Design and implement a programming (name it SimpleMath) that reads two floating-point numbers (say R and T) and prints out their values, sum, difference, and product on separate lines with proper labels. Comment your code properly and format the outputs following these sample runs.Sample run 1:R = 4.0T = 20.0R + T = 24.0R - T = -16.0R * T = 80.0Sample run 2:R = 20.0T = 10.0R + T = 30.0R - T = 10.0R * T = 200.0Sample run 3:R = -20.0T = -10.0R + T = -30.0R - T = -10.0R * T = 200.0

Answer:

Nothing just the video is series of photos together

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+ "]";  }  }

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:

import math

n = int(input("Enter the number of sides: "))

s = float(input("Enter the length of a side: "))

area = (n * s**2) / (4 * math.tan(math.pi/n))

print("The area is: " + str(area))

Explanation:

*The code is in Python.

Import the math to be able to compute the pi and tan

Ask the user to enter the number of sides and the length of a side

Calculate the area using the given formula

Print the area

Answer:

Storage devices tend to be built in/physical pieces of storage such as an SD Card or Hard Drive. Virtual storage is more-so like Cloud storage, where the files are hosted elsewhere.

Explanation:

Answer:

hardware

Explanation:

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

Answer:

keyboard

Explanation:

Answer:

The answer is d, because it is not reliable in the sense that it can be used for important and sensitive information when your collecting data.

Answer: provided in the explanation section

Explanation:

 Network’s vulnerabilities:

           Vulnerability is a weak spot in your network that might be exploited by a security threat. Risks are the potential consequences and impacts of unaddressed vulnerabilities. In other words, failing to do Windows Updates on your Web server is vulnerability.

           Regularly scheduled network vulnerability scanning can help an organization identify weaknesses in their network security before the bad guys can mount an attack. The goal of running a vulnerability scanner or conducting an external vulnerability assessments is to identify devices on your network that are open to known vulnerabilities without actually compromising your systems.

           The overall objective of a Vulnerability Assessment is to scan, investigate, analyze and report on the level of risk associated with any security vulnerabilities discovered on the public, internet-facing devices and to provide your organization with appropriate mitigation strategies to address those discovered vulnerabilities.

Network topology issues:

Coverage Topology

           Coverage problem reflects how well an area is monitored or tracked. The coverage and connectivity problems in networks have received considerable attention in the research community in recent years

Geographic routing

           Geographic routing uses geographic and topological information of the network to achieve optimal routing schemes with high routing efficiency and low power consumption

Port Numbers

           Port Numbers While IP addresses determine the physical endpoints of a network connection, port numbers determine the logical endpoints of the connection. Port numbers are 16-bit integers with a useful range from 1 to 65535.

Port numbers are assigned by an organization called IANA and ports are allocated to various needs to avoid confusion.

Ports are classified into 3 main categories.

Well Known Ports (Port numbers 0 - 1023)

           In a client-server application, the server usually provides its service on a well-known port number. Well-known port numbers are a subset of the numbers which are assigned to applications. According to RFC1700 [5], well-known port numbers are managed by the Internet Assigned Numbers Authority (IANA). They used to be in the range from 1 to 255, but in 1992 the range was increased up to 1023.

Registered Ports (Port numbers1024 - 49151)

Such ports are used by programs run by users in the system.

           In addition to the well-known ports below 1024 there are more port numbers assigned to applications but are located anywhere from 1024 to 65535.

Private or Dynamic Ports (Port numbers 49152 - 65535)

           Private ports are not assigned for any specific purpose.

Discuss the concepts of well-known ports and give a brief description of the most commonly used ports: 20, 21, 23, 25, 53, and 110.

Commonly Used Port Numbers

The following port numbers are unofficial list of commonly used for linux/unix based servers.

20&21               TCP                FTP (File server protocol)

           FTP is one of the most commonly used file transfer protocols on the Internet and within private networks. An FTP server can easily be set up with little networking knowledge and provides the ability to easily relocate files from one system to another. FTP control is handled on TCP port 21 and its data transfer can use TCP port 20 as well as dynamic ports depending on the specific configuration.

23                   TCP/UDP          Telnet

           Telnet is the primary method used to manage network devices at the command level. Unlike SSH which provides a secure connection, Telnet does not, it simply provides a basic unsecured connection. Many lower level network devices support Telnet and not SSH as it required some additional processing. Caution should be used when connecting to a device using Telnet over a public network as the login credentials will be transmitted in the clear.

25                   TCP/UDP         SMTP   (for sending outgoing emails)

           SMTP is used for two primary functions, it is used to transfer mail (email) from source to destination between mail servers and it is used by end users to send email to a mail system.

53         TCP/UDP        DNS Server (Domain name service for DNS request)

           The DNS is used widely on the public internet and on private networks to translate domain names into IP addresses, typically for network routing. DNS is hieratical with main root servers that contain databases that list the managers of high level Top Level Domains (TLD) (such as .com). T

110             TCP                 POP3 (For receiving emails)

           POP version 3 is one of the two main protocols used to retrieve mail from a server. POP was designed to be very simple by allowing a client to retrieve the complete contents of a server mailbox and then deleting the contents from the server.

