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Daniel [21]
3 years ago
9

. You need to access customer records in a database as you're planning a marketing campaign. What language can you use to pull t

he records most relevant to the campaign? a. FTP b. SQL C. SMTP d. TLS​
Engineering
1 answer:
Rzqust [24]3 years ago
3 0

SQL can be used to access customer records in a database most relevant to the campaign.

<h3>SQL</h3>

SQL (Structured Query Language) is a standardized programming language that's used to manage relational databases and perform various operations on the data in them.

  • SQL databases comprise a set of tables containing data in rows and columns.

SQL can be used to access customer records in a database most relevant to the campaign.

Find out more on SQL at: brainly.com/question/25694408

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Please view the file attached below and help me answer all the questions!!
evablogger [386]

Answe whats the question i can help you

Explanation:

8 0
3 years ago
Find the error in the following pseudo code
IRISSAK [1]

Pseudocodes are used as a prototype of an actual program.

The error in the pseudocode is that, the while loop in the pseudocode will run endlessly.

From the pseudocode, the first line is:

<em>Declare Boolean finished = false</em>

The while loop is created to keep running as long as <em>finished = false.</em>

So, for the while loop to end, the finished variable must be updated to true.

This action is not implemented in the pseudocode.

Hence, the error in the pseudocode is that, the while loop is an endless loop

Read more about pseudocodes at:

brainly.com/question/17442954

8 0
2 years ago
(a) A 10.0-mm-diameter Brinell hardness indenter produced an indentation 2.3 mm in diameter in a steel alloy when a load of 1000
vampirchik [111]

Answer:

(a)  We are asked to compute the Brinell hardness for the given indentation. for HB, where P= 1000 kg, d= 2.3 mm, and D= 10 mm.  

Thus, the Brinell hardness is computed as

HB=2P/\pi D{D-\sqrt{D^2-d^2}

=2*1000hg/\pi (10mm)[10mm-\sqrt{(1000^2-(2.3mm)^2} ]

(b)    This  part  of  the  problem  calls  for  us  to  determine  the  indentation diameter d which  will  yield  a  270  HB  when P=  500  kg.  

d=\sqrt{D^2-[D-\frac{2P}{(HB)\pi D} } ]^2\\=\sqrt{(10mm)^2-[10mm-\frac{2*500}{450( \pi10mm)} } ]^2

6 0
4 years ago
Storing parts outside doesn't cause any environmental risks as long as the items are covered.
telo118 [61]

Answer:

False

Explanation:

3 0
3 years ago
Read 2 more answers
1 // Lab 2 tryIt2A 2 #include 3 using namespace std; 4 5 int main() 6 { int x = 1, y = 3; 7 int X = 2, Y = 4; 8 9 cout &lt;&lt;
padilas [110]

Answer:

Here is the complete program:

#include <iostream>

 using namespace std;    

 int main()

 {  int x = 1, y = 3;  

 int X = 2, Y = 4;  

 cout << "tryIt 2A" <<endl;

   cout << x << y << endl;  

   cout << "x" << "y" << endl;  

   cout << X << " " << Y << endl;

   cout << 2 * x + y << endl;  

   cout << 2 * X + Y << endl;  

   //cout << x + 2*y << endl;  

   cout << "x = ";  

   cout << x;  

   cout << " y = ";  

   cout << y;        

   return 0;

   }

Explanation:

I will explain the code line by line in the comment with each line of code and the output of each cout statement.

  • int x = 1, y = 3;  

This statement assigns value 1 to integer variable x and 3 to int variable y

  • int X = 2, Y = 4;  

This statement assigns value 2 to integer variable X and 4 to int variable Y As C++ is a case sensitive language so variable x and y are different from variables X and Y.

  • cout << "tryIt 2A" <<endl;

This statement has cout which is used to display output on the screen. So the output displayed by this cout statement is:

tryIt2A

  • cout << x << y << endl;  

This statement will print the values stored in x and y variables. So output displayed by cout statement here is 1 and 3. As there is not space or next line specified in the statement so output displayed will look like this:

13

  • cout << "x" << "y" << endl;  

This statement will display x and y but these are not the variable x and y. They are enclosed in double quotation marks so they are treated as strings not variables so the output displayed is:

xy

  • cout << X << " " << Y << endl;

This statement will print the values stored in X and Y variables. So output displayed by cout statement here is 2 and 4. As there is  space " " specified in the statement so 2 and 4 are displayed with a space between them so the output displayed will look like this:

2 4

  • cout << 2 * x + y << endl;  

This statement has an arithmetic operation in which 2 is multiplied by the values stored in variable x and then the result is added by value of y. So  2*1 = 2 and 2 + 3 = 5. So the result produced by this cout statement is:

5

  • cout << 2 * X + Y << endl;  

This will work same as above cout statement but the only difference is that the values of capital X and Y variables are calculated here. So 2 * 2 = 4 and then 4 + 4 = 8. The result produced by this cout statement is:

8

  • //cout << x + 2*y << endl;  

This is a comment because before this statement // is written which is used for single line comment. So compiler ignores comments and will not compile this statement.

  •    cout << "x = ";  

This will display "x = " as it is not variable but it is treated as a line to be displayed on the screen. So cout statement displays:

x =

  • cout << x;

This will print the value stored in x variable as there are no double quotes around x so it is a variable which contains value 1. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen:

x = 1

  • cout << " y = ";

This will display "y = " as it is not variable but it is treated as a line to be displayed on the screen. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen

x = 1 y =

  • cout << y;    

This will print the value stored in y variable as there are no double quotes around y so it is a variable which contains value 3. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen:

x = 1 y = 3

So the output of the entire program along with the program is attached as screenshot.

6 0
4 years ago
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