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Save Write a program to implement Binary search using recursive algorithm. Submit a screenshot of your Java source codes and IDE/Comment line results in PDF file. Solution BinarySearchTest.java import java.util.Scanner; public class BinarySearchTest { public static void main(String[] args) { int[] arr1 = { 1,2,3,4,5,6,7,8,9,10}; Scanner scan = new Scanner(System.in); System.out.println(\"Enter the key: \"); int key = scan.nextInt(); int index = binarySearch(arr1,0,arr1.length,key); System.out.println(\"Index is \"+index); } public static int binarySearch(int[] array, int a, int e, int key) { if (a < e) { int mid = a + (e - a) / 2; if (key < array[mid]) { return binarySearch(array, a, mid, key); } else if (key > array[mid]) { return binarySearch(array, mid+1, e , key); } else { return mid; } } return -1; } } Output: Enter the key: 4 Index is 3.
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![Save Write a program to implement Binary search using recursive algorithm. Submit a
screenshot of your Java source codes and IDE/Comment line results in PDF file.
Solution
BinarySearchTest.java
import java.util.Scanner;
public class BinarySearchTest {
public static void main(String[] args) {
int[] arr1 = { 1,2,3,4,5,6,7,8,9,10};
Scanner scan = new Scanner(System.in);
System.out.println("Enter the key: ");
int key = scan.nextInt();
int index = binarySearch(arr1,0,arr1.length,key);
System.out.println("Index is "+index);
}
public static int binarySearch(int[] array, int a, int e, int key) {
if (a < e) {
int mid = a + (e - a) / 2;
if (key < array[mid]) {
return binarySearch(array, a, mid, key);
} else if (key > array[mid]) {
return binarySearch(array, mid+1, e , key);
} else {
return mid;
}
}
return -1;
}
}
Output:
Enter the key:
4](https://image.slidesharecdn.com/savewriteaprogramtoimplementbinarysearchusingrecursivealgo-230702161904-4586554a/85/Save-Write-a-program-to-implement-Binary-search-using-recursive-algo-pdf-1-320.jpg)
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import java.awt.LinearGradientPaint; public class searchComparis.pdf
import java.awt.LinearGradientPaint;
public class searchComparision {
/**
* @param args
*/
public static void linearSearch(int arr[],int n)
{
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{
break;
}
}
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}
static void binarySearch(int arr[],int start,int end,int n)
{
if(endn)
{
binarySearch(arr, start,mid-1,n);
}
else
{
binarySearch(arr,mid+1,end,n);
}
return;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int arr[] = {4,8,12,23,54,84,89,110};
int n=8;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=33;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=84;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
}
}
Solution
import java.awt.LinearGradientPaint;
public class searchComparision {
/**
* @param args
*/
public static void linearSearch(int arr[],int n)
{
for(int i =0;i=n)
{
break;
}
}
System.out.println(\"\");
}
static void binarySearch(int arr[],int start,int end,int n)
{
if(endn)
{
binarySearch(arr, start,mid-1,n);
}
else
{
binarySearch(arr,mid+1,end,n);
}
return;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int arr[] = {4,8,12,23,54,84,89,110};
int n=8;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=33;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=84;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
}
}.
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import java.awt.LinearGradientPaint; public class searchComparis.pdf
import java.awt.LinearGradientPaint;
public class searchComparision {
/**
* @param args
*/
public static void linearSearch(int arr[],int n)
{
for(int i =0;i=n)
{
break;
}
}
System.out.println(\"\");
}
static void binarySearch(int arr[],int start,int end,int n)
{
if(endn)
{
binarySearch(arr, start,mid-1,n);
}
else
{
binarySearch(arr,mid+1,end,n);
}
return;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int arr[] = {4,8,12,23,54,84,89,110};
int n=8;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
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linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=84;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
}
}
Solution
import java.awt.LinearGradientPaint;
public class searchComparision {
/**
* @param args
*/
public static void linearSearch(int arr[],int n)
{
for(int i =0;i=n)
{
break;
}
}
System.out.println(\"\");
}
static void binarySearch(int arr[],int start,int end,int n)
{
if(endn)
{
binarySearch(arr, start,mid-1,n);
}
else
{
binarySearch(arr,mid+1,end,n);
}
return;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int arr[] = {4,8,12,23,54,84,89,110};
int n=8;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=33;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
n=84;
linearSearch(arr,n);
binarySearch(arr,0,arr.length-1,n);
}
}.
