Once sales are​ forecasted, ________ must be generated to estimate required raw materials. A. a pro forma statement B. a production plan C. a cash budget

Final answer:

The answer explains the concept of clipping in geometry and how it relates to implementing clippers in a pipeline. It also discusses the proofs and theorems in geometry regarding segments, polygons, and constructions using straight-edge and transferer of segments.

Explanation:

Clipping against the top of the clipping rectangle can be done independently of the other sides due to the nature of line segments. This concept is essential in implementing a clipper as a pipeline of four simple clippers.

Proofs and theorems in geometry demonstrate the relationships and properties of segments, polygons, and shapes in a structured manner.

Understanding the construction using straight-edge and transferer of segments allows for solving a variety of geometrical problems by drawing straight lines and laying off segments.

Answer:

The severity and likelihood of harm

Explanation:

So that there is no room for a data breach.

Answer:

500kbps

Explanation:

Bandwidth is the actual amount of data that a network is capable of transferring theoretically.

Throughput is the actual amount of data passing through a connection. It is the rate (usually measured in bps- bits per sec or pps - packets per second) at which packets or bits are successfully delivered over a network channel

If R₁=500 kbps, R₂=2 Mbps, and R₃ = 1 Mbps.

Since there is no other traffic in the link from Host A to Host B, the Throughput is the minimum of the links between Host A and Host B. Therefore:

Throughput= Mininum(R₁,R₂,R₃)

=500kbps

Answer:

Hi Zoom! Good question. This is particularly useful for learning sorting functions in arrays and user input handling as well. Please find the implementation below.

Explanation:

I have implemented the code in Python 2+. You can save the below code into a file called ArrayDemo.py and then run it from the command line with the command "python ArrayDemo.py". I have declared the demo_array in the code to an array of 10 unsorted integer values to test this. You can change the code if needed to have this array input by user as well and it should work fine.

Note: if you're using Python 3+ then you'll need to change the "raw_input" in the code to just "input" to make it work.

ArrayDemo.py

demo_array = [2,1,3,4,5,7,6,8,10,9]

option = 0

while (option != 4):

 option = raw_input("Enter 1 to view list in ascending order, 2 to view list in descending order, 3 to choose specific poisition to view, 4 to quit the application: ")

 try:

   option = int(option)

   if option == 1:

     sort_by = 'asc'

     print(sorted(demo_array, reverse=False))

   elif option == 2:

     sort_by = 'desc'

     print(sorted(demo_array, reverse=True))

   elif option == 3:

     position = raw_input("Enter specific position to view (0-9): ")

     try:

       position = int(position)

       if position < 0 or position > 9:

         print("Position does not exist")

       else:

         print(demo_array[position])

     except ValueError:

       print("Position does not exist!")

   elif option == 4:

     break

   else:

     break

 except ValueError:

   print("Invalid input!")

Answer:

The actual methods used to protect against data loss are Technical controls, but the program that identifies which data to protect is a Procedural control.

Explanation:

In technical control, different software has been installed and protocols has been deployed to avoid any data loss in network security. As all the devices and protocols that we need while installing a network needs technical support. This is the reason, these things are called technical control of the system.

Another control in network security is procedural control. In this type of control it has been identified that, which data should be protected or which data should not be accessed to unauthorized users. This is called procedural control.

Answer:

184

Explanation:

Given the codes

The initial value of cost is 82.

The if condition in Line 3 will pass and therefore cost = 82 + 10 = 92

The if condition in Line 8 will pass and therefore cost = 92 * 2 = 184

The if condition in Line 13 will fail l and therefore the if block will be skipped.

At last, the we get cost = 184.

Answer:

The correct answer to the following question will be "Iterative procedure".

Explanation:

Iterative production or procedure seems to be a technique of distinguishing product progression from expansive implementation to littler bits. This including code in regurgitated processes is designed, developed and attempted.

Regularly iterative architecture is used in conjunction with gradual improvement where a certain more sketched-out process of code progress is composed of littler components building on each other.

This will improve observer performance, as each move is systematic and cheap to run.

