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3 years ago

The best way for employees to improve their skills online is to enroll in a digital

Computers and Technology

2 answers:

Maksim231197 [3]3 years ago

6 0

The best way for employees to improve their skills online is to enroll in a digital office.

office.

<u>Explanation:</u>

In the present innovation-driven world, upskilling is turning out to be progressively significant just in light of the fact that innovation changes so quick.

A key preferred position for workers in web-based learning is that representatives can finish their preparation voluntarily and as indicated by their own timetables.

Enrolling in a digital office would allow an individual to learn at their ease, pace and comfort. It will definitely help the individual to expand and acquire skills.

So now full-time workers or the individuals who essentially don't have the opportunity to return to college full-time have a lot of chances to discover low maintenance courses.

9966 [12]3 years ago

4 0

Answer:

To enroll in a digital office.

Explanation:

A digital office is a working environment in which information is stored and organised electronically. The office is one in which technology is used at the service of greater productivity and efficiency. It is a complete local and virtual experience where information are easily communicated to users.

Skill improvement and shapening is very paramount in the technologically driven society of today. Enrolling in a digital office would allow employees to improve their skills becaues it involves learning at ease and in comfort. This would give room for an individual to improve on personal skills and acquire new skills.

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kirill115 [55]

Answer:

// here is code in C++.

#include <bits/stdc++.h>

using namespace std;

void switch_fun(int input)

{

//array to store binary of input number

int bin[7];

// convert the number into binary

for(int i=0; input>0; i++)

{

bin[i]=input%2;

input= input/2;

}

// print the switch number and their status

cout<<"Switch number:\t 1\t 2\t 3\t 4\t 5\t 6\t 7 \n";

cout<<"status: ";

for(int j=6;j>=0;j--)

{

// if bit is 1 print "ON" else print "OFF"

if(bin[j]==1)

cout<<"\t ON";

else

cout<<"\t OFF";

}

int main()

{

int n;

cout<<"enter the number between 1-100 only:");

// read a number in decimal

cin>>n

// validate the input

while(n<1 ||n>100)

{

cout<<"wrong input!! Input must be in between 1-100:"<<endl;

cout<<"enter the number again :";

cin>>n;

}

// call the function with input parameter

switch_fun(n);

return 0;

}

Explanation:

Read a decimal number from user. If the number is not between 1 to 100 the it will again ask user to enter the number until the input is in between 1-100.Then it will call the fun() with parameter n. First it will convert the decimal to binary array of size 7.Then print the status of switch.If bit is 1 then print "ON" else it will print "OFF".

Output:

enter the number between 1-100 only:-5

wrong input!! Input must be in between 1-100:

enter the number again :125

wrong input!! Input must be in between 1-100:

enter the number again :45

Switch number: 1 2 3 4 5 6 7

Status: OFF ON OFF ON ON OFF ON

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Answer:

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Explanation:

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2 years ago

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Answer:

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Explanation:

If not then Im sorry

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2 years ago

A computing company is running a set of processes every day. Each process has its own start time and finish time, during which i

Helga [31]

Answer:

Above all else, let us show that this issue has the greedy choice property.

This implies that worldwide ideal arrangement can be gotten by choosing local ideal arrangements. Presently notice the following:

The global optimal solution of the issue requests us to track down the minimum number of times process_check module should be run.

It is likewise referenced that during each running interaction, process_check module should be called atleast once. Along these lines, the minimum possible number of process_check calls happens when process_check is made close to the quantity of the processes that are run.

Along these lines, we see that the global optimal solution is shaped by choosing optimal answer for the nearby advances.

Along these lines, the issue shows greedy choice property. Thus, we can utilize a greedy algorithm for this issue.

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Now repeat stage 2, till all processes are finished.

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Komok [63]

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