The function below takes a single string parameter: sentence. Complete the function to return everything but the middle 10 characters of the string. You can assume that the parameter always has at least twelve characters and is always an even number of characters long. Hint: To find the middle of the string, you can half the length of the string (floor division). Then slice out 5 less than the middle to 5 more than the middle.

Answer:

See explaination for the details.

Explanation:

Code:

def create_3D(H, W, D):

result = []

for i in range(D):

result.append([])

for j in range(H):

result[i].append([])

for k in range(W):

result[i][j].append(i+j+k)

return result

print(create_3D(2, 3, 4))

Check attachment for onscreen code.

RC4 stream cipher in C. User is able to enter any key that is 5 bytes to 32 bytes long

Explanation:

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#define N 256   // 2^8

void swap(unsigned char *a, unsigned char *b) {

   int tmp = *a;

   *a = *b;

   *b = tmp;

}

int KSA(char *key, unsigned char *S) {

   int len = strlen(key);

   int j = 0;

   for(int i = 0; i < N; i++)

       S[i] = i;

   for(int i = 0; i < N; i++) {

       j = (j + S[i] + key[i % len]) % N;

       swap(&S[i], &S[j]);

   }

   return 0;

}

int PRGA(unsigned char *S, char *plaintext, unsigned char *ciphertext) {

   int i = 0;

   int j = 0;

   for(size_t n = 0, len = strlen(plaintext); n < len; n++) {

       i = (i + 1) % N;

       j = (j + S[i]) % N;

       swap(&S[i], &S[j]);

       int rnd = S[(S[i] + S[j]) % N];

       ciphertext[n] = rnd ^ plaintext[n];

   }

   return 0;

}

int RC4(char *key, char *plaintext, unsigned char *ciphertext) {

   unsigned char S[N];

   KSA(key, S);

   PRGA(S, plaintext, ciphertext);

   return 0;

}

int main(int argc, char *argv[]) {

   if(argc < 3) {

       printf("Usage: %s <key> <plaintext>", argv[0]);

       return -1;

   }

   unsigned char *ciphertext = malloc(sizeof(int) * strlen(argv[2]));

   RC4(argv[1], argv[2], ciphertext);

   for(size_t i = 0, len = strlen(argv[2]); i < len; i++)

       printf("%02hhX", ciphertext[i]);

   return 0;

}

Answer:

Check the explanation

Explanation:

Here is the code for you:

#include <iostream>

#include <fstream>

using namespace std;

int main ()

{

int aNumber=0;

int numbers=0;

double sum=0.0;

double average=0.0;

ifstream randomFile;

randomFile.open("Random.txt");

if (randomFile.fail())

cout << "failed to read file.";

else

{

while (randomFile >> aNumber)

{

numbers++;

sum+=aNumber;

}

if (numbers>0)

average = sum/numbers;

else

average=0.0;

cout << "Number of numbers: " << numbers << "\n";

cout << "Sum of the numbers: " << sum << "\n";

cout << "Average of the numbers: " << average;

}

randomFile.close();

return 0;

}

In this exercise we have to use the computer language knowledge in C++ to write the code as:

the code is in the attached image.

In a more easy way we have that the code will be:

#include <iostream>

#include <fstream>

using namespace std;

int main ()

{

int aNumber=0;

int numbers=0;

double sum=0.0;

double average=0.0;

ifstream randomFile;

randomFile.open("Random.txt");

if (randomFile.fail())

cout << "failed to read file.";

else

{

while (randomFile >> aNumber)

{

numbers++;

sum+=aNumber;

}

if (numbers>0)

average = sum/numbers;

else

average=0.0;

cout << "Number of numbers: " << numbers << "\n";

cout << "Sum of the numbers: " << sum << "\n";

cout << "Average of the numbers: " << average;

}

randomFile.close();

return 0;

}

See more about C++ code at brainly.com/question/25870717

Answer:

Java Programming

Details explained below.

