Let U = {b1, b2, , bn} with n ≥ 3. Interpret the following algorithm in the context of urn problems. for i is in {1, 2, , n} do for j is in {i + 1, i + 2, , n} do for k is in {j + 1, j + 2, ..., n} do print bi, bj, bk How many lines does it print? It prints all the possible ways to draw three balls in sequence, without replacement. It prints P(n, 3) lines. It prints all the possible ways to draw an unordered set of three balls, without replacement. It prints P(n, 3) lines. It prints all the possible ways to draw three balls in sequence, with replacement. It prints P(n, 3) lines. It prints all the possible ways to draw an unordered set of three balls, without replacement. It prints C(n, 3) lines. It prints all the possible ways to draw three balls in sequence, with replacement. It prints C(n, 3) lines.

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Answer:

Bits

Explanation:

Answer:

Explanation:

The solution code is written in Python 3.

Firstly, we can create a Library class with one constructor (Line 2). This constructor won't take any input parameter value. There is only one instance variable, books, in the class (Line 3). This instance variable is an empty list.

To test our class, we can create an object lib1 (Line 5).  Next use that object to add the book item to the books list in the object (Line 6-8).  

Answer: Provided in the explanation segment

Explanation:

Below is the code to run this program.

we have that the

Program

import java.util.Scanner;

public class CharMatching {

  public static void main(String[] args) {

      //Scanner object for keyboard read

      Scanner scnr=new Scanner(System.in);

      //Variable for user input string

      String userInput;

      //Variable for firstletter input

      char firstLetter;

      //Read user input string

      userInput=scnr.nextLine();

      //Read first letter from user

      firstLetter=scnr.nextLine().charAt(0);

      //Comparison without case sensititvity and result

      if(Character.toUpperCase(firstLetter)==Character.toUpperCase(userInput.charAt(0))) {

          System.out.println("Found match: "+firstLetter);

      }

      else {

          System.out.println("No match: "+firstLetter);

      }

  }

}

Output

Hello

h

Found match: h

cheers i hope this helped !!!

Answer:

public class RectangleArea {

   public static void main(String[] args) {

       Rectangle myRectangle = new Rectangle(15.8, 7.9);

       double calArea = myRectangle.calcArea(15.8, 7.9);

       System.out.println("The area is "+ calArea);

   }

}

class Rectangle{

   private double length;

   private double width;

   //Defined Constructor

   public Rectangle(double length, double width) {

       this.length = length;

       this.width = width;

   }

   public double calcArea(double len, double wi){

       return len* wi;

   }

}

Explanation:

Answer:

public class SwitchCase {

   public static void main(String[] args) {

       int num = 0;

       int a = 10, b = 20, c = 20, d = 30, x = 40;

       switch (num){

           case 102: a += 1;

           case 103: a += 1;

           case 104: a += 1;

           case 105: a += 1;

           break;

           case 208: b += 1; x = 8;

           break;

           case 209: c = c * 3;

           case 210: c = c * 3;

           break;

           default: d += 1004;

       }

   }

}

Explanation:

Answer:

Check the explanation

Explanation:

Let the population mean time on task of original website be [tex]\mu_1[/tex] and that of new prototype be [tex]\mu_2[/tex]

Here we are to test

[tex]H_0:\mu_1=\mu_2\;\;against\;\;H_1:\mu_1<\mu_2[/tex]

The data are summarized as follows:

                                                       Sample 1             Sample 2

Sample size                                        n=20                    n=20

Sample mean                                 [tex]\bar{x}_1=243.4[/tex]             [tex]\bar{x}_2=253.4[/tex]

Sample SD                                     s1=130.8901          s2=124.8199

The test statistic is obtained from online calculator as

t=-0.23

The p-value is given by 0.821

As the p-value is more than 0.05, we fail to reject the null hypothesis at 5% level of significance and hence conclude that there is no statistically significant difference between the original website and the prototype developed by the team at 5% level of significance.

Answer: I would suggest you consider your audience and how you can connect to them. Is your presentation, well, presentable? Is whatever you're presenting reliable and true? Also, no more than 6 lines on each slide. Use colors that contrast and compliment. Images, use images. That pulls whoever you are presenting to more into your presentation.

