Answer:
yes the internet is a type of wan
Explanation:
hope this helps
Answer:
31 bits.
Explanation:
Given, total number of registers = 55
Total instructions = 60
Size of memory = 16 KB
Now, no of registers are 55. We find the next greater or equal power of 2 which is 64 = 26. Hence, 6 bits are required to represent a register operand.
Number of instructions = 60. We find the next greater or equal power of 2 which is 64 = 26. Hence, 6 bits are required to represent a instruction.
Size of memory = 64 KB = 26 * 210 * 23 bits = 219 bits. Hence, 19 bits are required to represent a memory location.
Now, an instruction has 2 parts, opcode and operand. As given there are only two address instructions which are memory operand and register operand.
Hence, total bits would be: 6 bits (opcode) + 6 bits (register operand) + 19 (memory operand) = 31 bits.
There are the secound hand , the hour hand, the minute hand, and the numbers on the face of the clock.
<span>public class ExampleMinNumber <span>{
</span></span><span> public static void main(String[] args) <span>{
</span></span>int a = 11<span>;
</span> int b = 6<span>;
</span><span> int c = minFunction(a, b);
</span><span> System.out.println("Minimum Value = " + c<span>);
}
</span></span><span>/** returns the minimum of two numbers */
</span><span>public static int minFunction(int n1, int n2) <span>{
</span></span>int min<span>;
</span>if (n1 > n2<span>)
</span>min = n2<span>;
else
min = n1;
return min;
}
}
</span>java 8.0