Answer:
Required memory size is 16k x 8
16k = 24 x 210 = 214
Hence, No. of address lines = 14
No. of data lines = 8
a) Size of IC 1024 x 1
Total number of ICs required = 16k x 8 / 1024 x 1 = 16 x 8 = 128
b) Size of IC 2k x 4
Total number of ICs required = 16k x 8 / 2k x 4 = 8 x 2 = 16
c) Size of IC 1k x 8
Total number of ICs required = 16k x 8 / 1k x 8 = 16 x 1 = 16
Explanation:
For a, 10 address lines from A0 to A9 are used to select any one of the memory location out of 1024 memory locations present in a IC.
For b, 11 address lines from A0 to A10 are used to select any one of the memory location out of 2k=2048 memory locations present in a IC.
For c, 10 address lines from A0 to A9 are used to select any one of the memory location out of 1k=1024 memory locations present in a IC.
Hi;
In the question, Robert gives the explanation that there is an error in the BIOS. A BIOS (Standing for Basic Input & Output System) is a ROM chip, and is vital for the computer to initialize devices such as RAM, the CPU, etc. If there is ever an error there, a computer simply cannot boot.
From the options given, your answer given would be C. ROM.
I hope this helps!
Researches at symantec have uncovered a version of the stuxnet computer virus that was used to attack irans nuclear program in November.
Answer: yes because it helps
Explanation: it shows everything