The shorter cycles of Iterative production affect the ability of the analyst to manage system’s requirements in a way that improve the observer performance as each move is systematic and cheap to run.

An Iterative production means a technique of distinguishing product progression from expansive implementation to littler bits.

An Iterative production which includes code in regurgitated processes is designed, developed and attempted.

In conclusion, the shorter cycles of Iterative production affect the ability of the analyst to manage system’s requirements in a way that improve the observer performance as each move is systematic and cheap to run.

Read more about Iterative production

brainly.com/question/6019026

Solution:

It will not fit on Lenovo Ideapad N580.

The Ideapad N580 will fit HDD of form factor 2.5" (2.5 Inch) but the form factor of Toshiba mk8009GAH is 1.8" hence won't fit in the slot.

The interface type specifies which type of connector used to connect the motherboard and the HDD and it should be the same in order to use on it.

Ideapad uses the SATA interface type to connect the HDD but the Toshiba mk8009GAH is having an ATA type interface on it hence they cannot be connected with each other.

No, the Toshiba MK8009GAH 80-GB 4200-RPM hard drive will not fit in the Lenovo IdeaPad N580 laptop because it is a 1.8-inch PATA (IDE) drive, while the laptop requires a 2.5-inch SATA drive.

To determine if the Toshiba MK8009GAH 80-GB 4200-RPM hard drive will fit in the Lenovo IdeaPad N580 laptop, we'll need to consider both the drive form factor and interface connector compatibility. Most laptops, including the Lenovo IdeaPad N580, use a 2.5-inch SATA hard drive. The Toshiba MK8009GAH, however, is an older 1.8-inch PATA (IDE) drive (as indicated by its 4200-RPM speed which is more common in older drives, and the MK series which was typically PATA drives).

The answer is d. No, the drive form factor and interface do not match. The physical size (form factor) of the Toshiba drive is smaller and the interface is IDE, not SATA. Therefore, this drive would not fit nor connect properly without an appropriate adapter, which typically is not a feasible solution for replacing a laptop hard drive.

Final answer:

The swapArrayEnds() method exchanges the positions of the first and last elements in an array. For an array with two or more elements, the method uses a temporary variable to hold one element during the swap. This method is a basic array manipulation technique in programming.

Explanation:

The method swapArrayEnds() swaps the first and last elements of an array. In a programming language like Java or Python, this task is typically performed by accessing the elements at their respective indices and then swapping their values. Here’s an example of how you might write this method in Java:

By checking if the array is not null and has more than one element, we ensure that we do not try to swap ends on an empty or single-element array, which would be unnecessary. This is a common operation in array manipulation, and understanding how to implement it is useful in many programming and computer science applications.

Answer:

top down approach

Explanation:

Top down approach is process of breaking down system into sub-system which helps in formulating the overall system. At the end every sub system can be refined with more details.

While bottom up approach is process of gathering systems to design large complex system.

Top down is better as basic information is present and resources can be utilized effectively without impacting maintennance and operations.

Final answer:

Choosing between a top-down approach and a bottom-up approach for developing a school registration system depends on the project goals and context. A top-down approach is hierarchically structured and policy-driven, while a bottom-up approach focuses on detailed, user-centric components. A hybrid approach may offer the best of both methodologies.

Explanation:

When developing a logical model of a registration system at a school or college, choosing between a top-down approach and a bottom-up approach depends on the specific context and goals of the project. A top-down approach generally starts with defining the high-level processes and structures of the system based on the policies and requirements of the institution. It is methodical, following a hierarchical path from the general to the specific, which can be beneficial when there are clear objectives and standards to adhere to.

On the other hand, a bottom-up approach builds the system from the lowest level of detailed requirements, often focusing on user needs and individual system components before integrating them into the full model. This can be advantageous when needing to understand complex details from the outset or when user feedback is critical in shaping the system's functionality.

