Answer:
low transaction fees and speedier processing
Explanation:
I think it the answer would be B but I'm not sure
it is called an attribute
Answer:
22
Explanation:
1. We are going to have at hand 32 Enqueue Operation, with 10 from front and 15 dequeue and 5 empty queue operation
2. You dequeued total number of 15-5 =10 elements since 5 dequeue did not change the state of the queue, so invariably 10 dequeue is done.
3. Next is to enqueued a total of 32 elements.
Enqueue Operation do not change the state(and Size) of the queue, and can be ignored.
4. To arrive at the Total Size of queue, we will have 32-10 = 22 at the end
Answer : 22 because its a 5 dequeue Operations.
Answer:
Opcode = 3
Mode =2
RegisterRegister =7
AR = 20
Explanation:
a) Number of addressing modes = 4 = 22 , So it needs 2 bits for 4 values
Number of registers = 65 = 1000001 in binary , So it needs 7 bits
AR = 20
Bits left for opcode = 32 -(2+7+20) = 3
