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guapka [62]
3 years ago
7

Time-management techniques work most effectively when performed in which order?

Computers and Technology
1 answer:
Sholpan [36]3 years ago
4 0

Answer:

study-time survey, project schedule, prioritize tasks, reward system.

Explanation:

Time management can be defined as a strategic process which typically involves organizing, planning and controlling the time spent on an activity, so as to effectively and efficiently enhance productivity. Thus, when time is properly managed, it avails us the opportunity to work smartly rather than tediously (hardly) and as such making it possible to achieve quite a lot within a short timeframe. Also, a good time management helps us to deal with work-related pressures and tight schedules through the process of properly allocating the right time to the right activity.

Hence, time-management techniques work most effectively when performed in the following sequential order; study-time survey, project schedule, prioritize tasks, and designing (creating) a reward system.

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Use computer magazines and/or the internet to investigate one of these DBMSs: DB2, SQL Server, MySQL, Oracle, or Sybase. Prepare
tigry1 [53]

Answer:

The definition for the problem is listed in the section below on the explanations.

Explanation:

<u>For SQL Server</u>

  • Data Recovery: Data recovery is done well on SQL Server. It is achieved by the restoration method. Behind this, a sysadmin must reconstruct a collection of SQL backups throughout a restore series that is technically right as well as relevant.
  • Data Security: There are several approvals as well as limits for accessing main data objects or databases. These data are accessed only by the approved clients.
  • Concurrency Control: This is essentially synchronization of the evidence or information accessed by multiple users as well as from separate locations.

Concurrency or Competitiveness control becomes accomplished with SQL Server Portable, using locking to further improve security. The locking regulates how multiple individuals can simultaneously navigate & alter shared storage, without interfering with one another.

8 0
3 years ago
HEY DO U LIKE TRAINS!
Triss [41]

Answer:

yes

Explanation:

yes

3 0
4 years ago
State three positive uses of the computer in the government sector​
WINSTONCH [101]

Well Here I got 4

1. Email Functions

2. Distributing Payments

3. Record Keeping

4. Direct Mail Promotions

<em>Have A great night!</em>

<em></em>

6 0
4 years ago
1)
spayn [35]

Answer:

1)

for(i = 0; i < NUM_VALS; ++i) {

  if(userValues[i] == matchValue) {

     numMatches++;  }    }

2)  

for (i = 0; i < NUM_GUESSES; i++) {

      scanf("%d", &userGuesses[i]);   }

  for (i = 0; i < NUM_GUESSES; ++i) {

         printf("%d ", userGuesses[i]);    }

3)

sumExtra = 0;

for (i = 0; i < NUM_VALS; ++i){

     if (testGrades[i] > 100){  

        sumExtra = testGrades[i] - 100 + sumExtra;    }       }

4)

for (i = 0; i < NUM_VALS; ++i) {

    if (i<(NUM_VALS-1))  

   printf( "%d,", hourlyTemp[i]);

    else  

    printf("%d",hourlyTemp[i]); }      

Explanation:

1) This loop works as follows:

1st iteration:

i = 0

As i= 0 and NUM_VALS = 4 This means for condition i<NUM_VALS  is true so the body of loop executes

if(userValues[i] == matchValue) condition checks if element at i-th index position of userValues[] array is equal to the value of matchValue variable. As matchValue = 2 and i = 0 So the statement becomes:

userValues[0] == 2

2 == 2

As the value at 0th index (1st element) of userValues is 2 so the above condition is true and the value of numMatches is incremented to 1. So numMatches = 1

Now value of i is incremented to 1 so i=1

2nd iteration:

i = 1

As i= 1 and NUM_VALS = 4 This means for condition i<NUM_VALS  is true so the body of loop executes

if(userValues[i] == matchValue) condition checks if element at i-th index position of userValues[] array is equal to the value of matchValue variable. As matchValue = 2 and i = 1 So the statement becomes:

userValues[1] == 2

2 == 2

As the value at 1st index (2nd element) of userValues is 2 so the above condition is true and the value of numMatches is incremented to 1. So numMatches = 2

Now value of i is incremented to 1 so i=2

The same procedure continues at each iteration.