In the implementation phase of a policy process, one might consider elements from both strategies. For example, important federal policies might require adherence to regulations (top-down), but the practical and on-the-ground impact of policies might be best understood by considering local circumstances (bottom-up). Thus, developing a registration system could potentially benefit from a hybrid approach, combining the structured, policy-driven aspects of the top-down method with the detailed, user-driven focus of the bottom-up method.

How does multithreading affect the throughput of a GetFile Server hosting many very large files?

Answer:

It create multiple threads of large file which eventually causes in slowing down the speed of server which results in slower response time.

How does multithreading affect the average response time for a GetFile server hosting a few small files?

Answer:

Multiple threads in smaller files result in rapid response time and enormous process speed in small files.

Final answer:

This answer provides C++ and Python code for a Mad Libs game that uses loops to generate sentences based on user inputs until 'quit 0' is entered.

Explanation:

This lab activity requires the creation of a Mad Libs game using loops in both C++ and Python programming languages. The game will prompt users to input a string and an integer, and use those inputs to construct and output a humorous sentence. The loop will terminate when the user inputs 'quit 0'. Below you'll find an example code snippets in C++ and Python that accomplish this task.

C++ Example

Write complete code in C++ to create a Mad Libs game that continues to ask for user inputs until 'quit 0' is entered:

#include
#include
using namespace std;

int main() {
   string word;
   int number;
   while (true) {
       cin >> word;
       if (word == "quit") {
           cin >> number;
           if (number == 0) {
               break;
           }
       }
       cin >> number;
       cout << "Eating " << number << " " << word << " a day keeps the doctor away." << endl;
   }
   return 0;
}

Python Example

Write complete code in Python to accomplish the same task:

while True:
   word, number = input().split()
   number = int(number)
   if word == 'quit' and number == 0:
       break
   print(f'Eating {number} {word} a day keeps the doctor away.')

Answer:

Explanation: see attachment below

Answer:

C++ code is explained

Explanation:

#include <iostream>

#include <string>

#include <sstream>

using namespace std;

int firstDigit(int n){

// Converting int to string

ostringstream convert;

convert << n;

string s = convert.str();

char first_char = s[0];

int first_int = first_char - '0';

return first_int;

}

int lastDigit(int n){

// Converting int to string

ostringstream convert;

convert << n;

string s = convert.str();

char last_char = s[s.length() - 1];

int last_int = last_char - '0';

return last_int;

}

int digits(int n){

// Converting int to string

ostringstream convert;

convert << n;

string s = convert.str();

int length = s.length();

return length;

}

int main() {

int number;

cout << "Enter integer: " << endl;

cin >> number;

cout << "The first digit of the number is " << firstDigit(number) << endl;

cout << "The last digit of the number is " << lastDigit(number) << endl;

cout << "The number of digits of the number is " << digits(number) << endl;

}

Explanation:

Security is an essential aspect in any company and for any software organization, through a better protected area or network, the company will be able to maintain important data and operations with access for only authorized users.

Currently, there is a network of hackers and intruders of IT systems whose intention is to steal relevant data and obtain financial benefits from the invasion of company systems, so the IT infrastructure must be a planned and well implemented step in companies, to avoid intruders and failures that could harm an organization's operations.

The ideal IT structure is one developed to meet the needs of a company and its professionals, so there is a need for research, testing and for the implementation of the system to be directed at strategies and organizational objectives.

Another important factor is the development of the policy that will guide the rules and conduct for using the IT system, it is necessary that the policy has the ideal flexibility for the company, so that all its development is focused on coherence and cohesion with the company's strategy must be comprehensive to establish usage rules for users and authentication procedures for specific networks.

The company must have ownership over its system, the data contained therein, its use and privacy, so that the system is always protected, updated and compliant to assist in the improvement of organizational techniques and processes.

You can get complete answer in attached document.please have a look.

Explanation:

(2) Without rolloff, as given fc =1kHz Let us assume it is a first order RC filter ats transfer function is H(s)-RC RC withou

Then the transfer function becom

and step response is flat zero for without rolloff..

