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makkiz [27]
2 years ago
13

Which access object cannot be used to enter or edit data

Computers and Technology
2 answers:
kap26 [50]2 years ago
6 0

Answer:

Should be B. table !

dem82 [27]2 years ago
4 0
It’s b I swARE tvyvubu y tho

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/* ELEN 1301 Programming Assignment #5. Name : Your name. Student ID : Your student ID #. Due date : Due date Purpose of the pro
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Answer:

Here is the C++ program:

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

#include<iomanip> //to use setprecision

using namespace std; //to identify objects cin cout

int main() { //start of main function

     int n1, n2, n3, n4, n5, n6, n7, min, max;  // declare variables for 7 numbers, minimum value and maximum value

     double sum= 0; //declare variable to hold sum of 7 numbers

     double average;  //declare variable to hold average of 7 numbers

     cout<<"Enter first number: "; //prompts user to enter 1st number

     cin>>n1; //reads first number from user

     max = n1; //sets the first number to maximum

     min=n1; //sets the first number to minimum

     cout<<"Enter second number: "; //prompts user to enter 2nd number

     cin>>n2; //reads second number from user

     if(n2<min){ //if second number is less than min

          min=n2;      } //sets min to n2

     if(n2>max){ //if n2 is greater than max

          max = n2;      } //sets max to n2

     cout<<"Enter third number: ";  //prompts user to enter 3rd number

     cin>>n3; //reads third number from user

     if(n3<min){ //checks if n3 is greater than min

          min=n3;      } //sets n3 to min

     if(n3>max){ //checks if n3 is greater than max

          max = n3;      }      //sets max to n3

    cout<<"Enter fourth number: ";//prompts user to enter 4th number

     cin>>n4; //reads fourth number from user

     if(n4<min){  //if n4 is less than min

          min=n4;      }  //sets min to n4

     if(n4>max){  //if n4 is greater than max

          max = n4;      }  //sets max to n4

     cout<<"Enter fifth number: "; //prompts user to enter 5th number

     cin>>n5; //reads fifth number from user

     if(n5<min){  //if n5 is less than min

          min=n5;     }  //sets min to n5

     if(n5>max){  //if n5 is greater than max

          max = n5;      }  //sets max to n5

     cout<<"Enter sixth number: "; //prompts user to enter 6th number

     cin>>n6; //reads sixth number from user

     if(n6<min){ // if n6 is less than min

          min=n6;      }  //sets min to n6

     if(n6>max){  //if n6 is greater than max

          max = n6;      }  //sets max to n6

     cout<<"Enter seventh number: ";//prompts user to enter 7th number

     cin>>n7; //reads seventh number from user

     if(n7<min){  //if n7 is less than minimum number

          min=n7;      }  //assigns n7 to min

     if(n7>max){  //if n7 is greater than the maximum number

          max = n7;      }  //assigns n7 to max

     sum = n1+n2+n3+n4+n5+n6+n7;  //adds 7 numbers

     average = sum/7;  //computes average of 7 numbers

     cout<<"The average is: "<<fixed<<setprecision(3)<<average<<endl; //displays average value up to show 3 digits below decimal point using setprecision method of iomanip library

     cout<<"The maximum number is: "<<max<<endl; //displays maximum number of 7 numbers

     cout<<"The minimum number is: "<<min<<endl;  //displays miimum number of 7 numbers    

     return 0; }

Explanation:

The program is well explained in the comments attached to each statement of program. For example if

n1 = 3

n2 = 9

n3 = 7

n4 = 6

n5 = 2

n6 = 5

n7 = 4

When n1 is read using cin then this number is set to max and min as:

min = 9

max = 3

Now when n2 is read, the first if condition checks if n2 is less than min and second if condition checks if n2 is greater than max. As n2 = 9 so it is not less than min so this if condition is false and n2 is greater than max i.e. 3 so this condition is true. So 9 is assigned to max.

min = 9

max = 9

Now when n3 is read the values of min and max become:

min = 7

max = 9

Now when n4 is read the values of min and max become:

min = 6

max = 9

Now when n5 is read the values of min and max become:

min = 2

max = 9

Now when n6 is read the values of min and max become:

min = 2

max = 9

Now when n7 is read the values of min and max become:

min = 2

max = 9

Now the statement       sum = n1+n2+n3+n4+n5+n6+n7;

executes which becomes:

sum = 3 + 9 + 7 + 6 + 2 + 5 + 4

sum = 36.0

Next program control moves to statement:

average = sum/7;

this becomes

average = 36/7

5.142857

Since this is to be displayed up to 3 decimal places so average = 5.143

the complete output of the program is attached.

3 0
3 years ago
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