Answer:

C

Explanation:

Answer:

multimedia keyboard i think

Explanation:

The keyboards that include all the keys found on a typical virtual keyboard including function and navigation keys are called;

Laptops

Read more at; https://brainly.in/question/11722276

Answer:

numbers = []

count = 0

while True:

   number = int(input("Enter a number: "))

   if number < 0:

       break

   else:

       numbers.append(number)

       count += 1

min_number = min(numbers)  

print(str(min_number) + " was the smallest among " + str(count) + " entered numbers.")

Explanation:

Initialize an empty array, to hold the entered numbers, and count as 0

Create a while loop that iterates until a specific condition is met

Inside the loop, ask the user to enter the numbers. When the user enters a negative number terminate the loop. Otherwise, add the entered number to the numbers array and increment the count by 1.

When the loop is done, find the smallest of the number in the numbers array using min method and assign it to the min_number.

Print the min_number and count

Answer:

Webcast

Explanation:

Webcast is the live broadcast of a video or audio feed from your event or conference harnessing the internet. Webcast is a media presentation of an event circulated over the internet by using media technology to distribute a single source to various viewers.

Generally,  webcast is briefly defined as broadcasting over the internet. The broadcast might not be specifically live  . One can capture the video or the audio he or she wants to publish and publish it online anywhere. Webcast allow your viewers to remain engage as if they are face to face with the broadcaster.

Webcasting is a a great tools that has been used by various multinational and company to reach wide range of audience . This web based broadcast can even provide opportunity for audience to ask question and even provide answers in real time.

Answer:

Webcast.

Explanation:

A Webcast is a form of Web conferencing that uses streaming media technologies to broadcast video and/or audio over the Internet from a single content source to many listeners or viewers simultaneously.

The Webcast is an internet conferencing tool that transmits and stores, corporate presentations, lectures, workshops, seminars, etc., with the advantage of being a simple operation tool that can bring together groups in real time offering dynamism and reducing geographical distance.

It is a widely used and useful tool for companies, as it can help when traditional communication channels are not sufficient to deal with extremely relevant issues for the company, such as training and presentation of materials for prospecting customers.

The Webcast is an advantageous technology for companies by reducing personnel displacement and reducing costs with facilities, infrastructure and space rental.

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:

The answer is void process Array (int arr {}, int n)

Explanation:

Solution

From the given question, the right choice answer i not listed  here.

However this is the right answer choice stated as follows: void process Array (int arr {}, int n)

The size of this array is not fixed. we need to pass or move the array as the second  argument to the function.

The option a and b is wrong.

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:

A central processing unit (CPU) is an important part of every computer. The CPU sends signals to control the other parts of the computer, almost like how a brain controls a body. The CPU is an electronic machine that works on a list of computer things to do, called instructions. hope that helps love!

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:

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: Provided in the explanation section

Explanation:

#include<iostream>

using namespace std;

class Money{

  private:

      int all_cents;

  public:

      Money();

      double get_value() const;

      Money(int,int);

      void printamount();

      friend Money operator+(const Money&, const Money&);

      friend ostream& operator<<(ostream&, const Money&);

      friend istream& operator>>(istream&, Money&);

};

Money::Money(){all_cents=0;}

double Money::get_value() const{return all_cents/100.0;}

Money::Money(int dollar ,int cents){

  all_cents = dollar*100+cents;

}

void Money::printamount(){

  cout<<"$"<<get_value()<<endl;

}

Money operator+(const Money& m1, const Money& m2){

  int total_cents = m1.all_cents + m2.all_cents;

  int dollars = total_cents/100;

  total_cents %=100;

  return Money(dollars,total_cents);

}

ostream& operator<<(ostream& out, const Money& m1){

  out<<"$"<<m1.get_value()<<endl;

  return out;

}

istream& operator>>(istream& input, Money& m1){

  input>>m1.all_cents;

  return input;

}

int main(){

  Money m1;

  cout<<"Enter total cents: ";

  cin>>m1;

  cout<<"Total Amount: "<<m1;

  Money m2 = Money(5,60);

  Money m3 = Money(4,60);

  cout<<"m2 = ";m2.printamount();

  cout<<"m3 = ";m3.printamount();

  Money sum = m2+m3;

  cout<<"sum = "<<sum;

}

cheers i hope this helped !!

Answer:

The program goes as follows

Comments are used to explain difficult lines

#include <iostream>

#include <string>

using namespace std;

int main()

{

string userinput;

getline(cin,userinput);

/* My Solution starts here*/

//Initialize sub string position to 0

int stringpos = 0;

// Declare an iterating variable to iterate through userinput

int startpos;  

//Declare and initialize a determinant variable

int k = 0;

while((startpos = userinput.find("darn",stringpos))!=string::npos)

{

 //If userinput contaings "darn", increase k by 1

k++;

// Go to next string position

stringpos = startpos+1;  

}

//If determinant variable k is p then, userinput does not contain "darn"

if(k==0)

{

 cout<<userinput;

}

//Otherwise, it contains "darn"

else

{

 cout<<"Censored";

}

/* My Solution ends here*/

return 0;

}

See attached for .cpp program file

Answer:

year = int(input("Enter a year: "))

if year >= 2101:

   print("Distant future")

elif year >= 2001:

   print("21st century")

elif year >= 1901:

   print("20th century")

else:

   print("Long ago")

Explanation:

*The code is in Python.

