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2 years ago

04. Suppose a computer program needs to sort a list of student records in ascending order

Computers and Technology

1 answer:

34kurt2 years ago

7 0

Answer:

D) ["Kathy Bones", "Jill Brewer", "Joe Schnook", "Tom Smith"]

Explanation:

The context of the problem explains a computer program that sorts names in "ascending order" (A to Z) since the ASCII table has capital A start a lower number and it increases from there to capital Z.

Notice how the attached file, which is a portion of the ASCII table, shows that letters after A are also higher in decimal value than the previous letter.

With this in mind, we know that all this program does is sort by last name alphabetical order. From there, just sort the given names using that same criteria, last name alphabetical order, and the correct answer is determined.

Please put "Brainliest" on my answer if it helped you out!

If you want to learn more about this subject, you can search:

- ASCII Table

- Lists in Programming

- Sorting Procedures

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3 years ago

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2 years ago

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3 years ago

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We will use linear interpolation in a C program to estimate the population of NJ between the years of the census, which takes pl

nikklg [1K]

Answer:

See explaination for program code

Explanation:

program code below:

#include<stdio.h>

int main()

{

//file pointer to read from the file

FILE *fptr;

//array to store the years

int years[50];

//array to store the population

float population[50];

int n = 0, j;

//variables for the linear interpolation formula

float x0,y0,x1,y1,xp,yp;

//opening the file

fptr = fopen("njpopulation.dat", "r");

if (fptr != NULL)

{

//reading data for the file

while(fscanf(fptr, "%d %f", &years[n], &population[n]) == 2)

n++;

//prompting the user

int year = -1;

printf("Enter the years for which the population is to be estimated. Enter 0 to stop.\n\n");

while(year != 0)

{

printf("Enter the year (1790 - 2010) : ");

scanf("%d", &year);

if (year >= 1790 && year <= 2010)

{

//calculating the estimation

xp = year;

for (j = 0; j < n; j++)

{

if (year == years[j])

{

//if the year is already in the table no nedd for calculation

yp = population[j];

}

else if (j != n - 1)

{

//finding the surrounding census years

if (year > years[j] && year < years[j + 1])

{

//point one

x0 = years[j];

y0 = population[j];

//point two

x1 = years[j + 1];

y1 = population[j + 1];

//interpolation formula

yp = y0 + ((y1-y0)/(x1-x0)) * (xp - x0);

}

printf("\nEstimated population for year %d : %.2f\n\n", year, yp);

}

else if (year != 0)

{

printf("\nInvalid chioce!!\n\n");

}

printf("\nAborting!!");

}

else

{

printf("File cannot be opened!!");

}

close(fptr);

return 0;

}

OUTPUT :

Enter the years for which the population is to be estimated. Enter 0 to stop.

Enter the year (1790 - 2010) : 1790

Estimated population for year 1790 : 184139.00

Enter the year (1790 - 2010) : 1855

Estimated population for year 1855 : 580795.00

Enter the year (1790 - 2010) : 2010

Estimated population for year 2010 : 8791894.00

Enter the year (1790 - 2010) : 4545

Invalid chioce!!

Enter the year (1790 - 2010) : 1992

Estimated population for year 1992 : 7867020.50

Enter the year (1790 - 2010) : 0

Aborting!!

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3 years ago

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