Explanation:

import java.util.Scanner;

public class DemoDriver {

public static void main(String[] args) {

// TODO Auto-generated method stub

Scanner sc=new Scanner(System.in);

Purse sophie=new Purse();

Purse sally=new Purse();

while(true)

{

System.out.println("Enter 1 for Sophie, 2 for Sally, or 0 to exit: \n");

int girl=sc.nextInt();

int ch;

int pence,shill;

if(girl==1)

{

System.out.println("Enter 1 to give pence, 2 to give shillings, 3 to query her purse:\n");

ch=sc.nextInt();

if(ch==1)

{

System.out.println("Enter the pence to give:\n");

pence=sc.nextInt();

sophie.incPence(pence);

}else if(ch==2)

{

System.out.println("Enter the shillings to give:\n");

shill=sc.nextInt();

sophie.incshilling(shill);

}

else

{

System.out.println("Purse has "+sophie.getPence()+" pence,"+sophie.getshilling()+" shillings\n");

}

}

else if(girl==2)

{

System.out.println("Enter 1 to give pence, 2 to give shillings, 3 to query her purse:\n");

ch=sc.nextInt();

if(ch==1)

{

System.out.println("Enter the pence to give:\n");

pence=sc.nextInt();

sally.incPence(pence);

}else if(ch==2)

{

System.out.println("Enter the shillings to give:\n");

shill=sc.nextInt();

sally.incshilling(shill);

}

else

{

System.out.println("Purse has "+sally.getPence()+" pence,"+sally.getshilling()+" shillings\n");

}

}else

{

break;

}

}

}

} END.

Explanation:

yeah yeah is that we go together on the watch remember write a timer called that depressed display of the first 27 value in the XM Sirius

Answer:

See explaination

Explanation:

a) Reorganizing the code

ADD $s2 , $t2, $t3

ADD $t5, $s2, $t6

SUB $t4, $s2, $t5

AND $t5,$t4,$t6

The reordering of the instruction leads to less hazards as compared to before.

b) with NOP

ADD $s2 , $t2, $t3

NOP

NOP

SUB $t4, $s2, $t5

NOP

ADD $t5, $s2, $t6

NOP

NOP

AND $t5,$t4,$t6

c. Pipeline with stalls

see attachment please

d. pipeline with forwarding

see attachment

McDonald's H&M McDonalds Inc is incorrect for name accuracy.

Explanation:

The correct Name is McDonald's. But Mc Donald's H&M and McDonalds inc is incorrect for name accuracy.

McDonald's is a fast-food company based in America. It has its retail outlets throughout the world. It is very famous for its burgers and french fries. It is the world's second-largest employer in the private sector. It is a symbol of globalization as it has its outlets throughout the world.

The name 'McDonald's H&M' is incorrect for name accuracy since it amalgamates two distinct brand identities. While 'McDonalds Inc' may simply be missing an apostrophe, it is possibly acceptable, but 'McDonald's H&M' clearly combines unrelated trademarks.

When evaluating the list of business names provided for name accuracy, it's important to consider the standard format of business names and trademarks. Typically, the name 'McDonald's' is associated with the well-known fast-food chain, and is correctly punctuated with an apostrophe to indicate the possessive form, likely referring to the founder's name (McDonald).

'McDonald's H&M' would be incorrect for name accuracy, as it combines the names of two distinct and unrelated brands, McDonald's and H&M, which are independently operated and have separate trademark rights. 'McDonalds Inc' is missing the possessive apostrophe, which could be seen as a minor error, but might still be legally operational depending on registration specifics.

Therefore, out of the options given, 'McDonald's H&M' would be rated incorrect for name accuracy due to the improper combination of two separate brand identities into one business name.

The groups_per_user function receives a dictionary, which contains group names with the list of users.

Explanation:

The blanks to return a dictionary with the users as keys and a list of their groups as values is shown below :

def groups_per_user(group_dictionary):

   user_groups = {}

   # Go through group_dictionary

   for group,users in group_dictionary.items():

       # Now go through the users in the group

       for user in users:

       # Now add the group to the the list of

         # groups for this user, creating the entry

         # in the dictionary if necessary

         user_groups[user] = user_groups.get(user,[]) + [group]

   return(user_groups)

print(groups_per_user({"local": ["admin", "userA"],

       "public":  ["admin", "userB"],

       "administrator": ["admin"] }))

The missing statements in the program are:

The first missing instruction is to iterate through the group_dictionary.

To do this, we make use of the following enhanced for loop:

for group,users in group_dictionary.items():

The above statement would iterate through the keys and the values of the dictionary.

