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Anarel [89]
2 years ago
9

Assume you have an Access database with five different tables, including various pieces of information about your client base. Y

ou want to pull information from Table 1, which includes your client names and addresses. From Table 2, you want to pull information on how much they've paid for your product in the past year. What is the best way to set this up for a mail merge?
A. Copy the two tables from Access to Excel, and then perform the mail merge from Word.
B. Export both tables from Access and import them into Word to complete the mail merge.
C. Create a query that includes the required information for the mail merge.
D. Manually retype the lists into Word to complete the mail merge.
Computers and Technology
1 answer:
Fofino [41]2 years ago
4 0

The answer is A. Copy the two tables

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Give me two reasons why return statements are used in code.
levacccp [35]
It can stop the function when it’s no longer needed to keep running

And it can give a certain value to send back so it can be used elsewhere in your code
8 0
3 years ago
Read 2 more answers
A lever has an effort arm that is 8 meters long and the residence (load) arm that is 1.5 meters long, how much effort is needed
andrew11 [14]

Answer:

1066.67 N

Explanation:

Given that two measurements of the arm and an input weight. To answer this problem,we need to balance the forces and use the lengths of the arms. 

Force × effort of arm distance= input weight × load distance

200 N * 8 m = x * 1.5 m

1600 = 1.5x

x = 1600/1.5

x = 1066.666 N

it takes 1066.67 N to lift the input weight

4 0
3 years ago
I would A lot of knowledge and education for computers and <br> Technology
borishaifa [10]

Answer:

Pay attention in your class.

Explanation:

If you paid attention in your class, you would have at least a basic understanding of the material.

7 0
2 years ago
Read 2 more answers
Write a program that will open the file random.txt and calculate and display the following: A. The number of numbers in the file
Rama09 [41]

Answer:

Here is the C++ program:

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

#include <fstream>  //to manipulate files

using namespace std;  //to identify objects like cin cout

int main(){  //start of main function

  ifstream file;   //creates an object of ifstream

   file.open("random.txt"); //open method to open random.txt file using object file of ifstream

       

    int numCount = 0;  //to store the number of all numbers in the file          

    double sum = 0;   //to store the sum of all numbers in the file

    double average = 0.0;   //to store the average of all numbers in the file

    int number ; //stores numbers in a file

                 

        if(!file){   //if file could not be opened

           cout<<"Error opening file!\n";    }   //displays this error message

       

       while(file>>number){   //reads each number from the file till the end of the file and stores into number variable

           numCount++; //adds 1 to the count of numCount each time a number is read from the file          

           sum += number;  }  //adds all the numbers and stores the result in sum variable

           average = sum/numCount;  //divides the computed sum of all numbers by the number of numbers in the file

     

       cout<<"The number of numbers in the file: "<<numCount<<endl;  //displays the number of numbers

       cout<<"The sum of all the numbers in the file: "<<sum<<endl;  //displays the sum of all numbers

       cout<<"The average of all the numbers in the file: "<< average<<endl;  //displays the average of all numbers

       file.close();     }  //closes the file    

   

Explanation:

Since the random.txt is not given to check the working of the above program, random.txt is created and some numbers are added to it:

35

48

21

56

74

93

88

109

150

16

while(file>>number) statement reads each number and stores it into number variable.

At first iteration:

35 is read and stored to number

numCount++;  becomes

numCount = numCount + 1

numCount = 1      

sum += number; this becomes:

sum = sum + number

sum = 0 + 35

sum = 35

At second iteration:

48 is read and stored to number

numCount++;  becomes

numCount = 1+ 1

numCount = 2    

sum += number; this becomes:

sum = sum + number

sum = 35 + 48

sum = 83

So at each iteration a number is read from file, the numCount increments to 1 at each iteration and the number is added to the sum.

At last iteration:

16 is read and stored to number

numCount++;  becomes

numCount = 9 + 1

numCount = 10    

sum += number; this becomes:

sum = sum + number

sum = 674 + 16

sum = 690

Now the loop breaks and the program moves to the statement:

       average = sum/numCount;  this becomes:

       average = 690/10;

       average = 69

So the entire output of the program is:

The number of numbers in the file: 10                                                                                                           The sum of all the numbers in the file: 690                                                                                                     The average of all the numbers in the file: 69

The screenshot of the program and its output is attached.

5 0
3 years ago
computer has a 32-bit instruction word broken into fields as follows: opcode, six bits; two register file address fields, five b
lina2011 [118]

Answer:

a.  2^6, or 64 opcodes.

b.  2^5, or 32 registers.

c. 2^16, or 0 to 65536.

d.  -32768 to 32768.

Explanation:

a. Following that the opcode is 6 bits, it is generally known that the maximum number of opcodes should be 2^6, or 64 opcodes.

b. Now, since the size of the register field is 5 bits, we know that 2^5 registers can be accessed, or 32 registers.

c. Unsigned immediate operand applies to the plus/minus sign of the number. Since unsigned numbers are always positive, the range is from 0 to 2^16, or 0 to 65536.

d. Considering that the signed operands can be negative, they need a 16'th bit for the sign and 15 bits for the number. This means there are 2 * (2^15) numbers, or 2^16. However, the numbers range from -32768 to 32768.

6 0
3 years ago
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