Consider the following method, sumRows, which is intended to traverse all the rows in the two-dimensional (2D) integer array num and print the sum of all the elements in each row.
public static void sumRows(int[][] num)
{
for (int[] r : num)
{
int sum = 0;
for (int j = 0; j < num.length; j++)
{
sum += r[j];
}
System.out.print(sum + " ");
}
}
For example, if num contains {{3, 5}, {6, 8}}, then sumRows(num) should print "8 14 ".
The method does not always work as intended. For which of the following two-dimensional array input values does sumRows NOT work as intended?
A. {{10, -18}, {48, 17}}
B. {{-5, 2, 0}, {4, 11, 0}}
C. {{4, 1, 7}, {-10, -11, -12}}
D. {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
E. {{0, 1}, {2, 3}}

Answer:

The programming language is not stated; so, I'll solve this question using Java programming language

Comments are used for explanatory purpose

//Begin of Program

import java . util.*;

import java. math . RoundingMode;

import java . text . DecimalFormat;

public class keyfreq

{

private static DecimalFormat df = new DecimalFormat("0.00");

public static void main(String [] args)

{

 Scanner input = new Scanner(System.in);

 //Declare variable

 float f0;

//Prompt user for input

System.out.print("Enter Initial Key Frequency: ");

f0 = input.nextFloat();

//Initialize number of keys

int numkey = 1;

//Print first key frequency

System.out.print("Key Frequencies: " + df.format(f0)+" ");  

while(numkey<=4)

{

 //Calculate next frequency

 f0*= Math.pow(2,(1.0/12.0));

 //Print Frequency

 System.out.print(df.format(f0)+" ");  

  //Iterate to next frequency

 numkey++;

}

}

}

//End of Program

Explanation:

See Comments in the above program

See Attachment for source file

Answer:

hardware

Explanation:

hardware is the part where we can touch and feel while software ,we cant

Answer:

keyboard

Explanation:

Answer:

The value EAX contain after the code has been executed is 1233

Explanation:

Solution

Now,

The below instruction describes the MOV operation.

The MOV esi, OFFSET idArray .

It shows the movement, the offset value of the idArray in the esi.

Thus,

Follow the below instruction

MOV eax, [esi+7*TYPE idArray]

It will move the address of the idArray at that position into eax.

Value at the 7th position is moved to eax.

Therefore, the value at eax will be 1233.

Answer:

prompt("Enter a value for one edge of a cube")

Store user's value into edgeCube

area = 6 * (edgeCube * edgeCube)

volume = edgeCube * edgeCube * edgeCube

print("One side of the cube is: " + edgecube);

print("The area is: " + area)

print("The volume is: " + volume)

Answer:

Explanation:

The solution code is written in Python.

Firstly, prompt user to input three numbers for red, green and blue (Line 1-3).

Next, create three similar if statements to check if the red, green and blue are within the range 0-255 by using logical operator "or" to check if the number is smaller than 0 or bigger than 255. If either one condition is met, display a message to indicate a particular color number is not correct (Line 5-12).

The program written in python 3 which displays the whether the range of RGB values inputted by the user is in range goes thus :

r = int(input('Enter red value between 0 - 255'))

g = int(input('Enter green value between 0 - 255'))

b = int(input('Enter blue value between 0 - 255'))

#accepts inputs for red, green and blue values from the user

vals = {'red':r, 'green':g, 'blue':b}

#read the values into a dictionary and assign the values inputted by the user as the values

for k, v in vals.items():

#iterate through the dictionary as key-value pairs

if(v < 0) or (v > 255):

#Check of the values inputted is within range (0 - 255)

print('%s is not correct' %k)

#for any value which is out of range, display the name of the color and state that it is incorrect.

A sample run of the program ls attached below.

