Consider the modules A and B of the previous question
A. | C (a + b) > Ca + Cb |
B. | C (a + b) < Ca +Cb |
C. | C ( a + b) = Ca + Cb |
D. | None of the above |
Option: A Explanation : Click on Discuss to view users comments. |
Consider the above modules Ca and Cb where a and b are the sizes. Then
A. | E (a + b) > Ea + Eb |
B. | E (a + b) < Ea +Eb |
C. | E (a + b) > =Ea + Eb |
D. | None of the above |
Option: A Explanation : Click on Discuss to view users comments. |
Variation in debugging : coding ability has been reported to be
A. | 1:10 |
B. | 1:1 |
C. | 1:2 |
D. | 1:5 |
Option: A Explanation : Click on Discuss to view users comments. |