Answer:a)Stand-alone
Explanation: Stand-alone application is the application that is found on the system of every client.In accordance with the IT section, Business intelligence can be transferred into stand-alone application .This helps in the development of the essence of the system at an independent level.
Other options are incorrect because supporting a certain factor will not make it independent, cannot act as the group of ISs technology or web system for gaining profit.Thus,the correct option is option(a).
Answer:
Time to Live (or TTL)
Explanation:
This counter field is initially set at some value that is decremented by one each time the packet "hops." When the counter reaches zero the packet is disposed of.
Some of the online activities among businesses are:
- SEO consultant. ...
- Web designer or web developer. ...
- Blogger. ...
- Virtual assistant. ...
- Affiliate marketer, etc
<h3>What are Online Activities?</h3>
This refers to the various activities that are done on the world wide web and is usually used by businesses to increase visibility, and in turn, revenue.
Hence, we can see that the use of online activities by businesses are important because the products and services for sale by businesses are advertised to target audiences through some of the aforementioned activities.
Read more about online activities among businesses here:
brainly.com/question/27172895
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This isn’t even a question it’s just instructions for a question. can you elaborate???
Answer:
Explanation:
The following code is written in Java. Both functions traverse the linkedlist, until it reaches the desired index and either returns that value or deletes it. If no value is found the function terminates.
public int GetNth(int index)
{
Node current = head;
int count = 0;
while (current != null)
{
if (count == index)
return current.data;
count++;
current = current.next;
}
assert (false);
return 0;
}
public int removeNth(int index)
{
Node current = head;
int count = 0;
while (current != null)
{
if (count == index)
return current.remove;
count++;
current = current.next;
}
assert (false);
return 0;
}
