D. DoS. DoS is a way to potentially lag or even crash different types of servers ranging from as small as a simple website to bigger websites and video games servers.
Answer:
Answer explained below
Explanation:
When disabling inherited permissions on an object, if you select Remove all inherited permissions from this object then you lose every user or group assigned to the folder.
This will delete all existing permissions, including administrator accounts, leaving you a blank slate to apply your own permissions to the folder.
The professional American organization for designers has set up many standards and guidelines that its members have to follow is known as ANSI.
<h3>What mean ANSI?</h3>
The American National Standards Institute (ANSI) is known to be a kind of private, non-profit organization that gives or coordinates the U.S. voluntary standards and the conformity in terms of assessment system.
Hence, The professional American organization for designers has set up many standards and guidelines that its members have to follow is known as ANSI.
Learn more about American organization from
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Answer:
The printout of the program is 33
Explanation:
Given the code as follows:
- int[][] values = {{3, 4, 5, 1}, {33, 6, 1, 2}};
- int v = values[0][0];
- for (int row = 0; row < values.length; row++)
- for (int column = 0; column < values[row].length; column++)
- if (v < values[row][column])
- v = values[row][column];
-
- System.out.print(v);
The code above will find the largest value from the two-dimensional array and print it out.
The logic of finding the largest value is as follows:
- Create a variable, <em>v</em>, to hold the largest number (Line 2). At the first beginning, we simply set the value from the first row and first column (values[0][0]) as our current largest value.
- Next, we need to compare our current largest value against the rest of the elements in the two dimensional array.
- To make the comparison, we need a two-layers for loops that will traverse through every row and column of the array and compare each of the element with the current largest value (Line 3-6).
- If the current largest value, v, is smaller than any element of the array, update the <em>v</em> to the latest found largest value (Line 5-6).
- After completion of the for-loop, the v will hold the largest number and the program will print it out (Line 8).