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avanturin [10]
3 years ago
13

For how long should a media piece hold the user's attention?

Computers and Technology
2 answers:
PIT_PIT [208]3 years ago
7 0
C, the others don't make sense.
Nastasia [14]3 years ago
5 0

Answer:

C) until the message has been communicated

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What OBD-II term applies when all enabling standards for a specific diagnostic trouble code (DTC) are met?
mash [69]

Answer:

Option C i.e., Trip criteria is correct

Explanation:

The concept Trip criteria are described that when the allowing requirements for its diagnostic code are reached.

In other words, Trip seems to be a key-on method where all the allowing requirements for such a specific diagnostic display are fulfilled as well as the diagnostic monitoring is powered. It is finished once the ignition switch becomes switched off.

  • Option A and Option B are not correct because they are the engines that come under the performance ignition system and they are not related to the following scenario.
  • Option D is incorrect because it is the type of data mining that is not related to the scenario.
3 0
3 years ago
Write a function with two parameters, prefix (a string, using the string class from ) and levels (an unsigned integer). The func
Allisa [31]

Answer:

Here is the program:

#include <iostream>  //to use input output functions

#include <string>  // to use functions to manipulate strings

using namespace std;  //to identify objects cin cout

void function(string prefix, unsigned int levels){  // function that takes two parameters, a string, using the string class and levels an unsigned integer

   if (levels == 0) {  //if number of levels is equal to 0

       cout << prefix << endl;  //displays the value of prefix

       return;    }  

  for (int i = 1; i <=9 ; i++){  //iterates 1 through 9 times

       string sections = (levels == 1 ? "" : ".");  //if the number of levels is equal to 1 then empty space in sections variable otherwise stores a dot

       string output = prefix +  std::to_string(i) + sections;   // displays the string prefix followed by the section numbers. Here to_string is used to convert integer to string

       function(output, levels - 1);   } }   //calls function by passing the resultant string and levels-1  recursively to print the string prefix followed by section numbers

int main() {  // start of main function

   int level = 2;  //determines the number of levels

   function("BOX", level);  } //calls function by passing the string prefix and level value

Explanation:

The program has a function named function() that takes two parameters, prefix (a string, using the string class from ) and levels (an unsigned integer). The function prints the string prefix followed by "section numbers" of the form 1.1., 1.2., 1.3., and so on. The levels argument determines how many levels the section numbers have. If the value of levels is 0 means there is 0 level then the value of prefix is printed. The for loop iterates '1' through '9' times for number of digits in each level. If the number of levels is 1 then space is printed otherwise a dot is printed. The function() calls itself recursively to print the prefix string followed by section numbers.

Let us suppose that prefix = "BOX" and level = 1

If level = 1 then the loop for (int i = 1; i <=9 ; i++) works as follows:

At first iteration:

i = 1

i <=9 is true because value of i is 1

string sections = (levels == 1 ? "" : "."); this statement checks if the levels is equal to 1. It is true so empty space is stored in sections variable so,

sections = ""

Next, string output = prefix +  std::to_string(i) + sections; statement has prefix i.e BOX plus value of i which is 1 and this int value is converted to string by to_string() method plus sections has an empty space. So this statement becomes

string output = BOX + 1  

So this concatenates BOX with 1 hence output becomes:

output = BOX1

At second iteration:

i = 2

i <=9 is true because value of i is 2

string sections = (levels == 1 ? "" : "."); is true so

sections = ""

Next, string output = prefix +  std::to_string(i) + sections; statement becomes

string output = BOX + 2  

So this concatenates BOX with 1 hence output becomes:

output = BOX2

At third iteration:

i = 3

i <=9 is true because value of i is 3

string sections = (levels == 1 ? "" : "."); is true so

sections = ""

Next, string output = prefix +  std::to_string(i) + sections; statement becomes

string output = BOX + 3  

So this concatenates BOX with 1 hence output becomes:

output = BOX3

Now at each iteration the prefix string BOX is concatenated and printed along with the value of i. So at last iteration:

At last iteration:

i = 9

i ==9 is true because value of i is 9

string sections = (levels == 1 ? "" : "."); is true so

sections = ""

Next, string output = prefix +  std::to_string(i) + sections; statement becomes

string output = BOX + 9  

So this concatenates BOX with 1 hence output becomes:

output = BOX9

After this the loop breaks at i = 10 because the condition i<=9 becomes false. So the output of the entire program is:

BOX1                                                                                                                                          BOX2                                                                                                                                          BOX3                                                                                                                                          BOX4                                                                                                                                          BOX5                                                                                                                                          BOX6                                                                                                                                          BOX7                                                                                                                                          BOX8                                                                                                                                          BOX9  

The program along with the output is attached.

4 0
3 years ago
Will give brainliest to whoever can answer all of these questions correctly.
babymother [125]

false

ppt

chips designed to preform a function

Apologize to the customer for the inconvenience and offer to refund his money or replace the item

OneDrive-CollegeWork-FreshmanYear-NameOfClass

true

Social networking

paragraph

true

complete fill

true


 

7 0
3 years ago
Read 2 more answers
How does a machine learning model is deployed
yan [13]

Answer:

Explanation:

Deployment is the method by which you integrate a machine learning model into an existing production environment to make practical business decisions based on data. It is one of the last stages in the machine learning life cycle and can be one of the most cumbersome.

7 0
3 years ago
What are the advantages and disadvantages of UTF-8 compared to ASCII?
Romashka [77]

Answer:

UTF-8 and ASCII both are the character encoding.

In a system,every character has some binary representation,these are the method to encode them.Earlier only ASCII was there, for every character it uses 8 bits to represent.In ASCII only 8 bytes were there i.e 2^8 that is 256.We can't represent number beyond than 127 so it generate a need for other encoding to get into,these drawbacks lead to Unicode,UTF-8.

As ASCII codes only uses a single byte,UTF-8 uses upto 6 bytes to represent the characters.So we can save characters which are as long as 2^48 characters. We can read this encoding easily by the help of shift operators and it is also independent of byte order.

As messages on internet were transferred over 7 bit ASCII messages,so many mail servers removed this encoding.    

6 0
3 years ago
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