Ask the user to enter a year

Use if-else statement to check the year and print the appropriate message given for each condition. If year is greater than or equal to 2101, print "Distant future". If year is greater than or equal to 2001, print "21st century". If year is greater than or equal to 1901, print "20th century". Otherwise, print "Long ago".

Answer:

yoooooooooooooooooooooooo

Explanation:

lrmgkrmbm mrm g

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:

Following is the program in the C language

#include <stdio.h> // header file  

int main() // main function

{

float weight1[10]; // array declaration  

float sum=0,max1,t; // variable declaration

for(int k = 0; k < 5; k++) //iterating the loop

{

printf("Enter weight %d: ",k+1);

scanf("%f",&weight1[k]); // Read the array by user

}

printf("\n");

printf("You entered: ");

max1=weight1[0];

for(int k = 0; k < 5 ; k++)

{

sum=sum+weight1[k];

if(max1 < weight1[k]) // check condition for highest element

{

max1=weight1[k];

}

printf("%.2lf ",weight1[k]);

}

t=sum/5.0; // find average

printf("\nTotal weight: %.2lf\n",sum); // displat total

printf("Average weight: %.2lf\n",t); // display Average

printf("Max weight: %.2lf\n",max1); // display maximum value

return 0;

}

Output:

Enter weight 1: 1

Enter weight 2: 2

Enter weight 3:3

Enter weight 4:4

Enter weight 5: 5

You entered:  1.00  2.00  3.00  4.00  5.00

Total weight: 15.00

Average weight:  3.00

Max weight:  5.00

Explanation:

Following are description of program :

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:The users would likely know if an upgrade would be necessary or even useful.

First hand user information is often ignored by developers, change managers etc. However, obtaining first hand user input has proven vastly cost effective ,productive and easier to apply . By having early input the actual working interface can be designed so that daily users find it works effectively for them and others they interact with. It can also allow users to effectively aim to break the functionality before a crisis occurs etc. Furthermore by having user input the users will make a greater effort in ensuring the upgrade works seamlessly.

Explanation:

Answer:

  generate a random number in a known range and map the result to the range you want

Explanation:

Use any of the usual methods to get a random number in the range 0-20, then subtract 9. Use an IF statement to test if the result is 0. If it is, replace the value with -10.

Answer:

The odds for the event that either Romance or Downhill wins is 11/4

Explanation:

The odds that Romance will win is 2: 3

The odds that Downhill will win is 1:2

[tex]P(event) = \frac{Odd(event)}{1 + Odd(event)}[/tex]

Where P(event) = Probability that an event occurs

Odd(event) = The odds that an event occurs

Calculate the Probability that Romance wins, P(R):

[tex]P(R) = \frac{2/3}{2/3 + 1} \\P(R) = 2/5[/tex]

Calculate the probability that Downhill wins, P(D):

[tex]P(D) = \frac{1/2}{1/2 + 1} \\P(D) = 1/3[/tex]

Calculate the Probability that either Romance or Downhill wins, P(R or D):

P(R or D) = P(R) + P(D)

P(R or D) = 2/5 + 1/3

P(R or D) = 11/15

[tex]P(R or D) = \frac{Odd(R or D)}{1 + Odd(R or D)} \\\\11/15 = \frac{Odd(R or D)}{1 + Odd(R or D)}\\[/tex]

[tex]\frac{11}{15} + \frac{11}{15}[ Odd(R or D)] = Odd(R or D)\\\\\frac{4}{15}[ Odd(R or D)] = \frac{11}{15}\\\\\\Odd(R or D) = \frac{11}{15} * \frac{15}{4}\\\\Odd(R or D) = 11/4[/tex]

Answer:

I do Stray kids and Tower Of God is good combo mannnn

Answer:

Yes

Explanation:

I dont comprehend how this question requires any intellectual answer but it's ok.

Answer:

Top Secret, (A,B,C)

Explanation:

Solution:

Some information are known to be sensitive, and has a higher level of classification.

Examples for such levels are: Secret, Top secret ,Unclassified and  Confidential, and Unclassified.  all users of the system has a  level of clearance. fro a user to be able to gain access to document, he or she must have posses a level of the document.

Now, this is governed by two rules which is stated below:

No user can have access to any information that is considered higher than their level of clearance.

Secondly, a user can write to files only that contains an equal level or lower of his clearance level, this is to stop what is called the write down effect.

It is important to note that ABC, categories determine what type of access is permitted.

Now, for the given example, since, Bob and charlie cannot read it, which shows that the file must be of top priority or importance, then Bob and Charlie needs clearance to the file,the file 2 is considered a top priority.

Therefore, the File 2 is classified as a Top Secret, (A,B,C)