Answer:
0010
Explanation:
Serially left shifted means that the left most bit will enter the register first. The left most bit already stored in the register will move out of the sequence. The "bold" bits mentioned below highlight these left most bits:
Initial State of the Register:
0000
Group of bits entering:
1011
<u>First Clock Cycle:</u>
0000 <em>(This bold bit will move out)</em>
1011 <em>(This bold bit will move in from right side, shifting the whole sequence one place to the left).</em>
The resulting Sequence:
0001
<u>Second Clock Cycle:</u>
0001 <em>(This bold bit will move out)</em>
1011 <em>(This bold bit will move in from right side, shifting the whole sequence one place to the left).</em>
The resulting Sequence:
0010 <em>(Final Answer)</em>
Answer:
Option d) B is 1.33 times faster than A
Given:
Clock time, 
No. of cycles per instructions, 
Solution:
Let I be the no. of instructions for the program.
CPU clock cycle, 
= 2.0 I
CPU clock cycle, 
= 1.0 I
Now,
CPU time for each can be calculated as:
CPU time, T = 
Thus B is faster than A
Now,
Performance of B is 1.33 times that of A
Use the following rules:
- The sum of currents that enter and exit a node (junction) is always zero. So if you have 3 wires that connect, through one flows 2A, the other 3A, then the third must deliver 5A (taking the direction into account!)
- The sum of voltages across different components should always add up. So if you have a battery of 10V with two unknown resistors, and over one of the resistors is 4V, you know the other one has the remaining 6V.
- With resistors, V=I*R must hold.
With these basic rules you should get a long way!
<span>ALL OF THE ABOVE. The benefits for a CAD program is accuracy, repeatability, simplicity.</span>
Well since it’s a chart based on a PivotTable prettyyyy sure it’s gonna be a PibltChart
