Answer:
x=arr[arr.length-2]; is the correct answer for the given question.
Explanation:
In the above statement firstly we calculate the length of the given array.The arr.length function is used to calculate the array length.As mention in the question we have to assign the next to last element of the array to the variable x, so we used arr[arr.length-2] and finally, we assign them to the variable x.
so the final statement is x=arr[arr.length-2];
Answer:
a
Explanation:
You know that if you searched brianly for the answer some one else has already answered the same question jut trynna help some one out to help save your points
<h2>
Answer:</h2>
Option b, c, d are true.
The correct statements are as follows:
<h3>b. Can restrict a computer from receiving network traffic
</h3><h3>c. Stops attackers when they are outside of the company's internal network
</h3><h3>d. Stop a process or application from launching</h3><h3 /><h2>
Explanation:</h2>
Firewall can be defined as a wall or checkpoint that checks each entity before it accesses to go inside or outside a network.
A firewall:
- makes sure that the data inside a private network is safe by building a wall around it.
- restricts the malicious data to go inside the network.
- always ask about launching an application which is blocked by it due to suspicious activity.
A firewall can not:
- prevent a system from being fingerprinted by port scans.
- Disable an account.
<h3>I hope it will help you!</h3>
Answer:
this say's "Which device group?" hope i helpped
While proceeding with an internet search I find google to be the best source considering its high accuracy. When asking your question keep it minimal yet descriptive. After having found a few sources cross reference them with each other and throw out any sources that may hold false information. After having narrowed them down look to see how recent they are because since their publication new information may have been brought to light. Look to see if they have links to where they got their information. All in all just be diligent.