Learn more :https://brainly.com/question/18802298

Answer: Provided in the explanation section

Explanation:

class philosopher extends Thread

{

public void run()

{

diningps.count++;

philosopher(diningps.count);

}

public void philosopher(int i)

{

take_forks(i);

System.out.println(i+"philosopher Eating");

put_forks(i);

}

public void take_forks(int i)

{

while(diningps.mutex<=0)

{

System.out.println(i+"philosopher has to be wait while other philosopher in testing");

}

diningps.state[i]="HUNGRY";

test(i);

diningps.mutex=1;

while(diningps.S[i]<=0)

{

System.out.println(i+"philosopher has to be wait while his left or right side philosopher eating.");

}

}

public void put_forks(int i)

{

while(diningps.mutex<=0)

{

System.out.println(i+"philosopher has to be wait while other philosopher in testing");

}

diningps.state[i]="THINK";

test((i+4)%5);

test((i+1)%5);

diningps.mutex=1;

}

public void test(int i)

{

if(diningps.state[i]=="HUNGRY" && diningps.state[(i+4)%5]!="EAT" && diningps.state[(i+1)%5]!="EAT")

{

diningps.state[i]="EAT";

diningps.S[i]=1;

}

}

}

class diningps

{

static int mutex=1;

static int S[]={0,0,0,0,0};

static String state[]={"THINK","THINK","THINK","THINK","THINK"};

static int count=-1;

public static void main(String ar[])

{

philosopher ob=new philosopher();

philosopher ob1=new philosopher();

philosopher ob2=new philosopher();

philosopher ob3=new philosopher();

philosopher ob4=new philosopher();

ob.start();

ob1.start();

ob2.start();

ob3.start();

ob4.start();

}

}

cheers i hope this helped !!

Answer:

Explanation:

import java.io.*;

import java.util.*;

public class GolfScores

{

    // *************************************************************

    // the purpose of this program is to read golf scores between

    // four players and figure out the winner with the lowest score

    // **************************************************************

    public static void main(String[] args) throws FileNotFoundException

    {

         final int HOLES = 28, VALUES_PER_LINE = 4;

         int hole = 0;

         int parscores[] = new int[HOLES];

         //declare a file object to handle the text file with scores

         File file1 = new File("golfscore.txt");

         int sum0=0;

         int sum1=0;

         int sum2=0;

         int sum3=0;

         int sum4=0;

         Scanner scan = new Scanner(file1);

         // total par and scores for all holes

         int scores[][] = new int[HOLES][VALUES_PER_LINE];

         //loop to store the contents of the file into arrays

         while (scan.hasNext())

         {

             int index = 0;

             parscores[hole] = scan.nextInt();

                  while (index < 4)

             {

                  scores[hole][index] = scan.nextInt();                                             // values in a line

                  index++;

             }

             //

             hole++;

         }

         for(int i=0;i<HOLES;i++)

         {

             sum0+=parscores[i];

         }

         int sumarray[]=new int[4];

         //loop to calculate the total score of each player

         for(int row=0;row<HOLES;row++)

         {

             sum1+=scores[row][0];

             sum2+=scores[row][1];

            sum3+=scores[row][2];

             sum4+=scores[row][3];

         }

         System.out.println();

         System.out.println("Par for the course"+sum0);

         System.out.println("Player 1:"+sum1);

         System.out.println("Player 2:"+sum2);

         System.out.println("Player 3:"+sum3);

         System.out.println("Player 4:"+sum4);

         sumarray[0]=sum1;

         sumarray[1]=sum2;

         sumarray[2]=sum3;

         sumarray[3]=sum4;

         int min=sumarray[0];

         //Loop to find out the minimum score

         for(int j=0;j<4;j++)

         {

             if(sumarray[j]<min)

                  min=sumarray[j];

         }

         System.out.print("The winner is ");

         //condition statement to determine the winner

         if(min==sum1)

         {

             System.out.print("Player1");

         }

             else if(min==sum2)

             {

                  System.out.print("Player2");

         }

             else if(min==sum3)

             {

                  System.out.print("Player3");

         }

             else

             {

                  System.out.print("Player4");

         }

    }

}

cheers i hope this helped !!

Answer:

c

Explanation:

because its inportant



Answer:

[tex]L=2^{25830}[/tex] is the correct answer .

Explanation:

bandwidth = 3000 Hz

SNR =3162

We know that

[tex]bitrate = bandwidth * log2(1 + SNR)[/tex]

Putting the value of bandwidth and SNR in the given equation .

