answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Rasek
1 month ago
9

JAVA Write a program that first asks the user to type a letter grade and then translates the letter grade into a number grade. L

etter grades are A, B, C, D, and F, possibly followed by + or –. Their numeric values are 4, 3, 2, 1, and 0. There is no F+ or F–. A + increases the numeric value by 0.3, a – decreases it by 0.3. However, an A+ has value 4.0. Use a class Grade with a method getNumericGrade. Also provide a tester class. Use -1 as a sentinel value to denote the end of letter grade inputs. The running results of your program should be like: Please enter a letter grade (enter -1 to end the input): B- The numeric value is 2.7. Please enter a letter grade (enter -1 to end the input):C The numeric value is 2.0. Please enter a letter grade (enter -1 to end the input):-1 The grade translation ends.
Computers and Technology
1 answer:
Amiraneli [1K]1 month ago
4 0

Answer:

The following code addresses this question:

import java.util.*;//import package for user input

class GradePrinter//defining class GradePrinter

{

double numericValue = 0;//defining double variable

String grade = "";//defining String variable

GradePrinter()//defining default constructor  

{

Scanner xb = new Scanner(System. in );//defining Scanner  class object

System.out.print("Enter Grade: ");//print message

grade = xb.nextLine();//input string value  

}

double getNumericGrade()//defining double method getNumericGrade

{

if (grade.equals("A+") || grade.equals("A"))//defining if block that check input is A+ or A

{

numericValue = 4.0;//using  numericValue variable that hold float value 4.0

}

else if (grade.equals("A-"))//defining else if that check grade equals to A-

{

numericValue = 3.7;//using  numericValue variable that hold float value 3.7

}

else if (grade.equals("B+"))//defining else if that check grade equals to B-

{

numericValue = 3.3;//using  numericValue variable that hold float value 3.3

}

else if (grade.equals("B"))//defining else if that check grade equals to B

{

numericValue = 3.0;//using  numericValue variable that hold float value 3.0

}

else if (grade.equals("B-"))//defining else if that check grade equals to B-  

{

numericValue = 2.7;//using  numericValue variable that hold float value 2.7

}

else if (grade.equals("C+"))//defining else if that check grade equals to C+  

{

numericValue = 2.3; //using  numericValue variable that hold float value 2.3

}

else if (grade.equals("C")) //defining else if that check grade equals to C  

{

numericValue = 2.0; //using numericValue variable that hold float value 2.0

}

else if (grade.equals("C-")) //defining else if that check grade equals to C-  

{

numericValue = 1.7;//using umericValue variable that hold float value 1.7

}

else if (grade.equals("D+"))//defining else if that check grade equals to D+  

{

numericValue = 1.3;//using umericValue variable that hold float value 1.3

}

else if (grade.equals("D"))//defining else if that check grade equals to D

{

numericValue = 1.0;//using umericValue variable that hold float value 1.0

}

else if (grade.equals("F"))//defining else if that check grade equals to F

{

numericValue = 0;//using umericValue variable that hold value 0

}

else//defining else block

{

System.out.println("Letter not in grading system");//print message

}

return numericValue;//return numericValue

}

}

class Main//defining a class main

{

public static void main(String[] args)//defining main method

{

GradePrinter ob = new GradePrinter();// creating class GradePrinter object

double numericGrade = ob.getNumericGrade();//defining double variable numericGrade that holds method Value

System.out.println("Numeric Value: "+numericGrade); //print Value numericgrade.

}

}

Output:

Enter Grade: B

Numeric Value: 3.0

Explanation:

The provided code defines a class named "GradePrinter" that includes a string and a double variable "grade and numericValue." The grade variable captures user input from the console.

After obtaining input, the getNumericGrade method employs multiple conditional checks to assign a floating value based on the given grade.

In the main class, an instance of "GradePrinter" is created, where a double variable "numericGrade" is defined to hold the method's value, followed by a print statement that outputs its value.

You might be interested in
You are given a string of n characters s[1 : : : n], which you believe to be a corrupted text document in which all punctuation
ivann1987 [1066]

Response: explained in the explanation section

Explanation:

Given that:

Assume D(k) =║ true if [1::: k] is a valid sequence of words, or false otherwise

  • In determining D(n)

the sub problem s[1::: k] is a valid sequence of words IFF s[1::: 1] is valid and s[ 1 + 1::: k] is a valid word.

