Answer:
A.O(1)
Explanation:
In the implementation of queue by using linked chain the performance of the enqueue operation is O(1).We have to maintain two pointers one head and the other tailand for enqueue operation we have to insert element to the next of the tail and then make that element tail.Which takes O(1) time.
Answer: Have you tried restarting your ipad or Shutting down? Or closing your ipad and waiting for a little bit so It can leave Or you can try Pressing it harder.
Explanation:
Thats all I have to help
Explanation:
A.)
we have two machines M1 and M2
cpi stands for clocks per instruction.
to get cpi for machine 1:
= we multiply frequencies with their corresponding M1 cycles and add everything up
50/100 x 1 = 0.5
20/100 x 2 = 0.4
30/100 x 3 = 0.9
CPI for M1 = 0.5 + 0.4 + 0.9 = 1.8
We find CPI for machine 2
we use the same formula we used for 1 above
50/100 x 2 = 1
20/100 x 3 = 0.6
30/100 x 4 = 1.2
CPI for m2 = 1 + 0.6 + 1.2 = 2.8
B.)
CPU execution time for m1 and m2
this is calculated by using the formula;
I * CPI/clock cycle time
execution time for A:
= I * 1.8/60X10⁶
= I x 30 nsec
execution time b:
I x 2.8/80x10⁶
= I x 35 nsec
Answer:
Running RECURSIVE-MATRIX-CHAIN is asymptotically more efficient than enumerating all the ways of parenthesizing the product and computing the number of multiplications of each.
the running time complexity of enumerating all the ways of parenthesizing the product is n*P(n) while in case of RECURSIVE-MATRIX-CHAIN, all the internal nodes are run on all the internal nodes of the tree and it will also create overhead.
Explanation:
I would say two car length rule. I am not sure what the official license rule is or if it has been changed, but originally the rule was 3 seconds usually depending on how fast the car is going. The faster you are going, the longer it takes to stop. So two-car length rule would probably be the best choice. Definitely not A.