[tex]bitrate = 3000 * log_{2} ^\ (1 + 3162) \\\\birate = 3000 * log_{2} ^\ {3163} \\\\birate =3000 * 11.\ 6\\\\bitrate=\ 34860 bps[/tex]

Now using the formula

[tex]BitRate = 2 * Bandwidth * log_{2} (L)[/tex]

Putting the value of bitRate and bandwidth we get

[tex]34860=2\ * 3000\ * \ log_{2} (L)\\34860=6000 * \ log_{2} (L)[/tex]

[tex]31860-6000=log_2L\\25830=log_2L\\L=2^{25830}[/tex]

Answer:

Computer literacy is the correct answer and following are working of computer is given below.

Explanation:

Computer literacy is characterized as the awareness and capacity to make the effective use through the software as well as the computer technology, The computer literacy including the skill sets of varying from simple use to the computer science as well as the specialized solving problems.

 Working of computer

What is computer knowledge called?

Computer literacy

Computer literacy is defined as the knowledge and ability to use computers and related technology efficiently, with skill levels ranging from elementary use to computer programming and advanced problem solving.

Answer:

Explanation:

Agile testing is software testing that follows the best practices of Agile development. For example, Agile development takes an incremental approach to design. Similarly, Agile testing includes an incremental approach to testing. In this type of software testing, features are tested as they are developed.

Answer: Provided in the explanation section

Explanation:

Provided is the code  to run this program

Source Code:

import java.util.Scanner;

class PrintJobCost

{

  public static void main(String[] args) {

      Scanner input=new Scanner(System.in);

      System.out.print("Enter print job info:");

      String info=input.nextLine();

      String size="",type="";

      int count=0,i=0,len=info.length();

      double cost=0.0;

      while(info.charAt(i)!=' '){

          size=size+info.charAt(i);  

          i++;

      }

      i++;

      while(info.charAt(i)!=' '){

          type=type+info.charAt(i);

          i++;

      }

      i++;

      while(i<len){

          count=count*10+Integer.parseInt(String.valueOf(info.charAt(i)));

          i++;

      }

      if(size.equals("Letter"))

          cost=cost+0.05;

      else if(size.equals("Legal"))

          cost=cost+0.06;

      else if(size.equals("A4"))

          cost=cost+0.055;

      else if(size.equals("A5"))

          cost=cost+0.04;

      if(type.equals("Grayscale"))

          cost=cost+0.01;

      else if(type.equals("Colored"))

          cost=cost+0.10;

      cost=cost*count;

      System.out.printf("print job Cost:$ %.2f\n",cost);

  }

}

cheers i hope this helped !!!

Answer:

a) 179 segments

b) 28 segments

Explanation:

Given data :

A) Round trip time = 100 ms, Timeout = 101 ms

number of segments  for round trip = 3000/100 = 30 segments

number of segments for timeout = 15000/100 = 150 segments

total number of segments = 150 + 30 = 180

segments that get re-transmitted under A

Total segments - timeout = 180 - 101 = 179 segments

B) Round trip = 100 ms Timeout = 152 ms

number of segments for round trip = 3000/100 = 30 segments

number of segments for timeout = 15000/100 = 150 segments

total segments = 150 + 30 = 180

segments that get re-transmitted under B

total segments - timeout = 180 - 152 = 28 segments

Answer:

A python code (Python recursion) was used for this given question

Explanation:

Solution

For this solution to the question, I am attaching code for these 2 files:

item.py

code.py

Source code for item.py:

class Item(object):

def __init__(self, name: str, weight: int, value: int) -> None:

  self.name = name

  self.weight = weight

  self.value = value

def __lt__(self, other: "Item"):

  if self.value == other.value:

if self.weight == other.weight:

  return self.name < other.name

else:

  return self.weight < other.weight

  else:

   return self.value < other.value

def __eq__(self, other: "Item") -> bool:

  if is instance(other, Item):

return (self.name == other.name and

self.value == other.value and

self.weight == other.weight)

  else:

return False

def __ne__(self, other: "Item") -> bool:

  return not (self == other)

def __str__(self) -> str:

  return f'A {self.name} worth {self.value} that weighs {self.weight}'

Source code for code.py:

#!/usr/bin/env python3

from typing import List

from typing import List, Generator

from item import Item

'''

Inductive definition of the function

fun3(0) is 5

fun3(1) is 7

fun3(2) is 11

func3(n) is fun3(n-1) + fun3(n-2) + fun3(n-3)

