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Tomtit [17]
3 years ago
6

Gini is a manager at a website design firm. She has a timeline prepared for the product to be delivered to the client. She needs

to mark the work completion on her milestone chart. Which basic management skill is essential for her to monitor the progress of the project?
A.
controlling
B.
planning
C.
directing
D.
organizing
E.
managing
Computers and Technology
2 answers:
Oksi-84 [34.3K]3 years ago
8 0

Answer:

i thought it was E. Managing

Grace [21]3 years ago
7 0

Answer:

The correct answer is letter "A": controlling.

Explanation:

There are four (4) basic management skills: <em>Planning, Controlling, Organizing, </em>and <em>Leading.  </em>

Controlling implies reviewing the objectives the firm has established to evaluate the progress. It also involves measuring the resources the firm has available for reaching its goals and verifying if they will be enough to get to the finish line, otherwise, the company should find more efficient forms of allocating resources and funding its operations.

In other words, <em>the controlling skill of management sets the standards of performance within the institution.</em>

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OLEGan [10]

Answer:

The recommendations for Jesse about creating an SCR help desk are, the user interface, sending of data request to servers, sending data files back to the server, receiving  data files from the server, running an application programs locally to process data.

File server handles several jobs which includes, sending data files to client, storing of data files, and generating traffic on the internet.

The client/server tiers  allows user interface on the client and data can be stored on the server. it's elements includes the following: output, input, processes, performance, control and security

Explanation:

Given that:

(1)The recommendations is stated as follows:

A file sharing architecture is recommended where the clients can handle the following:

  • The user interface
  • Sends data request to servers
  • Sends data files back to the server
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The file server handles the following tasks or jobs which is stated below:

  • Receives data request from clients
  • Sends data files to client
  • Stores data files
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The Client design:

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The disadvantage of this, is that it is more difficult to design

Note: Kindly find an attached copy of part of the solution to this given question below

4 0
3 years ago
Write a program that reads students’ names followed by their test scores. The program should output each student’s name followed
bekas [8.4K]

Answer:

See explaination

Explanation:

#include <iostream>

#include <iomanip>

#include <fstream>

#include <string>

using namespace std;

void readStudentsData();

void levels(int&,int&,int&);

int main()

{

string studentsName[50];

int grade[50];

int highestScore = 0;

int belowAverage = 50;

int countStudents = 0;

int sum = 0;

int score[5]=

double average = 0;

ifstream infile;

ofstream outfile;

void readStudentsData();

infile.open("c:\\Data.txt");

int numOfStudents = 0;

infile >> studentsName[numOfStudents] >> grade[numOfStudents];

While (infile && numOfstudents < 50)

{

numOfStudents++;

infile >> studentsName[numOfStudents] >> grade[numOfStudents};

}

if (!infile)

{

cout << "File does not open." << endl;

return 1;

}

outfile.open("c:\\average.out");

outfile << fixed << showpoint;

outfile << setprecision(2);

infile >> studentsName;

infile >> score1 >> score2 >> score3 >> score4 >> score5 >>;

outfile << left << setw(20) << "Students Name"

<< setw(5) << "Score 1" << setw(5) << "Score 2"

<< setw(5) << "Score 3" << setw(5) << "Score 4"

<< setw(5) << "Score 5" << setw(5) << "Grade"

<< setw(5) << "Average" << setw(5) << "Below average"

<< setw(5) << "Highest" << endl;

if (average >= 90)

cout << "A" << endl;

else if (average >= 80)

cout << "B" << endl;

else if (average >= 70)

cout << "C" << endl;

else if (average >= 60)

cout << "D" << endl;

else

cout << "F" << endl;

}

average = static_cast<double>(score1+score2+score3+score4+score5) /5;

void level(int &highestScore,int &belowAverage,int &sum);

{

int level,high = 0,low = 17;

infile >> level;

while(level!=-1)

{

sum = sum + level;

if (level > high)

{

high = level;

infile >> level;

}

else

if (level < low)

{

low = level;

infile >> level;

}

else

if infile >> level;

}

outfile << ' ' << high;

else

if (high > highestScore)

highestScore=high;

else

if (low < belowAverage)

belowAverage=low;

}

do

{

cout << "Students Name: ";

cin >> name

cout << "Input score: " << endl;

cin >> score1>> score2>> score3>> score4>> score5;

cout << Students Name << Highest score);

}

While (highest score !=0);

infile.close();

outfile.close();

return 0;

}

5 0
3 years ago
What is the answer ????​
zzz [600]

Answer:

august po

Explanation:

pabrainliest ty

5 0
3 years ago
Please help me !!
gavmur [86]

e. uhmmm

19+21=30

like....

easy.

What grade class is this? I wanna do it cause it easy AFef

6 0
3 years ago
Which of the following boxes should replace question mark
Rus_ich [418]

Answer:

There is nothing please put it then i will edit this answer.

Explanation:

6 0
3 years ago
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