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// Average.java
import java.util.Scanner;
import java.util.ArrayList;
public class Average
{
private double mean;
private int [] data;
public Average()
{
Scanner scan = new Scanner(System.in);
//Create an array with 5 elements.
data = new int[5];
//input the data into the array
for (int i = 0; i < data.length; i++)
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{
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}
}
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output:
Enter your score for test #1: 87
Enter your score for test #2: 90
Enter your score for test #3: 92
Enter your score for test #4: 77
Enter your score for test #5: 89
The data stored in array: 87 90 92 77 89
Average: 87.0
*/
Solution
// Average.java
import java.util.Scanner;
import java.util.ArrayList;
public class Average
{
private double mean;
private int [] data;
public Average()
{
Scanner scan = new Scanner(System.in);
//Create an array with 5 elements.
data = new int[5];
//input the data into the array
for (int i = 0; i < data.length; i++)
{
System.out.print(\"Enter your score for test #\" + (i +1) + \": \");
data[i] = scan.nextInt();
}
}
public void calculateMean()
{
double sum = 0; //Accumulator
//calculate the sum of the test scores in the data array.
for (int i = 0; i < data.length; i++)
sum += data[i];
mean = (Double)sum/data.length;
}
public String toString()
{
String result = \"\ The data stored in array: \";
for (int i = 0; i < data.length; i++)
result = result + data[i] + \" \";
result = result + \"\ \" + \"Average: \" + mean + \"\ \";
return result;
}
}
// AverageDriver.java
public class AverageDriver
{
public static void main(String[] args)
{
Average avg = new Average();
avg.calculateMean();
System.out.println(avg.toString());
}
}
/*
output:
Enter your score for test #1: 87
Enter your score for test #2: 90
Enter your score for test #3: 92
Enter your score for test #4: 77
Enter your score for test #5: 89
The data stored in array: 87 90 92 77 89
Average: 87.0
*/.
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import java.util.Scanner;
/**
* @author
*
*/
public class SelectionSort {
/**
* @param args
*/
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// TODO Auto-generated method stub
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}
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Enter 12 integers.
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Before Sorting:
4 23 65 34 84 37 12 17 24 36 82 51
After Sorting:
4 12 17 23 24 34 36 37 51 65 82 84
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import java.util.Scanner;
import java.util.ArrayList;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.ObjectInputStream;
import java.io.ObjectOutputStream;
import java.util.ArrayList;
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}
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}catch(IOException e){
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return;
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}
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if(c == \'C\') return \'G\';
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public void DrawDNA(String seq){
System.out.println(\"---+-----+------+------+------\");
for (char c : seq.toCharArray() ) {
System.out.print(c + \" \");
}
System.out.println(\"DNA Sequence \");
System.out.println(\"................................\");
for (char c : seq.toCharArray() ) {
System.out.print(compliment_base(c) + \" \");
}
System.out.println(\" Complement \");
System.out.println(\"---+-----+------+------+------\");
System.out.println(\"\");
}
public void Prefix(int n, String base){
String output = base.substring(0,n);
DrawDNA(output);
}
public static void main(String[] args) {
DNA d = new DNA();
d.DrawDNA(\"ATGC\");
d.Prefix(4,\"ACGTTGCA\");
}
}
---+-----+------+------+------
A T G C DNA Sequence
................................
T A C G Complement
---+-----+------+------+------
---+-----+------+------+------
A C G T DNA Sequence
................................
T G C A Complement
---+-----+------+------+------.
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B. Tendons surround the epimysium.
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Explanation:
At the ends of a muscle, the fibers of the Epimysium may form a cable-like tendon or a sheet-
like aponeurosis..
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A. RNA uses only purines
B. RNA uses a different five-carbon sugar
C. RNA has a contains different sized phosphate groups
D. RNA has multiple bases attached to the sugar
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The document discusses the new features and certification process for Java SE 7. It describes that the certification has been updated to require more advanced skills. There are now two certification levels: Oracle Certified Associate, Java SE 7 Programmer (OCA) and Oracle Certified Professional, Java SE 7 Programmer (OCP). The OCA exam covers basic Java topics in 90 multiple choice questions over 140 minutes. The OCP exam requires passing the OCA first and covers more advanced topics in 90 questions over 150 minutes. Developers with prior certifications can take an upgrade exam to certify for Java SE 7.
