Answer:
Worksheet ranges is a group of cells, and a value range like 25 to 340. It is used for data modeling. Like you can use it in the vlookup, Hlookup and Xlookup for looking up a column cell value in the mentioned worksheet range. Different worksheet's ranges of cells can also be used, and in any number from throughout the workbook, and that makes the analysis quite easy. Through work sheet ranges we get the control over all the cells as well as cells group. Whatever analysis we do, we do on the list of cells. And through the worksheet ranges we get the control over the cells. And thus, this is extremely useful definitely.
Explanation:
Like one worksheet range can be from top left corner to the bottom right corner. And this assigns whole worksheet as the worksheet range.
Its used in almost all the Excel formulas, and you must know it.
An example is:
Worksheets("Sheet2").Range(Cells(2, "B"), Cells(35, "B")).select
This selects the Range B2:B35 from Sheet 2.
With 8<span> bits used to represent each color value, one pixel requires </span>24<span> bits. The number of colors available in a graphic is referred to as color depth. I hope you understand! :-)</span>
Answer:
Option c is the correct answer for the above question.
Explanation:
- The vulnerability is a term that is used to state that any system has some limitation by which the system causes an error while performing the action.
- The above question states a software and asked the action which is used by the analyst to prevent the limitation. So the system will use the input validation.
- It is because this prevents the error. it is because this will help to proceed with the wrong or invalid information on the database and the database or system can not hold the wrong data.
- Hence option c is the correct answer while the other option is not correct because other options can not able to fix the limitation of the above system.
Answer:
boolean isEven = false;
if (x.length % 2 == 0)
isEven = true;
Comparable currentMax;
int currentMaxIndex;
for (int i = x.length - 1; i >= 1; i--)
{
currentMax = x[i];
currentMaxIndex = i;
for (int j = i - 1; j >= 0; j--)
{
if (((Comparable)currentMax).compareTo(x[j]) < 0)
{
currentMax = x[j];
currentMaxIndex = j;
}
}
x[currentMaxIndex] = x[i];
x[i] = currentMax;
}
Comparable a = null;
Comparable b = null;
if (isEven == true)
{
a = x[x.length/2];
b = x[(x.length/2) - 1];
if ((a).compareTo(b) > 0)
m = a;
else
m = b;
}
else
m = x[x.length/2];
Answer:
I don't think so. In today's computer era, many different solution directions exist for any given problem. Where OOP used to be the doctrine of choice, now you would consider it only when the problem at hand fits an object-oriented solution.
Reason 1: When your problem can be decomposed in many different classes with each many instances, that expose complex interactions, then an OO modeling is justified. These problems typically produce messy results in other paradigms.
Reason 2: The use of OO design patterns provides a standardized approach to problems, making a solution understandable not only for the creator, but also for the maintainer of code. There are many OO design patterns.
