Answer:
The while loop is used and the factors of the integer are computed by using the modulus operator and checking if the remainder of the number divided by i is 0. 3. Then the factors of the integer are then again checked if the factor is prime or not.
Explanation:
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Explanation:
the objective of ITIL service operations is to make sure that IT services are delivered effectively and efficiently. the service operation life cycle stage includes the fulfilling of user requests, resolving service failure fixing problems and also carrying out routine operational tasks
Answer:
This code will print: 4
Explanation:
Following is the step-by-step explanation for the given code:
- Given is the array of data type double named myList, it has entries, 1, 5, 5, 5,5, 1:
double[] myList = {1, 5, 5, 5, 5, 1};
- Now the first element of the array (1) with index 0 will be stored in the variable max (data type double).
double max = myList[0];
- A variable indexOfMax having datatype int will be initiated as 0.
int indexOfMax = 0;
- Now for loop will be used to find the maximum number of the array. The variable i will be put as index for each element to compare with first element. If the checked element is greater than or equal to the integer in max, it will be replaced. So at the end the variable max will have value 5 that will be at index i = 4.
for (int i = 1; i < myList.length; i++)
{ if (myList[i] >= max)
{ max = myList[i];
- Now the variable i that is the index for max value will be stored in the variable indexOfMax (indexOfMax = 4).
indexOfMax = i; }}
- At end the value stored in variable indexOfMax will be printed, so 4 will be printed as output.
System.out.println(indexOfMax);
i hope it will help you!
Answer:
Answer is B, see explanations.
Explanation:
Answer is (B) NP−complete∩P=ϕ
Since, P≠NP, there is at least one problem in NP, which is harder than all P problems. Lets take the hardest such problem, say X. Since, P≠NP,X∉P .
Now, by definition, NP−complete problems are the hardest problems in NP and so X problem is in NP−complete. And being in NP, X can be reduced to all problems in NP−complete, making any other NP−complete problem as hard as X. So, since X∉P, none of the other NP−complete problems also cannot be in P.