nly Q4 needed

Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Only Q4 needed
1. Convert
(i)
(16)16 = ( )10
(ii) (187)10 = ()2
(iii) (111111110010)2 = ( )16
(iv) (01011)2 = 10
(v) (1001.0010)2 = ( )10
(vi) (8B3F)16 =( )2
(vii) (17)8 = ( )10
(viii) (010111100)2 = ( )8
(ix) (534.25)10 = )2
(x) (1026.2)s = )10
(xi) (2AEA.D6)16 = ( )2
%3D
%3D
2. Multiply following
(i)
(ii)
(iii)
(iv)
(v)
1111 by 1100
100 by 011
110010 by 10101
1010 by 1010
(73)s by (13)s
3. Divide following
11010 by 11
1111001 by 1001
(i)
(ii)
4. Represent -136.235 as the floating-point number using 6-bit register.
Transcribed Image Text:1. Convert (i) (16)16 = ( )10 (ii) (187)10 = ()2 (iii) (111111110010)2 = ( )16 (iv) (01011)2 = 10 (v) (1001.0010)2 = ( )10 (vi) (8B3F)16 =( )2 (vii) (17)8 = ( )10 (viii) (010111100)2 = ( )8 (ix) (534.25)10 = )2 (x) (1026.2)s = )10 (xi) (2AEA.D6)16 = ( )2 %3D %3D 2. Multiply following (i) (ii) (iii) (iv) (v) 1111 by 1100 100 by 011 110010 by 10101 1010 by 1010 (73)s by (13)s 3. Divide following 11010 by 11 1111001 by 1001 (i) (ii) 4. Represent -136.235 as the floating-point number using 6-bit register.
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