Answer:

The solution is written in Python 3

Explanation:

Firstly, create a function maxTemp that take one input file name (Line 1). Next create an exception handling try and except block to handle the situation when the read file is not found or cannot be open, it shall return -1 (Line 2, Line 15 -16).

Next use the readlines method to read the temperature data from the input file (Line 4). Before we traverse the read data using for loop, we create a variable highest to hold the value of max temperature (Line 5). Let's set it as zero in the first place.

Then use for loop to read the temp data line by line (Line 6) and if there is any empty line (Line 7) just ignore and proceed to the next iteration (Line 7 -8). Otherwise, the current row reading will be parsed to float type (Line 10) and the parsed data will be check with the current highest value. If current highest value is bigger than the current parsed data (), the program will set the parse data to highest variable (Line 11 -12).

After finishing the loop, return the highest as output (Line 14).  

Answer:

The explanation of the question is described in the section below.

Explanation:

(a)

Go Back N :

A gives in a maximum of 9 pieces. Typically they will be sent back sections 1, 2, 3, 4, 5 and then re-sent segments 2, 3, 4 and 5.

B sends out 8 ACK's. They are 4 ACKS including 1 series and 4 ACKS with 2, 3, 4 and 5 series amounts.

Selective Repeat :

A sends in such max of 6 bits. Subsequently, segments 1, 2, 3, 4, 5 and earlier re-sent Segments 2 will be sent.

B assigns five ACKs. We are 4our ACKS with numbers 1, 3, 4, 5. However there is one sequence quantity 2 ACK.

TCP :

A assigns in a total of 6 bits. Originally, segments 1, 2, 3, 4, 5 and future re-sent Segments 2 have always been sent.

B sends five ACKs. There are 4 ACKS with either the number 2 series. There has been one ACK with a sequence of numbers 6. Remember that TCP always needs to send an ACK with a sequence number you anticipate.

(b)

This is although TCP utilizes convenient retransmission without searching for the time out.

So, it's the right answer.

Answer:

Tom Servo

Bob (prints inside Cyborg constructor)

Bob (prints inside Cyberman constructor)

Explanation:

The above type of inheritance is multilevel inheritance where Robot is super class and Cyborg and Cyberman are sub classes.

So in above statement

Cyberman one = new Cyberman("Bob"); it calls the Cyberman(String name) in which super(name) calls Cyborg(String name) but before executing the body inside Cyborg constructor it first calls super class default constructor (i.e., Robot()) so the output of the above statement is

Tom Servo

Bob (prints inside Cyborg constructor)

Bob (prints inside Cyberman constructor)

Answer:

Code:

    #include <iostream>

    #include <fstream>

    #include <cctype>

    #include <cmath>

    #include <vector>

    #include <float.h>

    using namespace std;

    typedef vector<vector<double> > C2DArray;

    void Matrix_Mult(bool* error_flag, const int mRows, const int nCols, const int pCols, const C2DArray A_Matrix, const C2DArray B_Matrix, C2DArray& C_Matrix) {      

     int i = A_Matrix.size(), j = A_Matrix[0].size(), k = B_Matrix[0].size();

     double dummy;

     if ((j != B_Matrix.size()) || (i != mRows) || (j != nCols) || (k != pCols))  *error_flag = true;

     if ((C_Matrix.size() != i) || (k != C_Matrix[0].size()))  *error_flag = true;

     if (!(*error_flag)){

     for (k = 0; k < mRows; k++) { // Do each row of A with each column of B

      for (i = 0; i < pCols; i++) {

       dummy = 0.0;

       for (j = 0; j < nCols; j++) {

        dummy += A_Matrix[k][j]*B_Matrix[j][i];

          } // End for j

          C_Matrix[k][i] = dummy;

      } // End for i

     } // End for k

     }

        return;

    } // End Matrix_Mult

    int main()

