You do not need to configure a certificate authority on your network to use EFS.
<em>EFS</em> is the short form for<em> Encryption File System</em>. With EFS, users can encrypt their files and folders and even the entire content of a given drive. By encrypting these files and folders, the access to them are restricted and thus increasing, improving and enhancing the security level of the users' data.
In other words, even though there are other ways to restrict access (such as using <em>logon authentication</em> and <em>NTFS file permissions</em>), EFS allows to add another layer of security to data.
To <em>encrypt</em> and <em>decrypt</em> data files and folders in EFS, a <em>certificate authority (CA)</em> could be used. This is however not a requirement. In the case where there is no certificate authority, EFS will sign a default certificate that will be used for encryption. In other words, <em>EFS will generate its own certificate</em> if none does not exist.
<em>The following are other things to note about EFS</em>
i. EFS uses a public and private key pair to encrypt data.
ii. Users do not need to enable EFS. It is enabled by default.
iii. For EFS to encrypt a file, the NTFS file system must be used.
Since a certificate authority is not required on your network to use EFS, the correct option is:
(b) False.
Read more at: brainly.com/question/10410730
Answer:
arte digital es una forma de arte que utiliza principios de ingeniería como paleta.
Explanation:
Answer:Speculative prediction
Explanation: Speculative prediction is the prediction method which can be optimizing in nature.This method usually runs in such a way that can lead ahead of time by predicting the next step or it's result.
This technique is also known as the dynamic execution which is usually found in the CPU devices and some other modern processing devices.Speculation prediction and execution works even when there is no necessity of the knowing about the next instruction.The result appearing can be correct or incorrect.
Answer:
oky will do. Have a good day
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
