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ss7ja [257]
3 years ago
11

A senior center would like to add a new computer to their library so that members can check their email and read book reviews

Computers and Technology
2 answers:
Airida [17]3 years ago
4 0

Answer:

A Budget computer

Explanation:

Reason because for checking your email's and online reviews on products doesn't take much strain on the computer. That is why a good budget computer would be great for this task!

jenyasd209 [6]3 years ago
4 0

Answer:

D: A budget computer

Explanation: Looked it up on Quizlet.

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Image files are grouped into two categories: _____.
EleoNora [17]
It is grouped according to raster and vector format. Raster format are images used in a computer or printed. Vector format stores data and is compressed. 

Examples of common raster formats usually used in a computer are; jpeg, png, bitmap, and gif. Vector formats are; CGM, SVG and 3D vector.
4 0
4 years ago
GIVING BRAINLIEST What does output allow a computer to do? Display information Receive information Do complex math problems Do m
kumpel [21]
The answer is display information
4 0
3 years ago
Read 2 more answers
Write a parent program to fork(2) n processes where n is passed tothe program on the command line, and n can be from 1 to 10. Al
MrRissso [65]

Answer:

Complete code is given below:

Explanation:

#include <stdio.h>

#include <sys/types.h>

#include <unistd.h>

#include <time.h>

#include <sys/wait.h>

int main(int argc , char *argv[]){

pid_t mypid, childpid;

int status;

int i,n;

time_t t;

     

mypid = getpid();

printf("pid is %d.\n", mypid);

childpid = fork();

if ( childpid == -1 ) {

perror("Cannot proceed. fork() error");

return 1;

}

 

 

if (childpid == 0) {

 

 

mypid = getpid();

 

 

childpid = fork();

if ( childpid == -1 ) {

perror("Cannot proceed. fork() error");

return 1;

}

 

if (childpid == 0) {

 

mypid = getpid();

printf("Child pid is %d.\n", getppid());

 

childpid = fork();

 

 

if ( childpid == -1 ) {

perror("Cannot proceed. fork() error");

return 1;

}

random((unsigned) time(&t));

printf(" parent id = %d : Random = %d\n",mypid , (int)(random()% 100 +1));

printf("child id = %d : Random = %d\n",getpid , (int)(random()% 100 +1));

 

if (childpid == 0) {

 

printf("Child 2: I hinerited my parent's PID as %d.\n", mypid);

 

mypid = getpid();

printf("Child 2: getppid() tells my parent is %d. My own pid instead is %d.\n", getppid(), mypid);

 

sleep(30);

return 12;

} else return 15;

} else {

 

while ( waitpid(childpid, &status,0) == 0 ) sleep(1);

 

if ( WIFEXITED(status) ) printf("Child 1 exited with exit status %d.\n", WEXITSTATUS(status));

else printf("Child 1: child has not terminated correctly.\n");

}

} else {

printf(" fork() is ok and child pid is %d\n", childpid);

wait(&status);

 

if ( WIFEXITED(status) ) printf(" child has exited with status %d.\n", WEXITSTATUS(status));

else printf(" child has not terminated normally.\n");

}

 

return 0;

}

Output:

pid is 24503.

Child   pid  is 24565.

parent id = 24566 : Random = 87

child id = 1849900480 : Random = 78

pid is 24503.

Child 1  exited with exit status 15.

pid is 24503.

fork() is ok and child pid is  24565

child has exited with status 0.

6 0
3 years ago
It takes 2 seconds to read or write one block from/to disk and it also takes 1 second of CPU time to merge one block of records.
Alexxx [7]

Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way  merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

<em>For optimal 4-way merging, we need one dummy run with size 0.</em>

  1. Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300
  2. Merge the run as made in step 1 with the remaining 3 runs bearing length 1000, 1200, 1500. The merged run length is 6000 and is calculated as follows

       L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000

<em>The resulting run has length 6000 records</em>.

Part b

<u><em>For step 1</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec

So the input/output time is 46 seconds for step 01.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec

So the CPU  time is 23 seconds for step 01.

Total time in step 01

T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\

Total time in step 01 is 69 seconds.

<u><em>For step 2</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec

So the input/output time is 120 seconds for step 02.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec

So the CPU  time is 60 seconds for step 02.

Total time in step 02

T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\

Total time in step 02 is 180 seconds

Merging Time (Total)

<em>Now  the total time for merging is given as </em>

T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\

Total time in merging is 249 seconds seconds

5 0
3 years ago
Serveral cheetas are growling at each other while hunting for animals.for which need are they competing?
SVEN [57.7K]

Answer:

prey

Explanation:

5 0
3 years ago
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