Solution 1: Straightforward but exponential

'''

def fun3_1(n: int) -> int:

result = None

if n == 0:

result = 5 # Base case

elif n == 1:

result = 7 # Base case

elif n == 2:

result = 11 # Base case

else:

result = fun3_1(n-1) + fun3_1(n-2) + fun3_1(n-3) # Recursive case

return result

''

Solution 2: New helper recursive function makes it linear

'''

def fun3(n: int) -> int:

''' Recursive core.

fun3(n) = _fun3(n-i, fun3(2+i), fun3(1+i), fun3(i))

'''

def fun3_helper_r(n: int, f_2: int, f_1: int, f_0: int):

result = None

if n == 0:

result = f_0 # Base case

elif n == 1:

result = f_1 # Base case

elif n == 2:

result = f_2 # Base case

else:

result = fun3_helper_r(n-1, f_2+f_1+f_0, f_2, f_1) # Recursive step

return result

return fun3_helper_r(n, 11, 7, 5)

''' binary_strings accepts a string of 0's, 1's, and X's and returns a generator that goes through all possible strings where the X's

could be either 0's or 1's. For example, with the string '0XX1',

the possible strings are '0001', '0011', '0101', and '0111'

'''

def binary_strings(string: str) -> Generator[str, None, None]:

def _binary_strings(string: str, binary_chars: List[str], idx: int):

if idx == len(string):

yield ''.join(binary_chars)

binary_chars = [' ']*len(string)

else:

char = string[idx]

if char != 'X':

binary_chars[idx]= char

yield from _binary_strings(string, binary_chars, idx+1)

else:

binary_chars[idx] = '0'

yield from _binary_strings(string, binary_chars, idx+1)

binary_chars[idx] = '1'

yield from _binary_strings(string, binary_chars, idx+1)

binary_chars = [' ']*len(string)

idx = 0

yield from _binary_strings(string, binary_chars, 0)

''' Recursive KnapSack: You are looking to rob a jewelry store. You have been staking it out for a couple of weeks now and have learned

the weights and values of every item in the store. You are looking to

get the biggest score you possibly can but you are only one person and

your backpack can only fit so much. Write a function that accepts a

list of items as well as the maximum capacity that your backpack can

hold and returns a list containing the most valuable items you can

take that still fit in your backpack. '''

def get_best_backpack(items: List[Item], max_capacity: int) -> List[Item]:

def get_best_r(took: List[Item], rest: List[Item], capacity: int) -> List[Item]:

if not rest or not capacity: # Base case

return took

else:

item = rest[0]

list1 = []

list1_val = 0

if item.weight <= capacity:

list1 = get_best_r(took+[item], rest[1:], capacity-item.weight)

list1_val = sum(x.value for x in list1)

list2 = get_best_r(took, rest[1:], capacity)

list2_val = sum(x.value for x in list2)

return list1 if list1_val > list2_val else list2

return get_best_r([], items, max_capacity)

Note: Kindly find an attached copy of the code outputs for python programming language below

Answer:

Option B:

Some people have a difficulty in managing risks

Explanation:

Riding a motorcycle is a lot riskier than driving a car. This is because motorcyclists are exposed, and any accident that occurs will most likely lead to bodily harm to the rider.

Some individuals, have problems managing such risks, and cannot comfortably control the motorbike on the highway because of the number of cars speeding by.

Such individuals are not expected to be riding as this lack of confidence when riding on the highways can lead to accidents.

Answer:

6 yards

Explanation:

if to convert 15 feet into yards it would 15/3 = 5

Answer:

1.    11.77 GHz

2.   11.11  GHz

Explanation:

1.

From the given information;

the highest resulting clock speed we could achieve in GHz is calculated as follows:

Using the formula:

highest resulting clock speed = 1/max (stage,stage,stage,stage) + register delay)

highest resulting clock speed = 1/max(65ps, 45ps, 60 ps) + 20ps)

highest resulting clock speed = 1/(65 + 20)ps

highest resulting clock speed =  1/85 ps

highest resulting clock speed = 0.01176470588  × 10¹² Hz

highest resulting clock speed = 1.17647059 × 10¹⁰ Hz

highest resulting clock speed =  1.177 × 10¹⁰ Hz

To Ghz; we have:

highest resulting clock speed = (1.177 × 10¹⁰/10⁹ )GHz

highest resulting clock speed = 11.77 Ghz

2.

