiplying each digit by its appropriate weight:
(a) 51 (b) 137 (c) 1492 (d) 106.58
SECTION 2-2 CHECKUP
1. What is the largest decimal number that can be represented in binary with eight bits?
2. Determine the weight of the 1 in the binary number 10000.
3. Convert the binary number 10111101.011 to decimal.
I
SECTION 2-3 CHECKUP
1. Convert each decimal number to binary by using the sum-of-weights method:
(a) 23 (b) 57 (c) 45.5
2. Convert each decimal number to binary by using the repeated division-by-2 method
(repeated multiplication-by-2 for fractions):
(a) 14
(b) 21
(c) 0.375
SECTION 2-4 CHECKUP
1. Perform the following binary additions:
(a) 1101 + 1010
(b) 1011101101
2. Perform the following binary subtractions:
(a) 1101 - 0100
(b) 1001 - 0111
3. Perform the indicated binary operations:
(a) 110 x 111
(b) 1100 011
Transcribed Image Text:SECTION 2-1 CHECKUP Answers are at the end of the chapter. 1. What weight does the digit 7 have in each of the following numbers? (a) 1370 (b) 6725 (c) 7051 (d) 58.72 2. Express each of the following decimal numbers as a sum of the products obtained by multiplying each digit by its appropriate weight: (a) 51 (b) 137 (c) 1492 (d) 106.58 SECTION 2-2 CHECKUP 1. What is the largest decimal number that can be represented in binary with eight bits? 2. Determine the weight of the 1 in the binary number 10000. 3. Convert the binary number 10111101.011 to decimal. I SECTION 2-3 CHECKUP 1. Convert each decimal number to binary by using the sum-of-weights method: (a) 23 (b) 57 (c) 45.5 2. Convert each decimal number to binary by using the repeated division-by-2 method (repeated multiplication-by-2 for fractions): (a) 14 (b) 21 (c) 0.375 SECTION 2-4 CHECKUP 1. Perform the following binary additions: (a) 1101 + 1010 (b) 1011101101 2. Perform the following binary subtractions: (a) 1101 - 0100 (b) 1001 - 0111 3. Perform the indicated binary operations: (a) 110 x 111 (b) 1100 011
ANSWERS
SECTION CHECKUPS
Section 2-1 Decimal Numbers
1. (a) 1370: 10 (b) 6725: 100
2. (a) 51 = (5 × 10) + (1 × 1)
(b) 137 (1 X 100) + (3 × 10) + (7 × 1)
(c) 7051: 1000 (d) 58.72: 0.1
(c) 1492 = (1 x 1000) + (4 × 100) + (9 × 10) + (2 × 1)
(d) 106.58 = (1 x 100) + (0 × 10) + (6 × 1) + (5 × 0.1) + (8 × 0.01)
Number Systems, Operations, and Codes
Section 2-2 Binary Numbers
1.28 - 1 = 255
2. Weight is 16.
3. 10111101.011 = 189.375
Section 2-3 Decimal-to-Binary Conversion
1. (a) 23 10111
(b) 57 111001
10101
2. (a) 14 = 1110
(b) 21
Section 2-4 Binary Arithmetic
1. (a) 1101 + 1010 10111
2. (a) 1101 - 0100
1001
101010
3. (a) 110 X 111 =
(c) 45.5
(c) 0.375
101101.1
0.011
(b) 10111 + 01101 = 100100
(b) 10010111 = 0010
(b) 1100 011 = 100
Transcribed Image Text:ANSWERS SECTION CHECKUPS Section 2-1 Decimal Numbers 1. (a) 1370: 10 (b) 6725: 100 2. (a) 51 = (5 × 10) + (1 × 1) (b) 137 (1 X 100) + (3 × 10) + (7 × 1) (c) 7051: 1000 (d) 58.72: 0.1 (c) 1492 = (1 x 1000) + (4 × 100) + (9 × 10) + (2 × 1) (d) 106.58 = (1 x 100) + (0 × 10) + (6 × 1) + (5 × 0.1) + (8 × 0.01) Number Systems, Operations, and Codes Section 2-2 Binary Numbers 1.28 - 1 = 255 2. Weight is 16. 3. 10111101.011 = 189.375 Section 2-3 Decimal-to-Binary Conversion 1. (a) 23 10111 (b) 57 111001 10101 2. (a) 14 = 1110 (b) 21 Section 2-4 Binary Arithmetic 1. (a) 1101 + 1010 10111 2. (a) 1101 - 0100 1001 101010 3. (a) 110 X 111 = (c) 45.5 (c) 0.375 101101.1 0.011 (b) 10111 + 01101 = 100100 (b) 10010111 = 0010 (b) 1100 011 = 100
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