    {char rflag = 0; //Readiness flag  

    cout << "                 Mat_Mul1 (7 June 2013)\n";

    cout << "=========================================================================== \n";

    cout << "This program multiplies two matrices, [A] and [B]: [A][B] = [C].\n";

    cout << "[A] is M X N, [B] is N X P, and [C] is M X P\n";

    cout << "The dimensions of the [A] and [B] matrices, M, N, and P, should \n";

    cout << "have been saved beforehand in a file named mulmatdat.\n";

    cout << "mulmatdat should be in the same folder as the Mat_Mul1 executable.\n";

    cout << "The next values of mulmatdat should be the entries for matrix [A],\n";

    cout << "with data for row 1 first, then row 2, then row 3, etc.\n";

    cout << "The entries for the [B] matrix should follow, in the same order.\n";

    cout << "\nThe data is assumed to be of type double. Variables used within this program\n";

    cout << "are type double.\n";

    cout << "\nThe output is written to the file mulmatout.txt.\n";

    cout << "\nIs everything ready (are you ready to continue?)? If yes, Enter y. \n";

    cout << "Otherwise Enter any other key. \n";

    cin >> rflag;

    if (toupper(rflag) == 'Y') {

    C2DArray A, B, C; // A, B, and C Matrices

    int i, j, mDim, nDim, pDim; // Array index variables and matrix dimensions

    bool erFlag = false;  // Error flag

    ifstream in("mulmatdat.txt", ios::in);

    if (!in) {

            cout << "\nCannot open the input file.\n";

         cout << "\nEnter any key to continue. \n";

         cin >> rflag;

            return 0;

    }

    in >> mDim >> nDim >> pDim; //Input the Matrices' dimensions from the file

    if ((mDim < 1) || (nDim < 1) || (pDim < 1)){

            cout << "\nInvalid dimension entered. Program terminated. \n";

         cout << "\nEnter any key to continue. \n";

         cin >> rflag;

            in.close(); //Close the input file before terminating

            return 0;

    }

    ofstream out("mulmatout.txt", ios::out);

    if (!out) {

            cout << "\nCannot open the output file. Program terminated.\n";

         cout << "\nEnter any key to continue. \n";

         cin >> rflag;

            in.close(); //Close the input file before terminating

            return 0;

    }

    // Beginning of try block, if vector re-sizing unsuccessful

    try {

           // Resize the arrays to the appropriate sizes

            A.resize(mDim); // M rows

         B.resize(nDim); // N rows

         C.resize(mDim); // M rows

            for (i = 0; i < mDim; i++){

               A[i].resize(nDim); // M columns

            C[i].resize(pDim); // P columns

            } // End for i

         for (i = 0; i < nDim; i++){

               B[i].resize(pDim); // P columns

            } // End for i

    } // End of try block

    catch (bad_alloc& xa) { // Catch block, for exceptions

            in.close();

            out.close();

            cerr << "\nIn catch block, so an exception occurred: " << xa.what() << "\n";

            cout << "\nEnter any key to continue. \n";

            cin >> rflag;

            return 0;

    } // End of catch block

    for (i = 0; i < mDim; i++){ //Input the A Matrix from the file

            for (j = 0; j < nDim; j++){

              in >> A[i][j];

            }//End for j

    }//End for i

    for (i = 0; i < nDim; i++){ //Input the B Matrix from the file

            for (j = 0; j < pDim; j++){

             in >> B[i][j];

            }//End for j

    }//End for i

    in.close(); //Close the input file

    Matrix_Mult(&erFlag, mDim, nDim, pDim, A, B, C);

    if (erFlag){

      cout << "Inconsistent data sent to Matrix_Mult routine. Matrix multiplication NOT performed. Check data before running again.\n";

    }

    else {

     out.precision(DBL_DIG);

     out << "\nThe A matrix follows:\n";

     out << "\n";

     for (i = 0; i < mDim; i++){

      for (j = 0; j < nDim; j++){

     out << A[i][j] << " ";

      } // End for j

      out << "\n";

     }//End for i

     out << "\nThe B matrix follows:\n";

     out << "\n";

     for (i = 0; i < nDim; i++){

      for (j = 0; j < pDim; j++){

     out << B[i][j] << " ";

      } // End for j

      out << "\n";