Using the same formula from above;

highest resulting clock speed = 1/max(65ps, 55ps, 70 ps) + 20ps)

highest resulting clock speed = 1/(70 + 20)ps

highest resulting clock speed =  1/90 ps

highest resulting clock speed = 0.01111111111  × 10¹² Hz

highest resulting clock speed = 1.111111111  × 10¹⁰ Hz

highest resulting clock speed =  1.111 × 10¹⁰ Hz

To Ghz; we have:

highest resulting clock speed = (1.111 × 10¹⁰/10⁹ )GHz

highest resulting clock speed = 11.11 Ghz

In this exercise we have to calculate the delay time and the time to archive something, so we have that these times will be:

1) 11.77 GHz

2) 11.11  GHz

1) First we will use the knowledge to calculate the delay, so the formula can be used is:

[tex]HRCS = 1/max (stage,stage,stage,stage) + register delay)\\HRCS = 1/max(65ps, 45ps, 60 ps) + 20ps)\\HRCS = 1/(65 + 20)ps\\HRCS= 1/85 ps\\HRCS = 0.01176470588* 10^{12}Hz\\HRCS = 1.17647059 * 10^{10} Hz\\HRCS = 1.177 * 10^{10} HZ\\HRCS = 11.77 Ghz[/tex]

2) through the first formula we can also calculate the time for the speed of archiving, like this:

[tex]HRCS = 1/max(65ps, 55ps, 70 ps) + 20ps)\\HRCS = 1/(70 + 20)ps\\HRCS = 1/90 ps\\HRCS = 0.01111111111 * 10^{12} Hz\\HRCS = 1.111111111 * 10^{10} Hz\\HRCS = 11.11 Ghz[/tex]

See  more about GHZ at brainly.com/question/13112545

Answer: Provided in the explanation section

Explanation:

Programs:

IncomeTax.java

import java.util.Scanner;

public class IncomeTax {

  // Method to get a value from one table based on a range in the other table

  public static double getCorrespondingTableValue(int search, int [] baseTable, double [] valueTable) {

  int baseTableLength = baseTable.length;

  double value = 0.0;

  int i = 0;

  boolean keepLooking = true;

  i = 0;

  while ((i < baseTableLength) && keepLooking) {

  if (search <= baseTable[i]) {

  value = valueTable[i];

  keepLooking = false;

  }

  else {

  ++i;

  }

  }

  return value;

  }

  public static int readInput(Scanner scan){

      System.out.println("\nEnter annual salary (0 to exit): ");

      int annualSalary = scan.nextInt();

      return annualSalary;

  }

  public static void main (String [] args) {

  Scanner scnr = new Scanner(System.in);

  int annualSalary = 0;

  double taxRate = 0.0;

  int taxToPay = 0;

  int [] salaryBase = { 20000, 50000, 100000, 999999999 };

  double [] taxBase = { .10, .20, .30, .40 };

  // FIXME: Change the input to come from a method

  annualSalary = readInput(scnr);

  while (annualSalary > 0) {

  taxRate = getCorrespondingTableValue(annualSalary, salaryBase, taxBase);

  taxToPay= (int)(annualSalary * taxRate); // Truncate tax to an integer amount

  System.out.println("Annual salary: " + annualSalary +

  "\tTax rate: " + taxRate +

  "\tTax to pay: " + taxToPay);

  // Get the next annual salary

  // FIXME: Change the input to come from a method

  annualSalary = readInput(scnr);

  }

  return;

  }

  }

Output:

Enter annual salary (0 to exit):

10000

Annual salary: 10000   Tax rate: 0.1   Tax to pay: 1000

Enter annual salary (0 to exit):

50000

Annual salary: 50000   Tax rate: 0.2   Tax to pay: 10000

Enter annual salary (0 to exit):

0

Answer:

The java program for the given scenario is as follows.

import java.util.Scanner;

import java.lang.*;

public class Test

{

   //variables to hold miles per gallon and cost per gallon

   static double miles_gallon, dollars_gallon;

   public static void main(String[] args)

