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Varvara68 [4.7K]
3 years ago
3

Universal Containers created a Time-Based Workflow rule that sends a follow-up email to the customer two days after a Case is cl

osed. The System Administrator wants to verify that the workflow functions correctly. Which queue should the System Administrator view to monitor pending Workflow Actions?
Computers and Technology
1 answer:
Allisa [31]3 years ago
5 0

Answer:

Time-Based Workflow Queue.

Explanation:

Workflow management systems can be defined as a strategic software application or program designed to avail companies the infrastructure to setup, define, create and manage the performance or execution of series of sequential tasks, as well as respond to workflow participants.

An example of a work management system is Salesforce.com.

In this scenario, Universal Containers created a Time-Based Workflow rule that sends a follow-up email to the customer two days after a Case is closed. The System Administrator wants to verify that the workflow functions correctly. Hence, the queue which the System Administrator should view to monitor pending Workflow Actions is the, Time-Based Workflow Queue because it will remain available to be viewed until the Workflow Action is executed or the record no longer matches the triggering criteria.

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Read 2 more answers
A programming language is used to tell a computer what to do to solve a problem?
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3 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
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