The last iteration is shown below:

5th iteration:

i = 4

As i= 4 and NUM_VALS = 4 This means for condition i<NUM_VALS  is false so the loop breaks

Next the statement: printf("matchValue: %d, numMatches: %d\n", matchValue, numMatches);  executes which displays the value of

numMatches = 3

2)

The first loop works as follows:

At first iteration:

i = 0

i<NUM_GUESSES is true as NUM_GUESSES = 3 and i= 0 so 0<3

So the body of loop executes which reads the element at ith index (0-th) index i.e. 1st element of userGuesses array. Then value of i is incremented to i and i = 1.

At each iteration each element at i-th index is read using scanf such as element at userGuesses[0], userGuesses[1], userGuesses[2]. The loop stops at i=4 as i<NUM_GUESSES evaluates to false.

The second loop works as follows:

At first iteration:

i = 0

i<NUM_GUESSES is true as NUM_GUESSES = 3 and i= 0 so 0<3

So the body of loop executes which prints the element at ith index (0-th) index i.e. 1st element of userGuesses array. Then value of i is incremented to i and i = 1.

At each iteration, each element at i-th index is printed on output screen using printf such as element at userGuesses[0], userGuesses[1], userGuesses[2] is displayed. The loop stops at i=4 as i<NUM_GUESSES evaluates to false.

So if user enters enters 9 5 2, then the output is 9 5 2

3)

The loop works as follows:

At first iteration:

i=0

i<NUM_VALS is true as NUM_VALS = 4 so 0<4. Hence the loop body executes.

if (testGrades[i] > 100 checks if the element at i-th index of testGrades array is greater than 100. As i=0 so this statement becomes:

if (testGrades[0] > 100

As testGrades[0] = 101 so this condition evaluates to true as 101>100

So the statement sumExtra = testGrades[i] - 100 + sumExtra; executes which becomes:

sumExtra = testGrades[0] - 100 + sumExtra

As sumExtra = 0

testGrades[0] = 101

So

sumExtra = 101 - 100 + 0

sumExtra = 1

The same procedure is done at each iteration until the loop breaks. The output is:

sumExtra = 8

4)

The loop works as follows:

At first iteration

i=0

i < NUM_VALS is true as  NUM_VALS = 4 so 0<4 Hence loop body executes.

if (i<(NUM_VALS-1))   checks if i is less than NUM_VALS-1 which is 4-1=3

It is also true as 0<3 Hence the statement in body of i executes

printf( "%d,", hourlyTemp[i]) statement prints the element at i-th index i.e. at 0-th index of hourlyTemp array with a comma (,) in the end. As hourlyTemp[0] = 90; So 90, is printed.

When the above IF condition evaluates to false i.e. when i = 3 then else part executes which prints the hourlyTemp[3] = 95 without comma.

Same procedure happens at each iteration unless value of i exceeds NUM_VAL.

The output is:

90, 92, 94, 95

The programs along with their output are attached.

4 0
4 years ago
There are several methods of updating information and data on a webserver. We must consider who performs those updates upfront w
Thepotemich [5.8K]

Answer:

RTMC pro, HTML server-sent and Java Software Development Kit

Explanation:

<u>RTMC pro</u>:  The setup of RTMC pro is done automatically and there is provision of several elements to upload and store data on the server using screens. Screens are usually saved as files and then integrated to an internal or external server.

<u>HTML server-sent</u> : This is a way for webpages to communicate with the webserver. In this method, for ever HTML page there is an associated script present on the server side that will continuously provide realtime data update. This scrip will have to be configured with the data you want to update.

<u>Java Software Development Kit</u> : JAVA SDK is used to build software/applications on Java. This permits programmers to develop applications or applets to perform data updating on the servers. The developped applets can be used with internal servers as well as third party servers.

5 0
3 years ago
Read 2 more answers
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