Answer:
Option (4) is the correct answer.
Explanation:
In Java programming language ,array collection starts from 0 index location and ends in a size-1 index location. So to access the last elements the user needs to use a[Size-1] statement. so to modify the value of the last location of the array the user needs to use "a[size-1]= element;".
But when the user wants to add some new value to the end of the array list collection then he needs to use the statement--
a.add(element); //where add is a function, element is a value and a is a array list object.
Another option is invalid because--
- Option 1 is not the correct because "a[3]=element;" modify the value of the 3rd element of the array.
- Option 2 gives a compile-time error because add functions bracts are not closed.
- Option 3 gives the error because a[4] gives the location of the 5th element of the array but the above question says that a is defined with 4 elements.
The most cost-effective way to configure a client-side virtualization solution is by using one (1) physical NIC, three (3) virtual NICs, and one (1) virtual switch.
<h3>What is virtualization?</h3>
Virtualization refers to the creation of an abstraction layer over computer hardware through the use of a software, in order to enable the operating system (OS), storage device, server, etc., to be used by end users.
In this scenario, the most cost-effective way to configure a client-side virtualization solution is by using one (1) physical network interface card (NIC), three (3) virtual network interface cards (NICs), and one (1) virtual switch.
Read more on virtualization here: brainly.com/question/14229248
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<span>Moving through a neighborhood trying to locate open wireless access points is called wardriving.
</span> The action wardriving is performed by a person in a moving vehicle (car, bike..) who is using a laptop or smartphone and wants to catch a Wireless LAN (WiFi) network. For this purpose there is software <span>freely available on the Internet.</span>
Hi!
Well, this isn't exactly a question - but rather just a request. However, I'm going to attempt to try and <em>describe </em>to you how to approach this problem, instead of just writing the code for you and sending you on your way.
So, what's our general base goal here? We want to take a string into a function, and then print it out backwards. Seems simple enough!
Right away, we already have an idea how to set this code up. We need a main method which will call <em>PrintBackwards(), </em>which will have to take a parameter of type string.
This would look something like <em>PrintBackwards(string baseString). </em>Inside this method, we'd have to do something so we can see each character in this string and then store it in a new string.
I encourage you to try and tackle this on your own, but I can give you an idea. We can have a new valueless variable called reversedString, which will store our baseString but backwards.
We could try looping through the baseString for each character it possesses, and then keep adding onto our reversedString by doing something like +=. What I mean, is we'd access the very last index of baseString, and then keep appending characters into it.
So our loop would look something like <em>for(int i = baseString.length; i > 0; i--) {}.
</em>I haven't used C++ in awhile, so you'll have to find the specific syntax requirements. But with that loop, i represents the index of each character in baseString. It starts with the last index, and keeps going down in reverse.
<em>
</em>Inside our loop, we could do something like reverseString += baseString.index(i); Again, I don't remember the specific syntax - so you'll have to do this on your own.
<em>
</em>Hopefully, this helps! =)<em>
</em>
Answer:
35
Explanation:
We will be going inside B array, because he was in second place in array E,and will be the first element of array B, so it's 35
E can be understanded as:
E=[[21, 'dog', 'red'],[35, 'cat', 'blue'],[12, 'fish', 'green']], so, you can see array E as array of arrays or so-called two-dimensional array