    }//End for i

     out << "\nThe product matrix, C, follows:\n";

     out << "\n";

     for (i = 0; i < mDim; i++){

      for (j = 0; j < pDim; j++){

       out << C[i][j] << " ";

      } // End for j

      out << "\n";

     }//End for i

     cout << "\nDone! The solution is in the text file mulmatout.txt \n";

    } // End else !erFlag

    out.close();

    }

    else cout << "\nNot ready. Try again when ready with information. \n";

    cout << "\nEnter any key to continue. \n";

    cin >> rflag;

    return 0;

    }

Answer:

3,159 bytes

Explanation:

When implemented as a macro, for each of the 35 executions, the program will require enough memory for the whole macro code (61 bytes). Since the program's data and executable code require 1,024 bytes, the number of bytes required by the entire program is:

[tex]M = 1,024+(35*61)\\M=3,159\ bytes[/tex]

The program requires 3,159 bytes if the new code is added as a macro.

Answer:

TCP convention is totally connection oriented, it searches for the information that was going in the system definitely and will guarantee an affirmation on the parcel sent, likewise if the uplayer convention is down, it won't permit to move the information, fundamentally it will guarantee there is no any bundle misfortune during the exchange.  

~ However UDP is connectionless convention, there is no any affirmation business in UDP convention, it regularly utilized where not many datagram misfortune doesn't a make a difference. It was nearly a light gauge convention.  

So underneath are sway we will confront, if TCP is expelled.  

1. There will association foundation required in information move, so the application will work minimal quicker.  

2. without appropriate information control and affirmations, there is gigantic possibility for information misfortune.  

3. Presently the application needs to depend totally on UDP, since UDP doesn't has blockage control, the application will get clogged.  

4. Likewise the communication with top layer convention will be a wreck, it won't look if the layers are down or not. so if any upper layer is down, consequently the information will stuck the system, can't capable arrive at the goal.

Explanation:

Switching to TCP Reno from Tahoe requires adapting to faster congestion recovery and retransmission strategies. Applications should adjust for improved throughput and handling of packet loss signals.

Switching the TCP algorithm from Tahoe to Reno in the operating system affects how your application interacts with the network, particularly in terms of congestion control and handling of packet loss. Here are the key changes you would need to consider and potentially address in your application:

1. Congestion Control Behavior: Reno TCP introduces Fast Recovery and Fast Retransmit mechanisms, which allow faster recovery from packet loss compared to Tahoe. Your application should be aware of these behaviors, especially in scenarios where it needs to manage large data transfers or real-time communications. It may need to adjust its own congestion control strategies or settings to optimize performance with Reno TCP.

2. Handling of Packet Loss: Reno TCP reacts differently to packet loss compared to Tahoe. Instead of immediately halving the congestion window (as Tahoe does), Reno enters Fast Recovery, retransmits the lost packet, and then adjusts its congestion window dynamically. Your application should be prepared for potentially different patterns of packet loss and recovery, which could affect application-level retransmission strategies or timeouts.

3. Performance Monitoring and Tuning: With the change to Reno TCP, monitoring and tuning network performance metrics becomes crucial. Your application may need to adapt its monitoring mechanisms to track parameters specific to Reno's congestion control algorithms, such as congestion window size, packet retransmission rates, and round-trip time estimates.

4. Compatibility and Testing: Ensure compatibility with both Tahoe and Reno TCP variants during transitional periods. Your application should be able to gracefully handle connections with both types of TCP algorithms to support users who have not yet upgraded their operating systems.

5.Documentation and User Communication: Update documentation and inform users about any changes in network behavior or performance optimizations that may result from switching to Reno TCP. Provide guidance on configuring your application to leverage Reno's improvements effectively.

By considering these factors and making necessary adjustments in your application, you can ensure optimal performance and compatibility with the new Reno TCP algorithm introduced in the operating system update.