Explanation:

Answer:

A. 0.0450

B. 4

C. 0.25

D. 37.68

E. 6Hz

F. -0.523

G. 1.5m/s

H. vy = ∂y/∂t = 0.045(-37.68) cos (25.12x - 37.68t - 0.523)

I. -1.67m/s.

Explanation:

Given the equation:

y(x,t) = 0.0450 sin(25.12x - 37.68t-0.523)

Standard wave equation:

y(x, t)=Asin(kx−ωt+ϕ)

a.) Amplitude = 0.0450

b.) Wave number = 1/ λ

λ=2π/k

From the equation k = 25.12

Wavelength(λ ) = 2π/25.12 = 0.25

Wave number (1/0.25) = 4

c.) Wavelength(λ ) = 2π/25.12 = 0.25

d.) Angular frequency(ω)

ωt = 37.68t

ω = 37.68

E.) Frequency (f)

ω = 2πf

f = ω/2π

f = 37.68/6.28

f = 6Hz

f.) Phase angle(ϕ) = -0.523

g.) Wave propagation speed :

ω/k=37.68/25.12=1.5m/s

h.) vy = ∂y/∂t = 0.045(-37.68) cos (25.12x - 37.68t - 0.523)

(i) vy(3.5m, 21s) = 0.045(-37.68) cos (25.12*3.5-37.68*21-0.523) = -1.67m/s.

Answer: Provided in the explanation segment

Explanation:

Below is the code to carry out this program;

/* C++ program helps prompts user to enter the size of the array. To display the array elements, sorts the data from highest to lowest, print the lowest, highest and average value. */

//main.cpp

//include header files

#include<iostream>

#include<iomanip>

using namespace std;

//function prototypes

void displayArray(int arr[], int size);

void selectionSort(int arr[], int size);

int findMax(int arr[], int size);

int findMin(int arr[], int size);

double findAvg(int arr[], int size) ;

//main function

int main()

{

  const int max=50;

  int size;

  int data[max];

  cout<<"Enter # of scores :";

  //Read size

  cin>>size;

  /*Read user data values from user*/

  for(int index=0;index<size;index++)

  {

      cout<<"Score ["<<(index+1)<<"]: ";

      cin>>data[index];

  }

  cout<<"(1) original order"<<endl;

  displayArray(data,size);

  cout<<"(2) sorted from high to low"<<endl;

  selectionSort(data,size);

  displayArray(data,size);

  cout<<"(3) Highest score : ";

  cout<<findMax(data,size)<<endl;

  cout<<"(4) Lowest score : ";

  cout<<findMin(data,size)<<endl;

  cout<<"(5) Lowest scoreAverage score : ";

  cout<<findAvg(data,size)<<endl;

  //pause program on console output

  system("pause");

  return 0;

}

/*Function findAvg that takes array and size and returns the average of the array.*/

double findAvg(int arr[], int size)

{

  double total=0;

  for(int index=0;index<size;index++)

  {

      total=total+arr[index];

  }

  return total/size;

}

/*Function that sorts the array from high to low order*/

void selectionSort(int arr[], int size)

{

  int n = size;

  for (int i = 0; i < n-1; i++)

  {

      int minIndex = i;

      for (int j = i+1; j < n; j++)

          if (arr[j] > arr[minIndex])

              minIndex = j;

      int temp = arr[minIndex];

      arr[minIndex] = arr[i];

      arr[i] = temp;

  }

}

/*Function that display the array values */

void displayArray(int arr[], int size)

{

  for(int index=0;index<size;index++)

  {

      cout<<setw(4)<<arr[index];

  }

  cout<<endl;

}

/*Function that finds the maximum array elements */

int findMax(int arr[], int size)

{

  int max=arr[0];

  for(int index=1;index<size;index++)

      if(arr[index]>max)

          max=arr[index];

  return max;

}

/*Function that finds the minimum array elements */

int findMin(int arr[], int size)

{

  int min=arr[0];

  for(int index=1;index<size;index++)

      if(arr[index]<min)

          min=arr[index];

  return min;

}

cheers i hope this help!!!

Answer:

The formula used in this question is called the probability of combinations or combination formula.

Explanation:

Solution

Given that:

Formula applied is stated as follows:

nCr = no of ways to choose r objects from n objects

  = n!/(r!*(n-r)!)

