Answer:
The correct option is A
Explanation:
In project management, earliest finish time for activity A refers to the earliest start time for succeeding activities such as B and C to start.
Assume that activities A and B comes before C, the earliest finish time for C can be arrived at by computing the earliest start-finish (critical path) of the activity with the largest EF.
That is, if two activities (A and B) come before activity C, one can estimate how long it's going to take to complete activity C if ones knows how long activity B will take (being the activity with the largest earliest finish time).
Cheers!
Answer:
a) EDI Semantic Layer
Explanation:
EDI is an acronym for electronic data interchange and it can be defined as a form of communication between interconnected computer systems and software applications with respect to business informations in standard digital formats.
This ultimately implies that, electronic data interchange (EDI) involves the transfer of business informations such as financial transactions between the computer systems of various organizations such as banks, companies, governmental agencies, etc. Thus, it avails businesses the ability to create strategic communications using computer to computer links to effectively and efficiently exchange business informations electronically or in digital formats.
Hence, the EDI Semantic layer describes the business application that is driving electronic data interchange (EDI).
Answer:
The errors in the loop condition such that it is not giving desired results or it is not running accordingly.There are different types of loop errors which are as following:-
1.Infinite loop:-When the is not able to stop then the error is called infinite loop. for ex:-
int i=1;
while(i!=0)
{
cout<<"I am King"<<endl;
i++;
}
2.Off by one error:-This error mostly happens in loop for arrays as indexing of the array is from 0 to size-1 .So looping over the array up to the size is a off by one error.
3.Equality v/s assignment operator error:-In this error the condition in the loop is like this d=f which is wrong since = is assignment operator it assigns the value of f to d while d==f checks that the value of d and f are equal or not.
4.&& v/s || loop error:- In this error we use and operator (&&) instead of or operator (||) and vice versa.
symptoms of loop errors are not the desired output.
8 kilobytes have addresses from hex 0x0000 to 0x1fff (=8191). You would need 13 bits for that. 2^13-1 = 8191. If you write the highest address in binary, you get:
0001 1111 1111 1111. Just count the number of 1's.
If you find this difficult to visualise, write down all logical addresses of a small page size of e.g. 8 bytes. You can see you can make all addresses with 3 bits.