The next missing statement is to iterate through the values in users list

To do this, we make use of the following enhanced for loop:

for user in users:

Lastly, the group is added to the user list using:

user_groups[user] = user_groups.get(user,[]) + [group]

Read more about dictionary and lists at:

https://brainly.com/question/14353514

Answer:

import java.util.Scanner;

public class BirthdayChecker

{

   public static void main(String[] args)

   {

       String sentinelValue = "ZZZ";

       final int size = 10;

       int count = 0;

       String name = null;

       String dateOfBirth = null;

       String[] namesArray = new String[size];

       String[] dateOfBirthsArray = new String[size];

     Scanner scanner = new Scanner(System.in);

       System.out.println("Enter name or enter ZZZ " +  "to quit");

       name = scanner.nextLine();

       while(!name.equals(sentinelValue) && count < 10)

       {

           System.out.println("Enter date of birth " +  "(dd-mm-yyyy)");

           dateOfBirth=scanner.nextLine();

           namesArray[count] = name;

           dateOfBirthsArray[count] = dateOfBirth;

           System.out.println("Enter name or enter " +

           "ZZZ to quit");

           name=scanner.nextLine();

           count++;

       }

       System.out.println("Count of namesArray = "+count);

       System.out.println("The entered namesArray are:");

       for(int i = 0; i < count; i++)  

       {

           System.out.println(namesArray[i]);

       }

       boolean criteria = true;

       boolean flag = false;

       while(criteria)

       {

           System.out.println("Enter name to " +

           "display date of birth or enter " +

           "ZZZ to quit");

           name = scanner.nextLine();

           if(name.equals(sentinelValue))

               criteria = false;

           else

           {

               for(int i = 0; i < count && !flag;

               i++)

               {

                   if(namesArray[i].equals(name))

                   {

                       flag = true;

                       dateOfBirth = dateOfBirthsArray[i];

                   }

               }

               if(flag)

               {

                   System.out.println("Date of Birth "

                   + "of " + name + " is "

                   + dateOfBirth);

               }

               else

               {

                   System.out.println("Date Of Birth "

                   + "of " + name + " is not flag");

               }

           }

           flag = false;

       }

   }

}

Explanation:

The `BirthdayReminder` Java app lets users enter names and birthdates, then lookup birthdates by name until "ZZZ" is entered.