The Data given :

Menu A : 5 appetizers and 3 main dishes

Menu B : 3 appetizers and 4 main dishes

Total appetizers - 6

Total main dishes - 5

Now,

Part A :

Total ways = No of ways to select menu A + no of ways to select menu B

          = (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

          = 6C5*5C3 + 6C3*5C4

          = 6*10 + 20*5

          = 160

Part B :

Since, we can select the same number of appetizers/main dish again so the number of ways to select appetizers/main dishes will be = (total appetizers/main dishes)^(no of appetizers/main dishes to be selected)

Total ways = No of ways to select menu A + no of ways to select menu B  

          = (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

          = (6^5)*(5^3) + (6^3)*(5^4)

          = 7776*125 + 216*625

          = 1107000

Part C :

No of ways to select same appetizers and main dish for all the options

= No of ways to select menu A + no of ways to select menu B  

= (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

=(6*5) + (6*5)

= 60

Total ways = Part B - (same appetizers and main dish selected)      

= 1107000 - 60

= 1106940

Answer:

Term describes the order of arrangement of files and folders on a computer would be ORGANIZATION.

:) Hope this helps!

Answer:

The term that describes the order of arrangement of files and folders on a computer is organization.

Explanation:

Answer:

A Program was written to carry out some set activities. below is the code program in C++ in the explanation section

Explanation:

Solution

CODE

#include <iostream>

using namespace std;

int main() {

string name; // variables

int number;

cin >> name >> number; // taking user input

while(number != 0)

{

// printing output

cout << "Eating " << number << " " << name << " a day keeps the doctor away." << endl;

// taking user input again

cin >> name >> number;

}

}

Note: Kindly find an attached copy of the compiled program output to this question.

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

The code can be found in the attached image.

In a very summarized way, we can describe the loop or looping in software as an instruction that keeps repeating itself until a certain condition is met.

To make it simpler we can write this code as:

#include <iostream>

using namespace std;

int main() {

string name; // variables

int number;

cin >> name >> number; // taking user input

while(number != 0)

{

// printing output

cout << "Eating " << number << " " << name << " a day keeps the doctor away." << endl;

// taking user input again

cin >> name >> number;

}

}

See more about C++ at brainly.com/question/19705654

Answer:

The java program for the given scenario is as follows.

import java.util.*;

//interface with method area

interface ObjectWithTwoParameters

{

double area (double d1, double d2);

}

class RectangleClass implements ObjectWithTwoParameters

{

//overriding area()

public double area (double d1, double d2)

{

return d1*d2;

}

}

class TriangleClass implements ObjectWithTwoParameters

{

//overriding area()

public double area (double d1, double d2)

{

return (d1*d2)/2;

}  

}

class CylinderClass implements ObjectWithTwoParameters

{

public double area (double d1, double d2)

{

return ( 2*3.14*d1*d1 + d2*(2*3.14*d1) );

}

}

public class Test

{

public static void main(String[] args)

{

//area displayed for all three shapes

ObjectWithTwoParameters r = new RectangleClass();

double arear = r.area(2, 3);

System.out.println("Area of rectangle: "+arear);

ObjectWithTwoParameters t = new TriangleClass();

double areat = t.area(4,5);

System.out.println("Area of triangle: "+areat);

ObjectWithTwoParameters c = new CylinderClass();

double areac = c.area(6,7);

System.out.println("Area of cylinder: "+areac);

}

}

OUTPUT

Area of rectangle: 6.0

Area of triangle: 10.0

Area of cylinder: 489.84

Explanation:

1. The program fulfils all the mentioned requirements.

2. The program contains one interface, ObjectWithTwoParameters, three classes which implement that interface, RectangleClass, TriangleClass and CylinderClass, and one demo class, Test, containing the main method.

3. The method in the interface has no access specifier.

4. The overridden methods in the three classes have public access specifier.

5. No additional variables have been declared.

6. The test class having the main() method is declared public.

7. The area of the rectangle, triangle and the cylinder have been computed as per the respective formulae.

8. The interface is similar to a class which can have only declarations of both, variables and methods. No method can be defined inside an interface.