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The document provides the top 20 Java programming interview questions for software development engineers in test (SDETs). It lists questions such as writing a program to find the second most frequent character in a string, reversing a string, sorting strings in lexicographical order, replacing a substring with another string, and checking if a string is a palindrome. It also provides sample code solutions for each question in Java and the expected output.
Static code analysis: what? how? why?
From code review to static code analysis. Static code analysis techniques. Some examples of defects in real projects. C, C++, C#, Java.
Java Simple Programs
The document contains code snippets from 3 weekly coding assignments:
1) A Java program to check if a string is a palindrome. It compares characters at the beginning and end of the string.
2) A Java program to sort a list of names in ascending order using string comparison and swapping.
3) A Java program to count the frequency of words in a given text by tokenizing, sorting, and printing the words.
arrays
This document contains sections on arrays in C programming including: declaring and initializing arrays; passing arrays to functions; sorting arrays; and computing statistics like mean, median, and mode from array data. Key points include: arrays allow grouping related data under one name; elements are accessed via subscript notation; arrays can be passed to functions by reference, allowing the function to modify the original array; sorting algorithms like bubble sort rearrange array elements into order; and common statistics like average, middle value, and most frequent value can be calculated from array data.
Static analysis: Around Java in 60 minutes
Theory
Code quality (bugs, vulnerabilities)
Methodologies of code protection against defects
Code Review
Static analysis and everything related to it
Tools
Existing tools of static analysis
SonarQube
PVS-Studio for Java what is it?
Several detected examples of code with defects
More about static analysis
Conclusions
Java Language fundamental
Java Language Fundamental, command line argument, control statement - if else, switch, loop, array etc.
Workshop "Can my .NET application use less CPU / RAM?", Yevhen Tatarynov
In most cases it’s very hard to predict the number of resources needed for your .NET application. But If you spot some abnormal CPU or RAM usage, how to answer the question “Can my application use less?”.
Let’s see samples from real projects, where optimal resource usage by the application became one of the values for the product owner and see how less resource consumption can be.
The workshop will be actual for .NET developers who are interested in optimization of .NET applications, QA engineers who involved performance testing of .NET applications. It also will be interesting to everyone who "suspected" their .NET applications of non-optimal use of resources, but for some reason did not start an investigation.
6_Array.pptx
This document provides information on arrays in Java. It begins by defining an array as a collection of similar data types that can store values of a homogeneous type. Arrays must specify their size at declaration and use zero-based indexing. The document then discusses single dimensional arrays, how to declare and initialize them, and how to set and access array elements. It also covers multi-dimensional arrays, providing syntax for declaration and initialization. Examples are given for creating, initializing, accessing, and printing array elements. The document concludes with examples of searching arrays and performing operations on two-dimensional arrays like matrix addition and multiplication.
ASP.NET
The document provides an overview of ASP.NET and C# programming concepts:
- It describes the .NET framework architecture including CLR, CLS, and CTS.
- It explains that C# is an object-oriented language developed by Microsoft to be used with the .NET framework.
- It provides examples of basic C# programs demonstrating variables, data types, conditional statements, loops, arrays, object-oriented concepts like encapsulation, inheritance and abstraction.
compiler design
This document provides an overview of compiler design and the first lecture in a compiler design course. It discusses the main components of a compiler including the lexical analyzer, parser, semantic analyzer, intermediate code generator, and target code generator. It provides examples of how a sample Java program is compiled to bytecode. The document also discusses the phases a compiler goes through to analyze, transform, and generate code from a source program. It notes that compilers are typically written in a high-level language and sometimes the same language being compiled. It outlines the course administration including a group compiler project, individual test, and course URL.
Unit 3
The document discusses arrays, strings, and functions in C programming. It begins by explaining how to initialize and access 2D arrays, including examples to take input from the user and store it in a 2D array. It then discusses initializing and accessing multidimensional arrays. Next, it covers array contiguous memory and the advantages and limitations of arrays. Finally, it discusses common programming errors related to array construction for real-time applications.