   {

       Scanner sc = new Scanner(System.in);

       System.out.print("Enter the car's mileage in miles per gallon: ");

       miles_gallon=sc.nextDouble();

       System.out.print("Enter the cost of gas in dollars per gallon: ");

       dollars_gallon=sc.nextDouble();

       System.out.printf("The cost of gas for 10 miles is $%.2f\n", drivingCost(10, miles_gallon, dollars_gallon));

       System.out.printf("The cost of gas for 50 miles is $%.2f\n", drivingCost(50, miles_gallon, dollars_gallon));

       System.out.printf("The cost of gas for 400 miles is $%.2f\n", drivingCost(400, miles_gallon, dollars_gallon));

   }

   //method to compute cost of gas

   public static double drivingCost(double drivenMiles, double milesPerGallon, double dollarsPerGallon)

   {

       double cost = ((drivenMiles/milesPerGallon)*dollarsPerGallon);

       return cost;

   }

}

OUTPUT

Enter the car's mileage in miles per gallon: 12

Enter the cost of gas in dollars per gallon: 23

The cost of gas for 10 miles is $19.17

The cost of gas for 50 miles is $95.83

The cost of gas for 400 miles is $766.67

Explanation:

1. Two double variables to hold user input for miles per gallon and cost of gas per gallon, are declared.

2. The variables are declared static, and at the class level (outside main()).

3. Inside main(), an object of Scanner class is created.

4. User input is taken for miles per gallon and cost of gas per gallon.

5. Next, the method, drivingCost() is called. This method takes three parameters - miles per gallon, cost of gas per gallon and miles driven. Based on these parameters and the given formula, the cost of gas is computed.

6. This value is then displayed to the user.

7. The method, drivingCost(), is called three times for 10 miles, 50 miles and 400 miles being driven.

8. The whole code is written inside a class since java is a purely object-oriented language.

9. The object of the class is not created since only a single class is used.

10. The program is saved as Test.java.

In this exercise we have to use the knowledge in computational language in JAVA to describe a code that best suits, so we have:

The code can be found in the attached image.

To make it simpler we can write this code as:

import java.util.Scanner;

import java.lang.*;

public class Test

{

  //variables to hold miles per gallon and cost per gallon

  static double miles_gallon, dollars_gallon;

  public static void main(String[] args)

  {

      Scanner sc = new Scanner(System.in);

      System.out.print("Enter the car's mileage in miles per gallon: ");

      miles_gallon=sc.nextDouble();

      System.out.print("Enter the cost of gas in dollars per gallon: ");

      dollars_gallon=sc.nextDouble();

      System.out.printf("The cost of gas for 10 miles is $%.2f\n", drivingCost(10, miles_gallon, dollars_gallon));

      System.out.printf("The cost of gas for 50 miles is $%.2f\n", drivingCost(50, miles_gallon, dollars_gallon));

      System.out.printf("The cost of gas for 400 miles is $%.2f\n", drivingCost(400, miles_gallon, dollars_gallon));

  }

  //method to compute cost of gas

  public static double drivingCost(double drivenMiles, double milesPerGallon, double dollarsPerGallon)

  {

      double cost = ((drivenMiles/milesPerGallon)*dollarsPerGallon);

      return cost;

  }

}

See more about JAVA at brainly.com/question/19705654

Answer:

C.

Explanation:

The component that t provides WiFi access for employees in home offices, branch offices, and on the corporate campus is Wireless access points (APs).

APs are generally a networking hardware device that allows other wifi devices to connect to wired network. It creates a wireless local area network, usually in offices or large buildings.

Answer: Provided in the explanation section

Explanation:

The below explanation will help to make this question simple to understanding.

For the questions presented, this analysis surfaces,

⇒ a) (10)*1 in this path we can go through A for C to B as 1010 exist from C to A then B also.

i.e. Option A is wrong , there is transition from c to A with 1.

b) 0 represent the path from B to A and from B to C also. So this is also not a solution.

i.e. Option B is wrong with input 1 there is path to both A and B from c

c) (1+11)0)*1 in this path for cover the 1+11 we have to go in two paths one is only 1 and second is 11. So 11 is the path which goes through the A.

i.e. Option c is wrong , with input 0 there is path to c input symbol 1 is not required

d) (01)+ is correct because we have + sign here not * so it not executes multiple times execute only 1 more time. So it is the path from B to C that does not through A.

cheers i hope this helped !!