9. The other classes use the methods of the interface by implementing the interface using the keyword, implements.

10. The object is created using the name of the interface as shown.

ObjectWithTwoParameters r = new RectangleClass();

Answer:

class Main {

  public static void main(String[] args) {

      char arr[] = {'T','E','D','R','W','B','S','V','A'};

      int n = arr.length;

      System.out.println("Selection Sort:");

      System.out.println("Iteration\tArray\tComparisons");

      long comp1 = selectionSort(arr);

      System.out.println("Total comparisons: "+comp1);

      System.out.println("\nInsertion Sort:");

      System.out.println("Iteration\tArray\tComparisons");

      long comp2 = insertionSort(arr);

      System.out.println("Total Comparisons: "+comp2);

      System.out.println("\nOverall Total Comparisons: "+(comp1+comp2));

  }

  static long selectionSort(char arr[]) {

      // applies selection sort for n/2 elements

      // returns number of comparisons

      int n = arr.length;

      long comparisons = 0;

      // One by one move boundary of unsorted subarray

      for (int i = n-1; i>=n-n/2; i--) {

              // Find the minimum element in unsorted array

              int max_idx = i;

              for (int j = i-1; j>=0; j--) {

                      // there is a comparison everytime this loop returns

                      comparisons++;

                      if (arr[j] > arr[max_idx])

                              max_idx = j;

              }

              // Swap the found minimum element with the first

              // element

              char temp = arr[max_idx];

              arr[max_idx] = arr[i];

              arr[i] = temp;

              System.out.print(n-1-i+"\t");

              printArray(arr);

              System.out.println("\t"+comparisons);

      }

      return comparisons;

  }

  static long insertionSort(char arr[]) {

      // applies insertion sort for n/2 elements

      // returns number of comparisons

      int n = arr.length;

      n = n-n/2;   // sort only the first n/2 elements

      long comparisons = 0;

      for (int i = 1; i < n; ++i) {

          char key = arr[i];

          int j = i - 1;

          /* Move elements of arr[0..i-1], that are

                  greater than key, to one position ahead

                  of their current position */

          while (j >= 0) {

              // there is a comparison everytime this loop runs

              comparisons++;

              if (arr[j] > key) {

                  arr[j + 1] = arr[j];

              } else {

                  break;

              }

              j--;

          }

          arr[j + 1] = key;

          System.out.print(i-1+"\t");

          printArray(arr);

          System.out.println("\t"+comparisons);

      }

      return comparisons;

  }  

  static void printArray(char arr[]) {

      for (int i=0; i<arr.length; i++)

          System.out.print(arr[i]+" ");

  }

}

Explanation:

Explanation is in the answer.

Answer:

Check the explanation

Explanation:

SDLC

SDLC stands for Software Development Life Cycle

It provides steps for developing a software product

It also checks for the quality and correctness of a software

The main aim of SDLC is to develop a software, which meets a customer requirements

SDLC involves various phases to develop a high-quality product for its end user

Phases of SDLC

           There are seven phases in a software development life cycle as follows,

Requirement Analysis   Feasibility Study   Design    Coding    Testing Deployment      Maintenance

kindly check the attached image below

Requirement analysis

This is the first stage in a SDLC.

In this phase, a detailed and precise requirements are collected from various teams by senior team members

It gives a clear idea of the entire project scope, opportunities and upcoming issues in a project

This phase also involves, planning assurance requirements and analyze risk involved in the project

Here, the timeline to finish the project is finalized

Feasibility Study

In this stage, the SRS document, that is “Software Requirement Specification” document is defined, which includes everything to be designed and developed during the life cycle of a project

Design

In this phase as the name indicates, software and system design documents are prepared, based on the requirement specification document

This enable us to define the whole system architecture

Two types of design documents are prepared in this phase as follows,

High-Level Design

Low-Level Design

Coding

In this phase, the coding of project is done in a selected programming language

Tasks are divided into many units and given to various developers

It is the longest phase in the SDLC

Testing

In this stage, the developed coding is tested by the testing team

The testing team, checks the coding in the user point of view

Installation/Deployment

In this stage, a product is released by the company to the client

Maintenance

Once the product is delivered to the client, the maintenance phase work will start by fixing errors, modifying and enhancing the product.

Employee Payroll System

Requirement analysis

Gather information from various team for requirement

Analyze the number of employees to handle the project

Timeline to finish the project, eg.1 month

Feasibility Study

The system requirements such as how many systems are required

Decide which software to be installed to handle the project.

For example, if we are going to develop the software using C language, then software for that has to be installed

The SRS document has to be prepared

Design

In this the HLD and LLD is prepared, the overall software architecture is prepared

The modules involved in coding are decided

Here the modules can be employee, payroll manager, Employee Log time, account details

The input and output are designed, such as employee name is the input and output is the salary for him, based on working hours

Coding

Here the coding is divided into various sections and given to 2 or more employees

One may code employee detail, one will code working hours of employees and one may code the banking details of employee

Testing

The coding is tested for syntax, declaration, data types and various kinds of error

The testing team will test the coding with various possible values

They may even check with wrong values and analyze the output for that

Installation

Now the software is installed in the client system and the client is calculating the payroll for an employee based on working hours in a month

Maintenance

If any error occurs, the team will clear the issue.

