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LUCKY_DIMON [66]
4 years ago
11

Which statement accurately describes DTP programs?

Computers and Technology
1 answer:
s2008m [1.1K]4 years ago
7 0

Answer:

O DTP programs are basically layout programs.

Explanation:

Desktop publishers are vector-based graphics software, and a perfect example of one such software is Microsoft Publisher. You can consider Adobe Illustrator, and Corel Draw as well, but they are huge software meant for many more things, and hence let's include only Microsoft publisher. And a long list follows as well. And they are in general a layout program used to prepare cards, advertisements, infographics, etc, and in fact the layout for them. And hence its more appropriate to say that it is basically a layout program that uses vector graphics to accomplish its goal. Various formats supported are like SVG, EPS, PDF,AI etc. And these are vector graphics based formats. The other type is Raster, and that includes Photoshop etc, with formats iike jpeg, gif etc, and which are pixel based.

You might be interested in
(3 marks)
muminat

Answer: 1.5 : 1

Explanation:

The average of 92 marks for the test is a weighted average of the proportion of students in the class.

Assume boys are in the proportion, x.

(90 * x) + (95 * (1 - x)) = 92

90x + 95 - 95x = 92

-5x = 92 - 95

-5x = -3

x = -3/-5

x = 0.6

Boys are 0.6 of the class

Girls are:

= 1 - 0.6

= 0.4

0.6 : 0.4

1.5 : 1

3 0
3 years ago
Jennifer is trying to install an anti-malware program on a computer that she believes might be infected. During the installation
sesenic [268]

Answer:

B. Install in Safe Mode.

Explanation:

Since Jennifer is using the administrator account, she should be able to install the program.

Hence, she should install the anti-malware software in Safe Mode.

In Computer science, Safe Mode is a mode in which the operating system loads only the bare minimum services, process and programs to boot or start up.

Hence, Safe Mode will ensure that when Jennifer is installing the anti-malware software, no other program or service is running which may interfere with her installation.

4 0
3 years ago
Which location on the ribbon contains the commands for adding a table to a document?
Maksim231197 [3]

Adding a table to a document can be done in Microsoft Word using the insert ribbon, the insert ribbon are mostly used for adding options to a document.

  • The insert ribbon is a multi functional ribbon which has up to about 10 different useful groups for adding elements to a document.

  • Some of the groups on the insert ribbon include ; Pages, Text, Header & Footer, Tables, illustrations, Links, Media and so on.

  • The Table group in the insert ribbon allows different table adding options such as inserting an already existing table, drawing a new table or importing an excel table.

Therefore, adding a table to a document is performed from the insert ribbon in Microsoft Word.

Learn more:brainly.com/question/21842366?referrer=searchResults

4 0
2 years ago
C++ question:
gogolik [260]
There's no "using namespace std;" statement. I see no reason for COLOR to be two dimensional or have a length of 5 with only four elements. The inside of the for loop with variable 'J' makes no sense.
3 0
3 years ago
Read 2 more answers
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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