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Answer:

See the pictures attached

Explanation:

Answer:

Explanation:

def the_perfect(n):

  try: #exception handling if n is a negative number

       n > 0

   except: #return -1 if the error is reached

       return -1

  else:

        total = 0

        for i in range(1, n): #for loop from 1 to the target number

           if n % i == 0:

               total += i

   return total == n  #should return true if number is perfect number

print(perfect_number(8))

Answer:

The memory and the processing unit communicate through the Memory Address Register (MAR) and the Memory Data Register (MDR).

Explanation:

Final answer:

The memory and processing unit in a computer communicates through the system bus, utilizing RAM for fast access to data. This interaction is optimized with techniques like caching and, when necessary, swapping to manage memory usage effectively.

Explanation:

The interface between the memory and the processing unit in a computer system facilitates communication and data exchange crucial for computational tasks. These two components communicate through a data pathway known as the system bus. This bus includes data lines for transferring data, address lines to specify where the data should be sent, and control lines to manage the timing and direction of data flow.

Random Access Memory (RAM) plays a pivotal role in this communication. When a process is initiated, data from the long-term storage (like a hard drive) is loaded into RAM, because RAM is much faster, though more expensive. The processing unit (CPU) then reads instructions and data directly from RAM, performs computations, and may write the results back to RAM or send them to output devices or long-term storage.

Another method to facilitate communication involves caching. CPU caches are small, faster memory locations that store copies of the data from frequently used main memory locations. These caches help reduce the time it takes for the CPU to access data from RAM, speeding up the computation process.

To manage and optimize memory use, operating systems may employ techniques like swapping, where data is moved back and forth between RAM and a hard drive, particularly when RAM is full. However, this can significantly slow down system performance due to the difference in speed between RAM and hard drives.

Answer:

Display random(1, 7).

Explanation:

In the following question, some details of the question are missing that is pseudocode.

//  Program shows three random numbers

// the range of 1 to 7.

Declare the Integer count

// Shows three random numbers.

For count = 1 To 3

Display random(7, 1)

End For

In the following pseudocode, it generates three random numbers from 1 to 7 because the for loop statement is starts from 1 and end at 3 so the loop will iterate three times and every time it generates one random number from 1 to 7. So, the following are the reason that describe the answer is correct according to the scenario.

The issue with the pseudocode might be related to the exclusive nature of the upper boundary in many random number generator methods. The programmer may need to add 1 to the function to generate values inclusive of 7.

Without seeing the actual pseudocode, it's tough to give a specific correction. However, the common error in generating random numbers is not considering the exclusive nature of the upper boundary in random number methods. Most methods generate numbers from 0 (inclusive) up to the specified upper limit number (exclusive).  To generate random numbers within the range from 1 to 7 (inclusive), the code should add 1 to the random output. Therefore, a possible correction would be to modify the random number generator function to include 7 and exclude 0.

For example, in many programming languages, one could use a function similar to '1 + Math.floor(Math.random() * 7)', ensuring that we generate random numbers between 1 and 7 inclusive. Kindly refer to the specific programming language practices when implementing.

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A battery has capacity (instance variable) which can be drained. Capacity is a double value measured in milliampere hours (mAh) -A typical AA battery has a capacity of 2000 to 3000 mAh

Explanation:

public class Battery

{

     private double fullCharge;

     private double batteryCapacity;

public Battery(doublecapacity)

{

     battery Capacity = capacity;

     full Charge = capacity;

}

public void drain(double amount)

{

    battery Capacity = battery  Capacity - amount;

}

public void charge()

{

     battery Capacity = fullCharge;

}

public double get RemainingCapacity()

{

    return battery Capacity;

}

}  

Answer:

value=int(input("Enter the value from where the user wants to starts: "))#Take the value from the user.

i=1#intialize the value of a variable by 1.

while(value<1000) and (i<9):#while loop which prints the value.

   print(value,end=" ")# print the value.

   value=value*2#calculate the value to print.

   i=i+1#calculate the value for the count.

Output:

If the user enter 5, then the output is : "5 10 20 40 80 160 320 640".

If the user enter 5, then the output is : "3 6 12 24 48 96 192 384".

Explanation:

Answer:

It will send ARP request broad cast because of security and pubilicity of data to the public.