Average.javaimport java.util.Scanner; import java.util.Arra.pdf
// Average.java
import java.util.Scanner;
import java.util.ArrayList;
public class Average
{
private double mean;
private int [] data;
public Average()
{
Scanner scan = new Scanner(System.in);
//Create an array with 5 elements.
data = new int[5];
//input the data into the array
for (int i = 0; i < data.length; i++)
{
System.out.print(\"Enter your score for test #\" + (i +1) + \": \");
data[i] = scan.nextInt();
}
}
public void calculateMean()
{
double sum = 0; //Accumulator
//calculate the sum of the test scores in the data array.
for (int i = 0; i < data.length; i++)
sum += data[i];
mean = (Double)sum/data.length;
}
public String toString()
{
String result = \"\ The data stored in array: \";
for (int i = 0; i < data.length; i++)
result = result + data[i] + \" \";
result = result + \"\ \" + \"Average: \" + mean + \"\ \";
return result;
}
}
// AverageDriver.java
public class AverageDriver
{
public static void main(String[] args)
{
Average avg = new Average();
avg.calculateMean();
System.out.println(avg.toString());
}
}
/*
output:
Enter your score for test #1: 87
Enter your score for test #2: 90
Enter your score for test #3: 92
Enter your score for test #4: 77
Enter your score for test #5: 89
The data stored in array: 87 90 92 77 89
Average: 87.0
*/
Solution
// Average.java
import java.util.Scanner;
import java.util.ArrayList;
public class Average
{
private double mean;
private int [] data;
public Average()
{
Scanner scan = new Scanner(System.in);
//Create an array with 5 elements.
data = new int[5];
//input the data into the array
for (int i = 0; i < data.length; i++)
{
System.out.print(\"Enter your score for test #\" + (i +1) + \": \");
data[i] = scan.nextInt();
}
}
public void calculateMean()
{
double sum = 0; //Accumulator
//calculate the sum of the test scores in the data array.
for (int i = 0; i < data.length; i++)
sum += data[i];
mean = (Double)sum/data.length;
}
public String toString()
{
String result = \"\ The data stored in array: \";
for (int i = 0; i < data.length; i++)
result = result + data[i] + \" \";
result = result + \"\ \" + \"Average: \" + mean + \"\ \";
return result;
}
}
// AverageDriver.java
public class AverageDriver
{
public static void main(String[] args)
{
Average avg = new Average();
avg.calculateMean();
System.out.println(avg.toString());
}
}
/*
output:
Enter your score for test #1: 87
Enter your score for test #2: 90
Enter your score for test #3: 92
Enter your score for test #4: 77
Enter your score for test #5: 89
The data stored in array: 87 90 92 77 89
Average: 87.0
*/.
Using xUnit as a Swiss-Aarmy Testing Toolkit
Modern Unit Testing practices act as a conduit for improved software designs that are more amenable to change and can be easily backed by automation for fast feedback on quality assurance. The necessity of reducing external dependencies forces us to design our modules with minimum coupling which can then be leveraged both at the module, component and subsystem levels in our testing. As we start to integrate our units into larger blocks and interface our resulting components with external systems we find ourselves switching nomenclature as we progress from Unit to Integration testing. But is a change in mindset and tooling really required?
The xUnit testing framework is commonly perceived as an aid to Unit Testing but the constraints that it imposes on the architecture mean that it is an excellent mechanism for invoking arbitrary code in a restricted context. Tests can be partitioned by categorisation at the test and fixture level and through physical packaging leading to a flexible test code structure. Throw in its huge popularity and you have a simplified learning curve for expressing more that just unit tests.
Using scenarios from his current system Chris aims to show how you can use a similar format and tooling for unit, component and integration level tests; albeit with a few liberties taken to work around the inherent differences with each methodology.
Java final lab
The document contains a list of 16 experiments to be performed in Java programming. It includes programs to check Armstrong numbers, sort strings, multiply matrices, calculate volume and surface area of a box, use copy constructors, create abstract classes for shapes and inherit subclasses, handle exceptions, create custom exceptions, create and run threads, use network classes, handle mouse events using applets, implement remote methods using RMI, create servlets, perform student mark list processing using JDBC, and create a basic text editor. It also provides sample code for 10 programs related to number swapping, multiplication tables, checking Armstrong numbers, calculating sum, checking palindromes, reversing numbers, finding largest of 3 numbers, printing prime numbers, Floyd
C# p9
The document discusses various aspects of arrays in C# including declaration, initialization, storing values, accessing values, passing arrays to methods, array properties and functions. It provides examples of one-dimensional and multi-dimensional arrays and covers topics like declaration, initialization, accessing values, passing arrays to methods and common array properties and functions.
