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Zarrin [17]
3 years ago
13

Post as a reply your example of "data, which is processed into information" case - examples should not necessarily be related to

computers.
Computers and Technology
1 answer:
taurus [48]3 years ago
8 0

Answer:

can u be more specific so that I can helo u?

You might be interested in
The permissions of a file in a Linux system are split into three sets of three permissions: read, write, and execute for the own
Reptile [31]

Answer:

Explanation:

def octal_to_string(octal):

   result = ''

   value_letters = [(4, 'r'), (2, 'w'), (1, 'x')]

   for c in [int(n) for n in str(octal)]:

       for value, letter in value_letters:

           if c >= value:

               result += letter

               c -= value

           else:

               result += '-'

   return result

print(octal_to_string(755))

print(octal_to_string(644))

print(octal_to_string(750))

print(octal_to_string(600))

**************************************************

7 0
3 years ago
Help! What is this graph and what does it represent?
Lunna [17]

Answer:

Explanation:

how much something had in each month in this graph

7 0
3 years ago
The address of the last cell of a memory RAM is 3FFFFh.a total capacity of the main memory and 4M bits and the data bus can tran
marysya [2.9K]

Hey there!:

Given the address of the last cell of a memory RAM is 3FFFFh then   :

There are 8 bits in a byte  

1024 bytes in a kilobyte  

 so 8*1024= 8192  

 so 8192 bits in a kb  

 32*8192 = 262144  

there are 262144 bits in 32KB .

a) 32 bit address registers must match 2^32 byte = 4 GB of physical memory.

However, with 32 Bit you can also address more than 4 GB, like Physical Address Extension (PAE) does.

The other way is that you can access 4 GB with less than 32 bit address register.

____________________________________________________

b) Main Memory = 4M × 32Kbits, RAM chips = 32K × 4bit.

For this memory we require 4 × 2 = 8 RAM chips.

Each chip requires 18 address bits (ie. 218 = 256K).

And 1M × 8 bits requires 20 address bits (ie. 220 = 1M )

_____________________________________________________

c) The size of the storage cells is known as the word size for the computer.

In some computers, the word size is one byte while in other computers the word size is two, four, or even eight bytes. In our 4M main memory the size of cell is 4*2^20 bytes.

Each storage cell in main memory has a particular address which the computer can use for storing or retrieving data.  

______________________________________________________

d) The size of MBR is the 4M bits of main memory.

_____________________________________________________

Hope this helps!

6 0
3 years ago
The frame work for storing records in database is a
andreev551 [17]
The frame work for storing records in database is a report. (b)
3 0
3 years ago
Write a program to process weekly employee time cards for all employees of an organization. Each employee will have three data i
Naddika [18.5K]

Answer:

The cpp program for the scenario is shown.

#include <iostream>

using namespace std;

 int main() {

 

     int count;

     int empNum[count];

     double work_hrs[count];

     double hrly_wage[count];

     double ot_wage[count];

     double hour = 40.00;

     double gross_pay[count];

     double tax=3.625;

     double total_pay = 0, avg_pay;

     

   

   cout<<"Enter the number of employees "<<endl;

   cin>>count;

   cout<<"Enter the details for the employees "<<endl;

   

   int i=0;

   

   while(i<count)

   {

       cout<<"Enter the id"<<endl;

       cin>>empNum[i];  

       

       cout<<"Enter the working hours"<<endl;

       cin>>work_hrs[i];

       

       cout<<"Enter the hourly pay"<<endl;

       cin>>hrly_wage[i];

       ot_wage[i] = hrly_wage[i]*1.5;

       

       i++;

       

   }

   

   cout<<"The payroll for the employees "<<endl;

   

   i=0;

   

   while(i<count)

   {

       if(work_hrs[i] > hour)

           gross_pay[i] = ( hour*hrly_wage[i] );

       else

           gross_pay[i] = ( hrly_wage[i]*work_hrs[i] );

       

       if(work_hrs[i] > hour)

           gross_pay[i] = gross_pay[i] + ( (work_hrs[i]-hour)*ot_wage[i] );

           

       gross_pay[i] = gross_pay[i]-( (gross_pay[i]*tax)/100 );

       

       total_pay = total_pay + gross_pay[i];

       i++;

       

   }

   

   avg_pay = total_pay/count;

   i=0;

   

   while(i<count)

   {

       cout<<"Gross pay of employee "<<empNum[i]<<" : "<<gross_pay[i]<<endl;

   

       i++;

   }

   cout<<"Average amount paid to all employees is "<<avg_pay<<endl;

   

   return 0;

 }

OUTPUT

Enter the number of employees                                                                                                                

2                                                                                                                                            

Enter the details for the employees                                                                                                    

Enter the id                                                                                                                                  

111                                                                                                                                          

Enter the working hours                                                                                                                      

46                                                                                                                                            

Enter the hourly pay                                                                                                                          

12                                                                                                                                            

Enter the id                                                                                                                                  

222                                                                                                                                          

Enter the working hours                                                                                                                      

50                                                                                                                                            

Enter the hourly pay                                                                                                                          

14                                                                                                                                            

The payroll for the employees                                                                                                          

Gross pay of employee 111 : 566.685                                                                                                          

Gross pay of employee 222 : 742.087                                                                                                          

Average amount paid to all employees is 654.386                                                                                              

Explanation:

1. User enters the number of employees.

2. User enters all pieces of information including identification number, hourly wage rate and number of hours worked.

3. Inside a while loop, user input is taken in the arrays.

4. Inside another while loop, the gross pay of each employee is computed. The gross pay of each employee is added to the variable, total_pay.

5. The value of the variable, avg_pay, is computed outside the loop.

6. All the while loops work over variable i till the value of i becomes 1 less than count.

7. The value of the variable, i, is made 0 before the loop begins.

8. The employee number and the gross pay of each employee is displayed followed by the average pay.

5 0
4 years ago
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