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Studentka2010 [4]

3 years ago

10

A company that provides finance services wants to shift to a mobile platform. What is the major advantage that the users of the

service will gain from this shift?

Computers and Technology

2 answers:

trapecia [35]3 years ago

8 0

The users will be able to quickly manage their finances, and track puchaces easier.

tekilochka [14]3 years ago

6 0

ACCESSIBLE ANYWHERE

on a mobile platform you can use it and see ir whenever as long as you have some kind of service.

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Draw a timeline for each of the following scheduling algorithms. (It may be helpful to first compute a start and finish time for

12345 [234]

Answer:

See explanation below

Explanation:

Previos concepts

First Come First Serve (FCFS) "is an operating system scheduling algorithm that automatically executes queued requests and processes in order of their arrival".

Shortest job next (SJN), or the shortest job first (SJF) or shortest "is a scheduling policy that selects for execution the waiting process with the smallest execution time".

Shortest remaining time (SRF) "is a scheduling method that is a preemptive version of shortest job next scheduling'".

Round robin (RR) is an algorithm where the time parts "are assigned to each process in equal portions and in circular order, handling all processes without priority"

Solution for the problem

Assuming the dataset given on the plot attached.

Part a

For this algorithm the result would be:

Job A 0-6

Job B 6-(6+3) = 6-9

Job C 9-(9+1) = 9-10

Job D 10-(10+4) = 10-14

Part b

For this algorithm the result would be:

Job A 0-6

Job C 6-(6+1) = 6-7

Job B 7-(7+3) = 7-10

Job D 10-(10+4) = 10-14

Part c

For this algorithm the result would be:

Job A 0-1 until 14

Job B 2-(2+3) = 2-5

Job C 3-(3+2) = 3-5

Job D 9-(9+5) = 9-14

Part d

For this algorithm the result would be:

Job A 0-2 , 7-9, 12-14

Job B 2-4, 9-10

Job C 4-(4+1) = 4-5

Job D 5-7, 10-12

8 0

3 years ago

Write a C++ program that reads students' names followed by their test scores. The program should output each students' name foll

Mashutka [201]

Answer:

#include<iostream>

#include<conio.h>

using namespace std;

struct studentdata{

char Fname[50];

char Lname[50];

int marks;

char grade;

};

main()

{

studentdata s[20];

for (int i=0; i<20;i++)

{

cout<<"\nenter the First name of student";

cin>>s[i].Fname;

cout<<"\nenter the Last name of student";

cin>>s[i].Lname;

cout<<"\nenter the marks of student";

cin>>s[i].marks;

}

for (int j=0;j<20;j++)

{

if (s[j].marks>90)

{

s[j].grade ='A';

}

else if (s[j].marks>75 && s[j].marks<90)

{

s[j].grade ='B';

}

else if (s[j].marks>60 && s[j].marks<75)

{

s[j].grade ='C';

}

else

{

s[j].grade ='F';

}

}

int highest=0;

int z=0;

for (int k=0;k<20; k++)

{

if (highest<s[k].marks)

{

highest = s[k].marks;

z=k;

}

}

cout<<"\nStudents having highest marks"<<endl;

cout<<"Student Name"<< s[z].Fname<<s[z].Lname<<endl;

cout<<"Marks"<<s[z].marks<<endl;

cout<<"Grade"<<s[z].grade;

getch();

}

Explanation:

This program is used to enter the information of 20 students that includes, their first name, last name and marks obtained out of 100.

The program will compute the grades of the students that may be A,B, C, and F. If marks are greater than 90, grade is A, If marks are greater than 75 and less than 90, grade is B. For Mark from 60 to 75 the grade is C and below 60 grade is F.

The program will further found the highest marks and than display the First name, last name, marks and grade of student who have highest marks.

6 0

3 years ago

What's the minimum storage for onedrive e2

cupoosta [38]

It should be 1 TB per user

3 0

3 years ago

Aloiza [94]

Answer:

#include<studio.h>

#include<conio.h>

void main()

{

char ch;

printf("Enter any letter");

scanf("%c",&ch);

3 0

3 years ago

Define a function UpdateTimeWindow() with parameters timeStart, timeEnd, and offsetAmount. Each parameter is of type int. The fu

NARA [144]

Answer:

Here is a UpdateTimeWindow() method with parameters timeStart, timeEnd, and offsetAmount

// the timeEnd and timeStart variables are passed by pointer

void UpdateTimeWindow(int* timeStart, int* timeEnd, int offsetAmount){

// this can also be written as *timeStart = *timeStart + offsetAmount;

*timeStart += offsetAmount; //adds value of offsetAmount to that of //timeStart

// this can also be written as *timeEnd = *timeEnd + offsetAmount;

*timeEnd += offsetAmount; } //adds value of offsetAmount to that of //timeEnd

Explanation:

The function has three int parameters timeStart, timeEnd, and offsetAmount.

First two parameters timeStart and End are passed by pointer. You can see the asterisk sign with them. Then in the body of the function there are two statements *timeStart += offsetAmount; and *End+= offsetAmount; in these statements the offsetAmount is added to the each of the two parameters timeStart and timeEnd.

8 0

3 years ago