Thus, we derive that D(k) is defined by the following recurrence relation:

D(k) = ║ false max(d[l] ∧ DICT(s[1 + 1::: k]) otherwise

Algorithm:

Valid sentence (s,k)

D [1::: k]             ∦ array of boolean variables.

for a ← 1 to m

do;

d(0) ← false

for b ← 0 to a - j

for b ← 0 to a - j

do;

if D[b] ∧ DICT s([b + 1::: a])

d (a) ← True

(b). Algorithm Output

      if D[k] == True

stack = temp stack            ∦stack assists in displaying the strings in order

c = k

while C > 0

stack push (s [w(c)]::: C] // w(p) denotes the index in s[1::: k] of the valid word // at position c

P = W (p) - 1

output stack

= 0 =

cheers, I hope this aids you!!!

8 0
2 months ago
Write the state of the elements of each of the following arrays after each pass of the outermost loop of the selection sort algo
Natasha_Volkova [1026]

Answer:

Below is the explanation provided.

Elaboration:

In the first array => 8, 5, -9, 14, 0, -1, -7, 3.

  • 8, 5, -9, 14, 0, -1, -7, 3 For the first iteration, find the smallest from the first to the eighth and reposition the third element in the first place.
  • -9, 8, 5, 14, 0, -1, -7, 3 In the second round, identify the smallest from the second to eighth and adjust the penultimate element in the second slot.
  • -9, -7, 8, 5, 14, 0, -1, 3 For the third pass, find the least from third to eighth and move the second last item from that segment.
  • -9, -7, -1, 8, 5, 14, 0, 3 For the fourth pass, find the minimum from the fourth to eighth and shift the last second item from that section.
  • -9, -7, -1, 0, 8, 5, 14, 3 For the fifth pass, seek for the smallest from the fifth to eighth and place the last element from that portion.
  • -9, -7, -1, 0, 3, 8, 5, 14 For the sixth pass, identify the smallest from the sixth to eighth and shift the last second element from that group.
  • -9, -7, -1, 0, 3, 5, 8, 14 are in order now.

In the second array => 15, 56, 24, 5, 39, -4, 27, 10.

  • 15, 56, 24, 5, 39, -4, 27, 10 In the first iteration, identify the smallest item from the first to the eighth and reposition the last third item in that section.
  • -4, 15, 56, 24, 5, 39, 27, 10 In the second round, find the minimum item from second to eighth and interchange the last fourth item in that segment.
  • -4, 5, 15, 56, 24, 39, 27, 10 For the third pass, select the smallest from the third to eighth and shift the last item from that set.
  • -4, 5, 10, 15, 56, 24, 39, 27 For the fourth stage, determine the smallest from the fourth to eighth and do not shift its position as it is already sorted.
  • -4, 5, 10, 15, 56, 24, 39, 27 For the fifth iteration, check for the smallest from the fifth to the eighth and intersperse the last third item from that section.
  • -4, 5, 10, 15, 24, 56, 39, 27 For the sixth phase, find the smallest from the sixth to the eighth and exchange the last item in the sixth.
  • -4, 5, 10, 15, 24, 27, 56, 39 are in order now.
4 0
1 month ago
A network administrator has received the IPv6 prefix 2001:DB8::/48 for subnetting. Assuming the administrator does not subnet in
8_murik_8 [964]

Answer:

subnets=65536

Explanation:

As per our knowledge,

--> the address's interface ID portion begins at 64

--> there exists a 48 bit network prefix

therefore,

the bits available for subnets are = 64-48=  16

thus, by utilizing 16 bits, the number of subnets can be calculated as = 2^16 = 65535

6 0
2 months ago
Other questions:
  • This program outputs a downwards facing arrow composed of a rectangle and a right triangle. The arrow dimensions are defined by
    15·1 answer
  • Tag groups can be nested up to ____ levels deep, with up to _______ tag subgroups under a single parent.
    14·1 answer
  • Modern operating systems decouple a process address space from the machine’s physical memory. List two advantages of this design
    15·1 answer
  • Write measurable performance objectives.Suppose that a trainer has identified as a generalgoal for a training module,"Able to fo
    13·1 answer
  • A developer writes a trigger on the Account object on the before update event that increments a count field. A workflow rule als
    12·1 answer
  • Consider a one-way authentication technique based on asymmetric encryption: A --> B: IDA B --> A: R1 A --> B: E(PRa, R1
    13·1 answer
  • 7.7 LAB: Using a while loop countdown Write a program that takes in an integer in the range 10 to 100 as input. Your program sho
    11·1 answer
  • Your computer uses 4 bits to represent decimal numbers (0, 1, 2, 3 and so on) in binary. What is the SMALLEST number for which a
    8·1 answer
  • Language levels have a direct influence on _______________
    10·1 answer
  • Of 500 sessions that occurred on a website during a one-week period, 200 of them started on the homepage; 100 of these 200 sessi
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!