Answer:
Explanation:
For the first iteration of i for loop 1 to n, the j for loop will run from 2 to n times. i.e. n-1 times.
For the second iteration of i for loop, the j for loop will run from 3 to n times. i.e. n-2 times.
From the third to the last iteration of i for loop, the j for loop will run n-1 to n times. i.e. 2 times.
From the second to the last iteration of i for loop, the j for loop will run from n to n times. i.e. 1 time.
For the last iteration of i for loop, the j for loop will run 0 times because i+1 >n.
Hence the equation looks like below:
1 + 2 + 3 + ...... + (n-2) + (n-1) = n(n-1)/2
So the number of total iterations is n(n-1)/2.
There are two operations per loop, i.e. Comparison and Multiplication, so the iteration is 2 * n(n-1)/2 = n ^2 - n
So f(n) = n ^ 2 - n
f(n) <= n ^ 2 for n > 1
Hence, The algorithm is O(n^2) with C = 1 and k = 1.
The answer to this question would be algorithm animation.
Since the algorithm is animate, then you will be able to watch it works. Watching the algorithm executed can give you much information, what the algorithm does and how the algorithm does it.
Normally you will not be able to do this since the algorithm only do what it needs to do without reporting the detail to you
It is true, because if you would try installing another OS after installing the first one, you would overwrite the OS which you installed before.
but in the second opinion an workstation is technically able to dual boot by installing differend operating systems on different drives. which approves that both answers can be right.
Answer:
B) 5 m/s
Explanation:
Took the test and this was the correct answer :)
