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skelet666 [1.2K]
3 years ago
14

What is Automation ? Please give an example ?

Computers and Technology
1 answer:
Tcecarenko [31]3 years ago
6 0

Answer:

Creation/application of technology to deliver goods/services with little human effort.

Explanation:

For example, manufacturing technology produces products without much need of human intervention.

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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
Why do people take off the keys on the school computers
beks73 [17]

Answer:

is this supposed to be a joke-

Explanation:

cuz i dont get it

8 0
2 years ago
Help ASAP!!! During which phase of a well-structured study session is each occurring?
gladu [14]

Answer:

1. drill

2. discuss

3. review

4. prepare

Explanation:

Your Welcome!

7 0
3 years ago
Read 2 more answers
Hardware and software work (blank) to allow user to form of function?
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The answer is Jointly.

<span>No hardware and software cannot work without each other. 
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Example:
<span>without a monitor, how can u install a software?</span>
8 0
4 years ago
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Refer to the exhibit. All workstations are configured correctly in VLAN 20. Workstations that are connected to switch SW1 are no
aleksandrvk [35]

Answer:

Allow VLAN 20 on the trunk link.

Explanation:

Enable DTP on both ends of the trunk by allowing VLAN20 on the trunk link.

From curriculum topic: 3.2.4

To Enable DTP on both switches is simply allowing the negotiation of trunking. The "Negotiation of Trunking" line in the graphic is what shows that DTP is already enabled. The graphic also shows how the native VLAN is 1, and the default VLAN for any Cisco switch is 1. The graphic shows the PCs are to be in VLAN 20.

This will remedy the problem.

8 0
3 years ago
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