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zavuch27 [327]
2 years ago
11

Suppose your cell phone carrier charges you a monthly fee of $30.00 for up to 300 minutes and $0.45 for each additional minute a

fter the first 300. Assuming you used your phone for x minutes with x > 300, the total monthly fee would be?
If you used your phone for <= 300 minutes then the charge is $30.00 Assume the user enters a positive value for the number of minutes
Computers and Technology
1 answer:
irakobra [83]2 years ago
6 0

Answer:

30.00 + x * .45 where x is the number of minutes

in code it would be:

import java.util.Scanner;

public class monthlyFee{

   final double MONTHLY_FEE = 30.00;

   final double ADDITIONAL_MINUTE_FEE = 0.45;

   double additionalMinute = 0;

   public monthlyFee(double minutes){

   additionalMinute = minutes;

   }

   public double calculate(){

       if(additionalMinute > 0){

       double TotalFee = MONTHLY_FEE + additionalMinute * ADDITIONAL_MINUTE_FEE;

       return TotalFee;

       }

       else

       return MONTHLY_FEE;

   }

   public static void main(String...args){

   Scanner input = new Scanner(System.in);

   System.out.println("Enter minutes: ");

   double minutes = input.nextDouble();

   monthlyFee obj1 = new monthlyFee(minutes);

   System.out.printf("$" + "%.2f", obj1.calculate());

   }

}

Explanation:

so whatever x is just multiple it by 0.45 to get the total for just the fee.

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Answer:

D. Selfishness

Explanation:

Selfishness is caring about yourself and not others

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3 years ago
You are given a string of n characters s[1 : : : n], which you believe to be a corrupted text document in which all punctuation
alexdok [17]

Answer: provided in the explanation section

Explanation:

Given that:

Assume D(k) =║ true it is [1 : : : k] is valid sequence  words or false otherwise

  • To determine D(n)

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

So, from here we have that D(k) is given by the following recorance relation:

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

Algorithm:

Valid sentence (s,k)

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

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 is used to print the strings in order

c = k

while C > 0

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

P = W (p) - 1

output stack

= 0 =

cheers i hope this helps !!!

8 0
3 years ago
Help me asap ill give brainliest
kobusy [5.1K]

Answer:

Give me brainliest thanks

6 0
3 years ago
4.2.3: Basic while loop expression. Write a while loop that prints userNum divided by 2 (integer division) until reaching 1. Fol
wlad13 [49]

Answer:

import java.io.*;

import java.util.Scanner;

class divide {

public static void main (String[] args) {

    Scanner num=new Scanner(System.in);//scanner object.

    int userNum=num.nextInt();

    while(userNum>1)//while loop.

    {

        userNum/=2;//dividing the userNum.

        System.out.print(userNum+" ");//printing the userNum.

    }

}

}

Input:-

40

Output:-

20 10 5 2 1

Input:-

2

Output:-

1

Input:-

0

Output:-

No Output

Input:-

-1

Output:-

No Output.

Explanation:

In the program While loop is used.In the while loop it divides the userNum by 2 in each iteration and prints the value of userNum.The inputs and corresponding outputs are written in the answer.

4 0
2 years ago
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What standards organization maintains the ethernet standard?.
Fofino [41]

Answer:

Ethernet standards are written and maintained by the IEEE, the Institute of Electrical and Electronic Engineers which has its corporate office in New York City and its operations center in Piscataway, New Jersey.

Explanation:

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