Answer:
So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %
Explanation:
Previous concepts
Input/output operations per second (IOPS, pronounced eye-ops) "is an input/output performance measurement used to characterize computer storage devices like hard disk drives (HDD)"
Solution to the problem
For this case since we have 4GB, but 512 MB are destinated to the operating system, we can begin finding the available RAM like this:
Available = 4096 MB - 512 MB = 3584 MB
Now we can find the maximum simultaneous process than can use with this:
And then we can find the maximum wait I/O that can be tolerated with the following formula:
The expeonent for p = 14 since we got 14 simultaneous processes, and the rate for this case would be 99% or 0.99, if we solve for p we got:
So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %
Agreed to all the terms mentioned in the contract
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Answer:
Following are the code:
Code:
total = 0; //assign value to total variable.
for (int i=0; i<x.length; i++) //for loop
{
total=total+x[i]; //add all array elements in total variable.
}
Explanation:
In the following question, it is defined that x and total is variable. Where variable x is an integer type array and total is an integer variable. we define some code for calculating the sum of the array element. In the above code, we use for loop that calculates sum of array elements that can be described as:
- To calculate the sum we use the total variable. In total variable, we assign value 0.
- Then we define for loop in loop we use total variables that add all array (x[]) elements.