Here's a Java application named `BirthdayReminder` that fulfills the specified requirements:

```java

import java.util.Scanner;

public class BirthdayReminder {

   public static void main(String[] args) {

       Scanner scanner = new Scanner(System.in);        

       String[] names = new String[10];

       String[] birthdates = new String[10];

       int count = 0;        

       // Input loop to enter names and birthdates

       while (count < 10) {

           System.out.print("Enter name (ZZZ to end): ");

           String name = scanner.nextLine();            

           if (name.equals("ZZZ")) {

               break; // Exit loop if sentinel value ZZZ is entered

           }            

           names[count] = name;            

           System.out.print("Enter birthdate (MM/DD/YYYY): ");

           String birthdate = scanner.nextLine();

           birthdates[count] = birthdate;            

           count++;

       }        

       System.out.println("\nNumber of names entered: " + count);

       System.out.println("List of names entered:");

       for (int i = 0; i < count; i++) {

           System.out.println(names[i]);

       }        

       // Lookup loop to find birthdate by name

       while (true) {

           System.out.print("\nEnter a name to lookup birthdate (ZZZ to end): ");

           String lookupName = scanner.nextLine();            

           if (lookupName.equals("ZZZ")) {

               break; // Exit loop if sentinel value ZZZ is entered

           }            

           boolean found = false;

           for (int i = 0; i < count; i++) {

               if (names[i].equals(lookupName)) {

                   System.out.println("Birthdate of " + lookupName + " is " + birthdates[i]);

                   found = true;

                   break;

               }

           }

           if (!found) {

               System.out.println("Name not found. Please enter a valid name.");

           }

       }        

       scanner.close();

   }

}

Explanation:

1. **Data Structures**: Two arrays `names` and `birthdates` are used to store up to 10 names and their corresponding birthdates.

2. **Input Loop**: The program prompts the user to enter names and birthdates in a loop until either 10 names are entered or the user inputs "ZZZ". Each name and birthdate pair is stored in the respective arrays.

3. **Output**: After input is finished, the program displays the count of names entered and lists the names that were stored.

4. **Lookup Loop**: The program then enters a lookup loop where it prompts the user to enter a name to retrieve the corresponding birthdate. It continues this until the user inputs "ZZZ". If a valid name is entered, it searches the `names` array for a match and displays the associated birthdate; if not found, it informs the user accordingly.

5. **End of Program**: The program closes the scanner once all operations are completed.

This Java application effectively manages user input, data storage, and retrieval as specified, providing a straightforward way to handle names and birthdates with user interaction and error handling.

Answer:

Explanation:

# Run Monte Carlo 10k

B <- 10000

results <- replicate(B, {

 winners <- sample(runners, 3)

 (winners[1] %in% "Jamaica" & winners[2] %in% "Jamaica" & winners[3] %in% "Jamaica")

})

mean(results)

Answer:

Check the explanation

Explanation:

Q1: SQL to list names of all the students whose GPA is above specific number:

select SName

from Student

where GPA >= 3.8;

Using the Student table we are able to get the details where the GPa is greater than the required value.

Q2: SQL to get list of names of all the students who applied to some college:

select SName from Student where SID in (

select distinct SID from Apply

);

Selecting the name of the students who are present in the Apply table.

Q3: SQL to get list of names all the students who did not apply to any college:

select SName from Student where SID not in (

select distinct SID from Apply

);

Selecting the name of the students who are not present in the Apply table.

Q4: SQL to get list of names and enrollment of all the colleges who received applications for a major involving bio:

select CName, Enrollment from College where CName in (

select CName from Apply where Major = 'Bio' and SID in (

select SID from Student

)

);

Final answer:

Calculations reveal that Vendor A's system would take 0.209 seconds, while Vendor B's system, despite having hardware 1000 times faster, would take an infeasible 10,973.839 seconds due to the different algorithms used. Thus, Vendor A's system is the superior choice for query speed.

Explanation:

To determine which database system to choose based on the query response times for 1,048,576 customers, we need to calculate the response times for both Vendor A and Vendor B given their respective algorithms and computer speeds.

For Vendor A, the time is proportional to 10nlog2n machine operations. Plugging in the number of customers we have:

TA = 10 * 1,048,576 * log2(1,048,576) * 10-9

We know that log2(1,048,576) = 20 because 220 = 1,048,576. So:

TA = 10 * 1,048,576 * 20 * 10-9 = 209,715.2 * 10-9 = 0.209 seconds (rounded to three decimal places)

For Vendor B, whose computer is 1000 times faster, the time is proportional to 10n2 machine operations.

TB = 10 * (1,048,576)2 * 10-9 / 1000

TB = 10 * 1,097,383,936,256 * 10-9 / 1000 = 1.097x1013 * 10-9 seconds = 10,973.839 seconds, which is not feasible for the system. Hence, Vendor A's system is clearly the better choice in terms of response time.

Answer:

Check the explanation

Explanation:

CODE:

#include <iostream>

#include <iomanip>

using namespace std;

double getLoanAmount() { // functions as per asked in question

double loan;

while(true){ // while loop is used to re prompt

cout << "Enter the car loan amount ($2,500-7,500): ";

cin >> loan;

if (loan>=2500 && loan <= 7500){ // if the condition is fulfilled then

break; // break statement is use to come out of while loop

}

else{ // else error message is printed

cout << "Error: $"<< loan << " is an invalid loan amount." << endl;

}

}

return loan;

}

double getMonthlyPayment(){ // functions as per asked in question

double monthpay;

while(true){ // while loop is used to re prompt

cout << "Enter the monthly payment ($50-750): ";

cin >> monthpay;

if (monthpay>=50 && monthpay <= 750){ // if the condition is fulfilled then

