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arlik [135]
4 years ago
13

A hardware compatibility list recommends striping with double parity as storage for an application. In a test environment, a tec

hnician realized that a different RAID level can be used to compensate for a limited number of available disks. Which of the following is the closest RAID level the technician should deploy while minimizing the number of disks and maintaining fault tolerance.
Computers and Technology
1 answer:
riadik2000 [5.3K]4 years ago
4 0

Answer:

RAID level 5 can be used to compensate a limited number of available disks.

Explanation:

There are two type of RAID

  • Software RAID
  • Hardware RAID

Software RAID

deliver services form the host.

Hardware RAID

provides hardware services.

RAID has levels

0, 1, 5, 6, and 10

RAID 0, 1, and 5 work on both HDD and SSD media,

4 and 6 also work on both media.

RAID 0 :Striping

In this level minimum of two disks,RAID 0 split the file strip the data.Multiple hard drive are used to split the data.

RAID 1 : Mirroring

In this level Minimum two disk require and provide data tendency.

RAID 5 :Stripping with parity

Parity is a binary data.RAID system calculate the value which system used to recover the data.

Most RAID system with parity function store parity blocks.

RAID 5 combines the performance of RAID 0 with redundancy of RAID 1.

RAID 5 level should minimize the fault tolerance.

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podryga [215]

Answer:

Following are the program in the C++ Programming Language.

#include <iostream> //header file

#include <cmath> //header file

using namespace std; //namespace

//define function

void quadratic (double a, double b, double c)

{

 double x,y; //set double type variables

 //check a==0

 if (a==0)

 { //then print message

   cout << "no solution for a=0";

 }

 //check if solution is less than 0

 else if ((b*b - 4*a*c) < 0)

 { //print message

   cout << "no real solutions";

 }

 //otherwise

 else

 {//solve the following quadratic equation

   x = (-b + sqrt( b*b - 4*a*c)) /(2*a);//in term of x

   y = (-b - sqrt(b*b - 4*a*c)) / (2*a);//in term of y

   //check x is greater than y

   if (x > y){

     cout<<x; //print the value of x

   }

   //otherwise

   else{

     cout<<y;//print the value of y

   }

 }

}

//define main method

int main() {

 //set double type variables and assign their value

 double x=10, y=50, z=-205;

 //call the function

 quadratic(x,y,z);

}

<u>Output</u>:

2.67204

Explanation:

Here, we define a function "quadratic" and pass three double data type variables i.e., "a", "b" and, "c" inside the function.

  • Set two double type variable i.e., "x", "y".
  • Set the if conditional statement and check the conditions is the variable a is equal to the 0 then, print the message or check another condition is the solution of the equation "b*b - 4*a*c" is less than 0, then print the message.
  • Otherwise, solve the quadratic equation in terms of the variable x or the variable y.
  • Again check the condition inside the else part is the variable x is greater than the variable y then, print the value of x otherwise print the value of y.

Finally, we define the main function inside it we set three double type variables i.e., "x", "y" and, "z" and initialize value 10, 50 and, -250 then, call the function and pass the variables x y z in its parameter.

6 0
4 years ago
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