The Memory Address Register (MAR) holds the address of the current instruction or data that is to be fetched from or transferred to memory.
CloseFirst, evaluate each part of the expression separately:
The NOR operation returns true only if both operands are false. Therefore, false NOR true is false (0).
The value of the expression is 0.
Close(a)
(b)
Truth table for the logic expression A NAND (B NOR C):
A | B | C | B NOR C | A NAND (B NOR C) |
---|---|---|---|---|
0 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 0 | 1 |
0 | 1 | 0 | 0 | 1 |
0 | 1 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 0 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 0 |
Cache memory plays a crucial role in enhancing system performance by:
(a)
(b)
Truth table for the logic expression (X XOR Y) AND Z:
X | Y | Z | X XOR Y | (X XOR Y) AND Z |
---|---|---|---|---|
0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 0 |
A | B | C | N | M | O | Z |
---|---|---|---|---|---|---|
0 | 0 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 1 | 0 | 1 |
0 | 1 | 0 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 0 | 1 | 0 |
1 | 0 | 0 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 1 | 1 | 0 |
1 | 1 | 1 | 0 | 0 | 1 | 0 |
The answer can include the following examples:
The answer can include the following examples:
First, evaluate each part of the expression separately:
The XOR (exclusive OR) operation returns true if exactly one of the operands is true. Therefore, false XOR false is false (0).
The value of the expression is 0.
CloseThe answer can include the following examples:
The value of the expression is 0.
CloseThe answer can include the following examples:
The answer can include the following examples:
3 in hexadecimal is 0011 in binary.
C in hexadecimal is 1100 in binary.
3C in hexadecimal is thus 0011 1100 in binary.
A | B | C | A OR B | NOT (A OR B) | NOT (A OR B) NAND C |
---|---|---|---|---|---|
0 | 0 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 0 | 1 | 0 |
0 | 1 | 0 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 0 | 1 |
1 | 1 | 0 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 0 | 1 |
Hexadecimal uses 16 possible values for each digit. Thus, the total number of different combinations that can be formed with two hexadecimal characters is the product of the possible values for each digit, which is \( 16 \times 16 = 256 \).
(a)
(b)
(c)
10011100 can be divided into 1001 and 1100.
1001 in binary is 9 in hexadecimal.
1100 in binary is C in hexadecimal.
The hexadecimal equivalent of the binary number 10011100 is 9C.
a) Each color uses 8 bits, so: \( 8 \times 3 = 24 \)
b) We have 256 possibilities for 3 colors, so: \( 256^3 = 16777216 \)
A | B | C | X | Y | Z |
---|---|---|---|---|---|
0 | 0 | 0 | 1 | 0 | 1 |
0 | 0 | 1 | 1 | 0 | 1 |
0 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 1 |
1 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 1 | 1 |
A | B | C | A OR B | NOT (A OR B) | NOT (A OR B) NAND C |
---|---|---|---|---|---|
A | B | C | X | Y | Z |
---|---|---|---|---|---|