Answer:
There is no table, so I can only comment on the statements:
The binary value of decimal 10 is A. ==> False, however A is a hexadecimal representation of 10.
The binary value of decimal 13 is 1001 ==> False, 13 would be 1101.
The binary value of decimal 15 is 1111. ==> True.
The binary value of decimal 14 is E. ==> Again E is a hexadecimal representation of 14.
Lsass.exe /................................................
Answer:
The correct answer to the following question will be Error-detection.
Explanation:
Error-detection: The detection of errors caused during the transmission from the transmitter to the receiver by damage and other noises, known as Error-detection. This error-detection has the ability to resolute if something went wrong and if any error occurs in the program.
There are mainly three types of error-detection, these types can be followed:
- Automatic Repeat Request (ARQ)
- Forward Error Correction
- Hybrid Schemes
There are two methods for error-detection, such as:
- Single parity check
- Two-dimensional parity check
The answer is the team. Teams are self-establishing which no one not even the scrum master tells the team how to turn product backlog into augmentations of stoppable functionality. Each team member put on his or her knowledge to all of the complications. The interaction that outcomes progresses the entire scrum team’s general competence and usefulness. The optimal size for a scrum team is seven people, plus or minus two. When there are fewer than five team members, there is less interface and as a result less productivity improvement. The product owner and scrum master roles are not comprised in this count. Team arrangement may variation at the end of a sprint. Every time Team membership is altered, the productivity increased from self-organization is reduced.
Answer:
#include <iostream>
using namespace std;
void divide(int numerator, int denominator, int *quotient, int *remainder)
{
*quotient = (int)(numerator / denominator);
*remainder = numerator % denominator;
}
int main()
{
int num = 42, den = 5, quotient=0, remainder=0;
divide(num, den, "ient, &remainder);
return 0;
}
Explanation:
The exercise is for "Call by pointers". This technique is particularly useful when a variable needs to be changed by a function. In our case, the quotient and the remainder. The '&' is passing by address. Since the function is calling a pointer. We need to pass an address. This way, the function will alter the value at the address.
To sum up, in case we hadn't used pointers here, the quotient and remainder that we set to '0' would have remained zero because the function would've made copies of them, altered the copies and then DELETED the copies. When we pass by pointer, the computer goes inside the memory and changes it at the address. No new copies are made. And the value of the variable is updated.
Thanks! :)
