Answer:
18, 13, 19
Explanation:
Number of computer programmers proficient only in Java = 45 - ( 1+1+6) = 37
Number of computer programmers proficient only in C++ = 30 - (6+1+5) = 18
Number of computer programmers proficient only in python = 20 - ( 1+1+5) = 13
Number of computer programmers are not proficient in any of these three languages = 100 - ( 37 + 18 + 13 + 1+ 1+ 5+ 6 ) = 100 - 81 = 19
Answer:
FIPS 140-2 es el acrónimo de Federal Information Processing Standard (estándares federales de procesamiento de la información), publicación 140-2, es un estándar de seguridad de ordenadores del gobierno de los Estados Unidos para la acreditación de módulos criptográficos. ... El Instituto Nacional de Estándares y Tecnología (NIST)
Explanation:
<span>SSL is used to process certificates and private/public key information. </span>SSL stands for Secure Sockets Layer. It is cryptographic protocols <span> that creates a trusted environment and </span>provides communications security over a computer network. The SSL protocol is used for establishing encrypted links between a web server and a browser.
Answer:
The method definition to this question can be given as:
Method definition:
public void clear(int[] arr, int num) //define method clear.
{
if (num == 0) //if block
{
return 0; return value.
}
else //else block
{
arr[num - 1] = 0; //assign value in arr.
return arr[]; //return value.
}
}
clear(arr, num - 1); //calling
Explanation:
The description of the above method definition as follows:
- Firstly we define a method that is "clear" that does not return any value because its return type is "void". This method accepts two integer variables that are "arr[] and num" where arr[] is an array variable and num is an integer variable.
- Inside a method, we use a conditional statement in if block we check that num variable value is equal to 0. if this condition is true so, it will return 0 otherwise it will go to else block in else block it will assign value in variable arr[num-1] that is "0" and return arr value.
<span>Shared memory buffering would work best. This would give the ports the best allocation of resources, using only those that are the most active and best allocated for the size of the frames being transmitted in the current traffic. In addition, any port can store these frames, instead of being specifically allocated as per other types of memory buffering.</span>