When a new employee joins, then the employee data will added to the database and the database is updated

Also if the client asks for any new features, it will done in this phase.

Explanation:

See the attached image for The interface (.h file) of a class Counter

Answer:

see explaination

Explanation:

#include <iostream>

#include <string>

#include <vector>

using namespace std;

vector<string> split(string, string);

int main()

{

vector<string> splitedStr;

string data;

string delimiter;

cout << "Enter string to split:" << endl;

getline(cin,data);

cout << "Enter delimiter string:" << endl;

getline(cin,delimiter);

splitedStr = split(data,delimiter);

cout << "\n";

cout << "The substrings are: ";

for(int i = 0; i < splitedStr.size(); i++)

cout << "\"" << splitedStr[i] << "\"" << ",";

cout << endl << endl;

cin >> data;

return 0;

}

vector<string> split(string target, string delimiter)

{

unsigned first = 0;

unsigned last;

vector<string> subStr;

while((last = target.find(delimiter, first)) != string::npos)

{

subStr.push_back(target.substr(first, last-first));

first = last + delimiter.length();

}

subStr.push_back(target.substr(first));

return subStr;

}

Here is the complete question.

In this lab, you use the flowchart and pseudocode found in the figures below to add code to a partially created C++ program. When completed, college admissions officers should be able to use the C++ program to determine whether to accept or reject a student, based on his or her test score and class rank.

start input testScore,

classRank if testScore >= 90 then if classRank >= 25 then output "Accept"

else output "Reject" endif else if testScore >= 80

then if classRank >= 50 then output "Accept" else output "Reject" endif

else if testScore >= 70

then if classRank >= 75 then output "Accept"

else output "Reject"

endif else output "Reject"

endif

endif

endif

stop

Study the pseudocode in picture above. Write the interactive input statements to retrieve: A student’s test score (testScore) A student's class rank (classRank) The rest of the program is written for you. Execute the program by clicking "Run Code." Enter 87 for the test score and 60 for the class rank. Execute the program by entering 60 for the test score and 87 for the class rank.

[comment]: <> (3. Write the statements to convert the string representation of a student’s test score and class rank to the integer data type (testScore and classRank, respectively).)

Function: This program determines if a student will be admitted or rejected. Input: Interactive Output: Accept or Reject

*/ #include using namespace std; int main() { // Declare variables

// Prompt for and get user input

// Test using admission requirements and print Accept or Reject

if(testScore >= 90)

{ if(classRank >= 25)

{ cout << "Accept" << endl; }

else

cout << "Reject" << endl; }

else { if(testScore >= 80)

{ if(classRank >= 50)

cout << "Accept" << endl;

else cout << "Reject" << endl; }

else { if(testScore >= 70)

{ if(classRank >=75) cout << "Accept" << endl;

else cout << "Reject" << endl; }

else cout << "Reject" << endl; } } } //End of main() function

Answer:

Explanation:

The objective here is to use the flowchart and pseudocode found in the figures below to add code to a partially created C++ program. When completed, college admissions officers should be able to use the C++ program to determine whether to accept or reject a student, based on his or her test score and class rank.

PROGRAM:

#include<iostream>

using namespace std;

int main(){

// Declare variables

int testScore, classRank;

// Prompt for and get user input

cout<<"Enter test score: ";

cin>>testScore;

cout<<"Enter class rank: ";

cin>>classRank;

// Test using admission requirements and print Accept or Reject

if(testScore >= 90)  

{ if(classRank >= 25)

{ cout << "Accept" << endl; }

else

cout << "Reject" << endl;

}

else { if(testScore >= 80)

{ if(classRank >= 50)

cout << "Accept" << endl;

else cout << "Reject" << endl; }

else { if(testScore >= 70)

{ if(classRank >=75) cout << "Accept" << endl;

else cout << "Reject" << endl;

}

else cout << "Reject" << endl; }

}

return 0;

} //End of main() function

OUTPUT:

See the attached file below:

Answer:

