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stellarik [79]
3 years ago
15

Write a program that prompts the user to enter two characters and display the corresponding major and year status. The first cha

racter indicates the major. And the second character is a number character 1, 2, 3, 4, which indicates whether a student is a freshman, sophomore, junior, or senior. We consider only the following majors:
B (or b): Biology

C (or c): Computer Science

I (or i): Information Technology and Systems

Note that your program needs to let the user know if the major or year is invalid. Also, your program should be case-insensitive: your program should tell "Biology" either user type ‘b’ or ‘B’.

Here are three sample runs:

Sample 1:

Enter two characters: i3

Information Technology and Systems

Junior

Sample 2:

Enter two characters: B5

Biology

Invalid year status

Sample 3:

Enter two characters: t2

Invalid major

Sophomore

What to deliver?

Your .java file including:

1. (Code: 5 points, Comments: 3 points) Your code with other appropriate comments.

2. (2 points) Four sample runs with the following four input:
(1) i3

(2) T2

(3) c2

(4) F0

Note that you need to run your program 4 times. Each time you run it, copy and paste the program output to the top of your program. After you finished these 4 runs, comment out the output you pasted so that you program would continue to run without any issue.
Computers and Technology
1 answer:
nydimaria [60]3 years ago
4 0

Answer:

  1. import java.util.Scanner;  
  2. public class Main {
  3.    public static void main(String[] args) {
  4.        Scanner input = new Scanner(System.in);
  5.        System.out.print("Please enter two characters: ");
  6.        String inputStr = input.nextLine();
  7.        if(inputStr.charAt(0) == 'B' || inputStr.charAt(0) == 'b'){
  8.            System.out.println("Biology");
  9.        }
  10.        else if(inputStr.charAt(0) == 'C' || inputStr.charAt(0)== 'c'){
  11.            System.out.println("Computer Science");
  12.        }
  13.        else if(inputStr.charAt(0) == 'I' || inputStr.charAt(0) == 'i')
  14.        {
  15.            System.out.println("Information Technology and Systems");
  16.        }
  17.        else{
  18.            System.out.println("Invalid major");
  19.        }
  20.        int num = Character.getNumericValue(inputStr.charAt(1));
  21.        if(num >= 1 && num <= 4){
  22.            switch (num){
  23.                case 1:
  24.                    System.out.println("freshman");
  25.                    break;
  26.                case 2:
  27.                    System.out.println("sophomore");
  28.                    break;
  29.                case 3:
  30.                    System.out.println("junior");
  31.                    break;
  32.                case 4:
  33.                    System.out.println("senior");
  34.                    break;
  35.            }
  36.        }
  37.        else{
  38.            System.out.println("Invalid year status");
  39.        }
  40.    }
  41. }

Explanation:

The code consists of two main parts. The part 1 is to validate the input major and print out the major according to the input character (Line 10 -22). If the input character is not matched with the target letters, a message invalid major will be displayed.

The part 2 is to validate the year status to make sure it only fall within the range of 1-4  (Line 26 -45). If within the range, the program will display the year major accordingly. If not a message invalid year status will be displayed.

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Assume user_name equals "Tom" and user_age equals 22. What is printed on the console when the following statement is executed? c
12345 [234]

Answer:

The ouput of the given code is :

22

is "Tom's age.

Explanation:

Here in this code the variable user_name and user_age are initialized to  "Tom" and 22 respectively as statement is given in the question i.e  cout << user_age << " \nis " + user_name << "'s age.";.This line will print the user_age i.e 22 after that the control moves to the next line and print is "Tom's age.

Following are the code in c++

#include <iostream> // header file

#include <string>

using namespace std;

int main() // main function

{

   string user_name="Tom";

   int user_age= 22;

cout << user_age << " \nis " + user_name << "'s age.";

return 0;

}

Output:

22

is "Tom's age.

8 0
3 years ago
- Consider the relation R = {A, B, C, D, E, F, G, H, I, J} and the set of functional dependencies F = { {B, C} -&gt; {D}, {B} -&
Olenka [21]

Answer:

The key of R is {A, B}

Explanation:

A key can be seen as a minimal set of attributes whose closure includes all the attributes in R.

Given that the closure of {A, B}, {A, B}+ = R, one key of R is {A, B} But in this case, it is the only key.

In order for us to to normalize R intuitively into 2NF then 3NF, we have to make use of these approaches;

First thing we do is to identify partial dependencies that violate 2NF. These are attributes that are

functionally dependent on either parts of the key, {A} or {B}, alone.

We can calculate

the closures {A}+ and {B}+ to determine partially dependent attributes:

{A}+ = {A, D, E, I, J}. Hence {A} -> {D, E, I, J} ({A} -> {A} is a trivial dependency)

{B}+ = {B, F, G, H}, hence {A} -> {F, G, H} ({B} -> {B} is a trivial dependency)

To normalize into 2NF, we remove the attributes that are functionally dependent on

part of the key (A or B) from R and place them in separate relations R1 and R2,

along with the part of the key they depend on (A or B), which are copied into each of

these relations but also remains in the original relation, which we call R3 below:

R1 = {A, D, E, I, J}, R2 = {B, F, G, H}, R3 = {A, B, C}

The new keys for R1, R2, R3 are underlined. Next, we look for transitive

dependencies in R1, R2, R3. The relation R1 has the transitive dependency {A} ->

{D} -> {I, J}, so we remove the transitively dependent attributes {I, J} from R1 into a

relation R11 and copy the attribute D they are dependent on into R11. The remaining

attributes are kept in a relation R12. Hence, R1 is decomposed into R11 and R12 as

follows: R11 = {D, I, J}, R12 = {A, D, E} The relation R2 is similarly decomposed into R21 and R22 based on the transitive

dependency {B} -> {F} -> {G, H}:

R2 = {F, G, H}, R2 = {B, F}

The final set of relations in 3NF are {R11, R12, R21, R22, R3}

4 0
3 years ago
Please hurry Arrange the steps of the engineering design process in the correct sequence.
AnnZ [28]

Answer:The Nine steps  of the engineering design process

  1. <u>Identify the Problem</u>-Defining the problem
  2. Finding solutions through Brainstorming technique
  3. Conducting a background research/Survey
  4. Developing the solution-Creating an array of solutions
  5. Selecting the best solution
  6. Building a prototype
  7. Testing and redesigning
  8. Improve the design-Specifying the requirement
  9. Communicating the result

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3 years ago
Which statement is true? Select 3 options.
irina1246 [14]

Answer:

The statements which are true are;

  • A user-defined data type can include other user-defined data types
  • A user-defined data type is defined using a class
  • A user-defined data type can include a list

Explanation:

A user-defined data type (UDT) is a datatype that is defined and derived by the use of the datatypes which preexist including existing user-defined datatypes and the built-in datatypes

It is therefore true that a user-defined data type can include other user-defined data types

A class is a user-defined data type that contains both its member data and member functions, that can be used when an instance of the class is first created

Therefore, a user-defined data type is defined using a class

In a user-defined data type, a variable has actual data within it which can include an array or list

Therefore a user-defined data type can include a list.

3 0
2 years ago
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Preciso de ajudar para resolver esse exercício, é para amanhã cedo!!<br><br> Em Dev C++
LenKa [72]
Bbbbbbbbbbbbbbbbbbbbbb
3 0
3 years ago
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