Answer:

Explanation:

The solution code is written in Python 3.

Firstly, prompt user to input a mileage (Line 1).

Next, create an if statement to check if a mileage greater than 100 (Line 3). If so, apply the formula to calculate amount with mileage above 100 (Line 4) otherwise,  calculate the amount just by multiplying the mileage by 0.25.

At last, print the total amount (Line 8).

Answer:

hardware-initiated reset

Explanation:

In order to properly switch it back to real address mode operations you need to perform a hardware-initiated reset. This is because once in protected mode the 80286 will remain there until the hardware itself is reset. Therefore manually reseting the hardware will allow the user to switch back the microprocessor into a real address mode operations.

Answer:

This is because when there is a mistake, misunderstanding or trouble and one need's evidence, the documented business rule will be available as your evidence.

Explanation:

Sorry if I'm wrong hope it helps

Answer:

For me this is because, if I don't accept them they won't let me see what I came there for and my megabyte will be wasting.

Explanation:

Answer:

Class Locker

public class Locker {  

int lockno;  

String student;

int booksno;

private CombinationLock comblock = new CombinationLock();

public Locker() {

 lockno = 0;

 student= "No Name";

 booksno = 0;

 comblock.reset();  }

public Locker(int lockno, String student, int booksno,

  CombinationLock comblock) {

 super();

 this.lockno= lockno;

 this.student= student;

 this.booksno= booksno;

 this.comblock= comblock;  }

public int getLockNo() {

 return lockno;   }

public void setLockNo(int lockno) {

 this.lockno= lockno;  }

public String getName() {

 return student;  }

public void setName(String student) {

 this.student = student;  }

public int getBooksNumber() {

 return booksno;   }

public void setBooksNumber(int booksno) {

 this.booksno= booksno;  }

public String getComblock() {

 return comblock.toString();  }

public void setComblock(int no1, int no2, int no3) {

 this.comblock.setNo1(no1);  

 this.comblock.setNo2(no2);  

 this.comblock.setNo3(no3);   }

public String printValues() {

 return "Locker [lockno=" + lockno+ ", student="

   + student+ ", booksno=" + booksno

   + ", comblock=" + comblock+ "]";  }  }

The class Locker has attributes lockno for lock number, student for students names and bookno for recording number of books. Locker class also an attribute of type CombinationLock named as comblock. Locker class includes a constructor Locker() with no input argument and also a constructor Locker() that requires input arguments for all attributes lockno, student, booksno and comblock. super() calls on immediate parent class constructor. Class Locker contains the following set methods: setLockNo, setName, setBooksNumber and setComblock. Class Locker contains the following get methods: getLockNo, getName, getBooksNumber and getComblock. printValues() method displays the values of attributes.

Explanation:

Class CombinationLock

*/ CombinationLock has attributes no1, no2 and no3. It has a  constructor CombinationLock() with no input argument and also a constructor CombinationLock() that requires input arguments for attributes no1 no2 and no3. Class CombinationLock contains the following set methods: setNo1, setNo2 and setNo3. The class contains the following get methods: getNo1, getNo2 and getNo3. The class includes a method printValues() to print out all attributes.

public class CombinationLock {

int no1;

int no2;

int no3;  

public CombinationLock() {

 this.no1= 0;

 this.no2= 0;

 this.no3= 0;   }  

public CombinationLock(int no1, int no2, int no3) {

 super();

 this.no1= no1;

 this.no2= no2;

 this.no3= no3;  }

public int getNo1() {

 return no1;  }

public void setNo1(int no1) {

 this.no1= no1;  }

public int getNo2() {

 return no2;  }

public void setNo2(int no2) {

 this.no2= no2;  }

public int getNo3() {

 return no3;  }

public void setNo3(int no3) {

 this.no3= no3;  }

public void reset() {

 this.no1=0;

 this.no2=0;

 this.no3=0;  }

public String printValues() {

 return "CombinationLock [no1=" + no1+ ", no2=" +  no2

   + ", no3=" + no3+ "]";  }  }