Following are the code to this question:

def pairSum(a,x): #find size of list

   s=len(a) # use length function to calculte length of list and store in s variable

   for x1 in range(0, s):  #outer loop to count all list value

       for x2 in range(x1+1, s):    #inner loop

           if a[x1] + a[x2] == x:    #condition check

               print(x1," ",x2) #print value of loop x1 and x2  

pairSum([12,7,8,6,1,13],13) #calling pairSum method

Output:

0   4

1   3

Explanation:

Description of the above python can be described as follows:

Answer:

check if its pluged in

Explanation:

Answer:

See explaination

Explanation:

//ReadFile.java

import java.io.File;

import java.io.FileNotFoundException;

import java.util.Scanner;

public class ReadFile

{

public static void main(String[] args)

{

int elementsCount=0;

//Create a File class object

File file=null;

Scanner fileScanner=null;

String fileName="sample.txt";

try

{

//Create an instance of File class

file=new File(fileName);

//create a scanner class object

fileScanner=new Scanner(file);

//read file elements until end of file

while (fileScanner.hasNext())

{

double value=fileScanner.nextInt();

elementsCount++;

}

//print smallest value

System.out.println(elementsCount+" data values are read");

}

//Catch the exception if file not found

catch (FileNotFoundException e)

{

System.out.println("File Not Found");

}

}

}

Sample output:

sample.txt

10

20

30

40

50

output:

5 data values are read

Answer:

(a) the number of times the value is performs is up to four cycles. and as such the integer i is executed up to 5 times.  (b)The point version of the floating point can have CPE of 3.00, even when the multiplication operation required is either 4 or 5 clock.

Explanation:

Solution

The two floating point versions can have CPEs of 3.00, even though the multiplication operation demands either 4 or 5 clock cycles by the latency suggests the total number of clock cycles needed to work the actual operation, while issues time to specify the minimum number of cycles between operations.

Now,

sum = sum + udata[i] * vdata[i]

in this case, the value of i performs from 0 to 3.

Thus,

The value of sum is denoted as,

sum = ((((sum + udata[0] * vdata[0])+(udata[1] * vdata[1]))+( udata[2] * vdata[2]))+(udata[3] * vdata[3]))

Thus,

(A)The number of times the value is executed is up to 4 cycle. And the integer i performed up to 5 times.

Thus,

(B) The floating point version can have CPE of 3.00, even though the multiplication operation required either 4 or 5 clock.

Answer:

Check the explanation

Explanation:

--Query 1)

SELECT ename, sal, age

FROM Emp;

--Query 2)

SELECT did

FROM Dept

WHERE floot = 10 AND budget<15000;

Answer:

Check the explanation

Explanation:

solution a:

def Rotate(string) :

n = len(string)

temp = string + string

for i in range(n) :

for j in range(n) :

print(temp[i + j], end = "")

print()

string = ("abcde")

Rotate(string)

solution b:

nums = [3,9,5,8,2,4,7]

res = list(filter(lambda n : n < 5, nums))

print(res)

Answer:

Hewo, Here are some ways in which apps earn money :-

hope it helps!

Answer:

Pay raise is an example of the rewards & consequences characteristic of an organization.

Answer: (a). 11.3137

(b). 22.849

Explanation:

Provided below is a step by step analysis to solving this problem

(a)

clc;close all;clear all;

a=2;x=3.5;

E=10;n=0;k=1;sn1=0;

while E >0.000001

cn=((log(a))^n)*(x^n)/factorial(n);

sn=sn1+cn;

E=abs((sn-sn1)/sn1);

sn1=sn;

n=n+1;

k=k+1;

end

fprintf('2^3.5 from tailor series=%6.4f after adding n=%d terms\n',sn,n);

2^3.5 from tailor series=11.3137 after adding n=15 terms

disp('2^3.5 using calculator =11.3137085');

Command window:

2^3.5 from tailor series=11.3137 after adding n=15 terms

2^3.5 using calculator =11.3137085

(b)

clc;close all;clear all;

a=6.3;x=1.7;

E=10;n=0;k=1;sn1=0;

while E >0.000001

cn=((log(a))^n)*(x^n)/factorial(n);

sn=sn1+cn;

E=abs((sn-sn1)/sn1);

sn1=sn;

n=n+1;

k=k+1;

end

fprintf('6.3^1.7 from tailor series=%6.4f after adding n=%d terms\n',sn,n);

disp('6.3^1.7 using calculator =22.84961748');

Command window:

6.3^1.7 from tailor series=22.8496 after adding n=16 terms

6.3^1.7 using calculator =22.84961748

cheers i hope this helped !!!