Answer:
O(N!), O(2N), O(N2), O(N), O(logN)
Explanation:
N! grows faster than any exponential functions, leave alone polynomials and logarithm. so O( N! ) would be slowest.
2^N would be bigger than N². Any exponential functions are slower than polynomial. So O( 2^N ) is next slowest.
Rest of them should be easier.
N² is slower than N and N is slower than logN as you can check in a graphing calculator.
NOTE: It is just nitpick but big-Oh is not necessary about speed / running time ( many programmers treat it like that anyway ) but rather how the time taken for an algorithm increase as the size of the input increases. Subtle difference.
Answer:
The answer is below
Explanation:
The amount of power dissipated by a processor is given by the formula:
P = fCV²
Where f = clock rate, C = capacitance and V = Voltage
For the old version of processor with a clock rate of f, capacitance C and voltage of V, the power dissipated is:
P(old) = fCV²
For the new version of processor with a clock rate of 20% increase = (100% + 20%)f = 1.2f, capacitance is the same = C and voltage of 20% increase = 1.2V, the power dissipated is:
P(new) = 1.2f × C × (1.2V)² = 1.2f × C × 1.44V² =1.728fCV² = 1.728 × Power dissipated by old processor
Hence, the new processor is 1.728 times (72.8% more) the power of the old processor
That is an interactive question.
The best possible solution for the technician to do is to go into the disk management and find out what exactly is going on. The technician should check whether there is partition that has unallocated space. It is 100% the case that the rest of the 500 GB is in the unallocated space.
The techie need to grow his partition. Possible option for a scenario like this is delete the unallocated 500 GB space using NTFS. He can then recreate the available 500 GB free space as 1TB partition.
