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 %
Answer:
Miguel y Maru están muy cansados. - Miguel y Maru están cansadísimos
Felipe es muy joven. - Felipe es jovencísimo
Jimena es muy inteligente. - Jimena es inteligentísima
La madre de Marissa está muy contenta. - La madre de Marissa está contentísima
Estoy muy aburrido. - Estoy aburridísimo
Explanation:
In this activity we have to switch the statements to the absolute superlative of the expressions. In Spanish we can add the suffix -ísimo to an adjective to refer to the highest degree of something. It can be translated in ENglish to "really, extremely, super or quie". The statements in English are:
- Miguel and Maru are very tired - Miguel and Mary are extremely tired
- Felipe is very young - Felipe is super young
- Jimena is very smart - Jimena is really smart
- Marissa´s mother is very happy - Marissa´s mother is extremely happy
- I´m very bored - I´m super bored
Answer:
// here is code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int n,no_open=0;
cout<<"enter the number of lockers:";
// read the number of lockers
cin>>n;
// initialize all lockers with 0, 0 for locked and 1 for open
int lock[n]={};
// toggle the locks
// in each pass toggle every ith lock
// if open close it and vice versa
for(int i=1;i<=n;i++)
{
for(int a=0;a<n;a++)
{
if((a+1)%i==0)
{
if(lock[a]==0)
lock[a]=1;
else if(lock[a]==1)
lock[a]=0;
}
}
}
cout<<"After last pass status of all locks:"<<endl;
// print the status of all locks
for(int x=0;x<n;x++)
{
if(lock[x]==0)
{
cout<<"lock "<<x+1<<" is close."<<endl;
}
else if(lock[x]==1)
{
cout<<"lock "<<x+1<<" is open."<<endl;
// count the open locks
no_open++;
}
}
// print the open locks
cout<<"total open locks are :"<<no_open<<endl;
return 0;
}
Explanation:
First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.
Output:
enter the number of lockers:9
After last pass status of all locks:
lock 1 is open.
lock 2 is close.
lock 3 is close.
lock 4 is open.
lock 5 is close.
lock 6 is close.
lock 7 is close.
lock 8 is close.
lock 9 is open.
total open locks are :3
computer hardware engineer and database architecture
