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:
1. Declaration: the return type, the name of the function, and parameters (if any)
2. Definition: the body of the function (code to be executed)
Explanation:
Manage the computer's resources, such as the central processing unit, memory, disk drives, and printers, (2) establish a user interface, and (3) execute and provide services for applications software.
Answer:
The sum of all positive even values in arr
Explanation:
We have an array named arr holding int values
Inside the method mystery:
Two variables s1 and s2 are initialized as 0
A for loop is created iterating through the arr array. Inside the loop:
num is set to the ith position of the arr (num will hold the each value in arr)
Then, we have an if statement that checks if num is greater than 0 (if it is positive number) and if num mod 2 is equal to 0 (if it is an even number). If these conditions are satisfied, num will be added to the s1 (cumulative sum). If num is less than 0 (if it is a negative number), num will be added to the s2 (cumulative sum).
When the loop is done, the value of s1 and s2 is printed.
As you can see, s1 holds the sum of positive even values in the arr
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