Essentials of Computer Organization and Architecture
5th Edition
ISBN: 9781284123036
Author: Linda Null
Publisher: Jones & Bartlett Learning
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Chapter 4, Problem 18E
Program Description Answer
The 2048 bytes with 64×8 RAM chips must have “11 address bits” which has “5 bits” for chip selector and “6 bits” for address on chip.
Hence, the correct diagram is option “E” which contains the appropriate bit address and chip address for given memory.
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I need help understanding how could I got the IP Address for the 1st Host on this Subnet, the IP Address for the Last Host on this Subnet & the Broadcast Address for this Subnet
My Explanation:8 bits --> 17 & 0
128 64 32 16 | 8 4 2 1 0 0 0 1 0 0 0 1 -->17+ 0 0 0 0 0 0 0 0 --> 0 0 0 0 0 0 0 0 0 --> 0 (The Major (Base) Network Address): 142.20.33.0
To find the broadcast address, we can set all the bits in the host portion of the address to 1, which gives us 142.20.33.255. Based on the subnet mask 255.255.255.0, no bits were borrowed from the host portion, meaning 8 bits remain in the host portion. Using the formula 2^h – 2 (where h represents the number of host bits), we get 2^8 - 2 = 254 hosts.
8 bits --> 17 & 252
128 64 32 16 | 8 4 2 1 0 0 0 1 0 0 0 1 --> 17+ 1 1 1 1 1 1 0 0 --> 252 0 0 0 1…
Which of the following opens when you click the launcher in the Size group on the Ribbon?
Question 19Select one:
a.
Size dialog box
b.
Layout dialog box
c.
Width and Height dialog box
d.
Format dialog box
How do you resize a graphic object horizontally while keeping the center position fixed?
Question 20Select one:
a.
Drag a side sizing handle.
b.
Press [Ctrl] and drag a side sizing handle.
c.
Press [Alt] and drag a side sizing handle.
d.
Press [Shift] and drag a side sizing handle.
Chapter 4 Solutions
Essentials of Computer Organization and Architecture
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