Answer:
5,10; 6,12; 7,14
Explanation:
We will demonstrate the iteration of the loop:
First iteration: Number = 7, Count = 5 at the beginning. We will check if Count <= Number? Since it is correct, prints 5,10. Increment the Count by 1.
Second iteration: Number = 7, Count = 6. We will check if Count <= Number? Since it is correct, prints 6,12. Increment the Count by 1.
Third iteration: Number = 7, Count = 7. We will check if Count <= Number? Since it is correct, prints 7,14. Increment the Count by 1.
Forth iteration: Number = 7, Count = 8. We will check if Count <= Number? Since it is not correct, the loop stops.
Answer: True
Explanation:
Yes, the given statement is true that the due to the very low feasibility the IT (Information technology) department are not aware of the hidden backlog that basically contain projects.
The feasibility is the main factor in the IT department as it helps to study whole objective of the project and then also uncover all the strength and weakness of the project so that we can easily go through the details to make the project more efficient and reliable.
Answer:
The statement is as follows:
print("{0:,.1f}".format(number))
Explanation:
Required
Statement to print 1234567.456 as 1,234,567.5
To do this, we make use of the format keyword, and we set the print format in the process.
To round up number to 1 decimal place, we use the following format:
"{0:,.1f}"
To include comma in the thousand place, we simply include a comma sign before the number of decimal place of the output; i.e. before 1
"{0:,.1f}"
So, the print statement is:
print("{0:,.1f}".format(number))
Marla can enter Food for examble for A11 on the exl page