break; // break statement is use to come out of while loop

}

else{ // else error message is printed

cout << "Error: $"<< monthpay << " is an invalid monthly payment." << endl;

}

}

return monthpay;

}

double getInterestRate(){ // functions as per asked in question

double rate;

while(true){ // while loop is used to re prompt

cout << "Enter the annual interest rate (1-6%): ";

cin >> rate;

if (rate>=1 && rate <= 6){ // if the condition is fulfilled then

break; // break statement is use to come out of while loop

}

else{ // else error message is printed

cout << "Error: "<< rate << "% is an invalid annual interest rate." << endl;

}

}

return rate;

}

int main() {

cout << setprecision(2) << fixed; // to print with 2 decimal places

int month = 0; // initializing month

// calling functions and storing the returned value

double balance= getLoanAmount();

double payment= getMonthlyPayment();

double rate= getInterestRate();

rate = rate /12 /100; // as per question

// printing as per required in question

cout << "Month Balance($) Payment($) Interest($) Principal($)"<< endl;

while(balance>0){ // while the balance is more than zero

month = month + 1; // counting Months

// calculations as per questions

double interest = balance * rate;

double principal = payment - interest;

balance = balance - principal;

// printing required info with proper spacing

cout <<" "<< month;

cout <<" "<< balance;

cout <<" "<< payment;

cout <<" "<< interest;

cout <<" "<< principal << endl;

}

cout << "Months to repay loan: " << month << endl; // displaying month

cout << "End of Banker Bisons";

Kindly check the output in the attached image below.

Answer:

See explaination

Explanation:

class Episode{

//Private variables

private String title;

private int seasonNumber;

private int episodeNumber;

//Argumented constructor

public Episode(String title, int seasonNumber, int episodeNumber) {

this.title = title;

this.seasonNumber = seasonNumber;

this.episodeNumber = episodeNumber;

}

//Getter and setters

public String getTitle() {

return title;

}

public void setTitle(String title) {

this.title = title;

}

public int getSeasonNumber() {

return seasonNumber;

}

public void setSeasonNumber(int seasonNumber) {

this.seasonNumber = seasonNumber;

}

public int getEpisodeNumber() {

return episodeNumber;

}

public void setEpisodeNumber(int episodeNumber) {

this.episodeNumber = episodeNumber;

}

public boolean comesBefore(Episode e){

//Check if titles' match

if(this.title.equals(e.getTitle())){

//Compare season numbers

if(this.seasonNumber<e.seasonNumber){

return true;

}

//Check if season numbers match

else if(this.seasonNumber==e.getSeasonNumber()){

return this.episodeNumber<e.getEpisodeNumber();

}

}

return false;

}

atOverride // replace the at with at symbol

public String toString() {

return title+", season "+seasonNumber+", episode "+episodeNumber;

}

}

class Main{

public static void main(String[] args) {

Episode e = new Episode("Friends",1,1);

System.out.println(e);

Episode e2 = new Episode("The Wire",3,5);

System.out.println(e2);

System.out.println(e.getTitle()+" comes before "+e2.getTitle()+" = "+e.comesBefore(e2));

}

}

class Animal:

   def __init__(self, arms, legs):

       self.arms = arms

       self.legs = legs

   def limbs(self):

       return self.arms + self.legs

# Creating an instance of Animal named spider with 4 arms and 4 legs

spider = Animal(4, 4)

# Calling the limbs method on the spider instance and saving the result to spidlimbs

spidlimbs = spider.limbs()

print("Number of limbs of the spider:", spidlimbs)

Answer:

What

Explanation:

what

what

Answer:

Explanation:

1: The three most popular data systems that make use of Big Data file systems approach are:

The HDFS (Hadoop Distributed File System), Apache Spark, and Quantcast File System(QFS).

HDFS is the most popular among these and it makes use of the MapReduce algorithm to perform the data management tasks. It can highly tolerate faults and can run on low-cost hardware. It was written in Java and it is an open-source software.

Apache Spark makes use of Resilient Distributed Data (RDD) protocol. It is much faster and lighter than the HDFS and it can be programmed using a variety of languages such as Java, Scala, Python, etc. Its main advantage over HDFS is that it is highly scalable.

While QFS was developed as an alternative to the HDFS and it is also highly fault-tolerant and with space efficient. It makes use of the Reed-Solomon Error Correction technique to perform the task of data management.

2: The NewSQL databases were developed as a solution to the scalability challenges of the monolithic SQL databases. They were designed to allow multiple nodes in the context of an SQL database without affecting the replication architecture. It worked really well during the starting years of the cloud technology. Some of the databases that make use of New SQL technology are Vitess, Citus, etc.

Vitess was developed as an automatic sharding solution to the MySQL. Every MySQL instance acts as a shard of the overall database and each of these instances uses standard MySQL master-slave replication to ensure higher availability.

While, Citus is a PostgreSQL equivalent of the Vitess. It ensures transparent sharding due to which it accounts for horizontal write scalability to PostgreSQL deployments.

NoSQL database technology was developed to provide a mechanism for the storage and retrieval of data that is modeled in a way other than the tabular relations used in the traditional databases (RDBMS). The most popular database that makes use of the NoSQL technology is MongoDB. It functions as a cross-platform document-oriented database. It is known for its ability to provide high availability of replica sets. A replica set is nothing but a bundle of two or more copies of the data

Answer:

40 ns would be the maximum MIPS value to be achieved for this computer.

Answer:

MIPS = 10.53

Explanation:

To find the maximum allowable MIPS, we need to obtain the total maximum time for an instruction to be executed by the CPU.

Total time to complete 1 instruction for the computer will entail the instruction passing through the 3 stages.

Total time to complete 1 instruction = 40 + 26 + 29 = 95 ns = (95 × 10⁻⁹) s

MIPS = millions instructions per second and it is given as

(Number of instruction)/(execution time × 1,000,000)

MIPS = 1/(95 × 10⁻⁹ × 10⁶)

MIPS = 10.53

Hope this Helps!!!