とある断片の超動的言語
This document contains links and code snippets related to Groovy and GitHub Gists. It includes examples of using Groovy for tasks like playing MIDI notes, generating musical scores, adding nodes to a Jenkins server, and calculating factorials in different styles from a newbie to an expert programmer. The document also references a Greasemonkey script for running a Groovy web console in Firefox.
(JAVA NetBeans) Write a Java program able to perform selection sort-So.docx
(JAVA NetBeans) Write a Java program able to perform selection sort.
Solution
import java.util.Scanner;
/**
* @author
*
*/
public class SelectionSort {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int arr[] = setArray();
System.out.println(\"\ Before Sorting:\");
printArray(arr);
arr = selectionSort(arr);
System.out.println(\"\ After Sorting:\");
printArray(arr);
}
/**
* reads and stores the data in the one-dimensional array
*/
/**
* @return
*/
public static int[] setArray() {
int arr[] = new int[12];
System.out.println(\"Enter 12 integers.\");
Scanner scanner = new Scanner(System.in);
for (int i = 0; i < arr.length; i++) {
arr[i] = scanner.nextInt();
}
return arr;
}
/**
* prints the data in the one-dimensional array
*/
public static void printArray(int[] arr) {
System.out.println();
for (int i = 0; i < arr.length; i++) {
System.out.print(arr[i] + \" \");
}
System.out.println();
}
/**
* method will sort the array in ascending order
*/
public static int[] selectionSort(int[] arr) {
for (int i = 0; i < arr.length - 1; i++) {
int index = i;
for (int j = i + 1; j < arr.length; j++)
if (arr[j] < arr[index])
index = j;
int smallerNumber = arr[index];
arr[index] = arr[i];
arr[i] = smallerNumber;
}
return arr;
}
}
OUTPUT:
Enter 12 integers.
4 23 65 34 84 37 12 17 24 36 82 51
Before Sorting:
4 23 65 34 84 37 12 17 24 36 82 51
After Sorting:
4 12 17 23 24 34 36 37 51 65 82 84
.
TestR: generating unit tests for R internals
TestR is a tool for automatically generating unit tests for R internals written in C. It works by instrumenting GNU-R to capture calls to built-in functions along with their arguments and return values. These calls are then used to generate test cases. Test cases are filtered to only include those that increase code coverage as measured by gcov. Experimental results found TestR was able to generate a test suite covering 80% of the GNU R test suite with only 4695 tests, and also uncovered errors in other R virtual machines.
Use arrays to store data for analysis. Use functions to perform the .pdf
Use arrays to store data for analysis. Use functions to perform the analysis described below. All
output is to be written to an output file.
Solution
import java.util.Scanner;
import java.util.ArrayList;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.ObjectInputStream;
import java.io.ObjectOutputStream;
import java.util.ArrayList;
public class DNA{
private String dna;
public DNA(){
}
public void read_store(String filename){
/*try {
Scanner in;
//in = new Scanner(new FileReader(filename));
dna = in.next();
}catch(IOException e){
System.out.println(\"file not there \" + e);
return;
}*/
}
public char compliment_base(char c){
if(c == \'A\') return \'T\';
if(c == \'T\') return \'A\';
if(c == \'G\') return \'C\';
if(c == \'C\') return \'G\';
return \' \';
}
public void DrawDNA(String seq){
System.out.println(\"---+-----+------+------+------\");
for (char c : seq.toCharArray() ) {
System.out.print(c + \" \");
}
System.out.println(\"DNA Sequence \");
System.out.println(\"................................\");
for (char c : seq.toCharArray() ) {
System.out.print(compliment_base(c) + \" \");
}
System.out.println(\" Complement \");
System.out.println(\"---+-----+------+------+------\");
System.out.println(\"\");
}
public void Prefix(int n, String base){
String output = base.substring(0,n);
DrawDNA(output);
}
public static void main(String[] args) {
DNA d = new DNA();
d.DrawDNA(\"ATGC\");
d.Prefix(4,\"ACGTTGCA\");
}
}
---+-----+------+------+------
A T G C DNA Sequence
................................
T A C G Complement
---+-----+------+------+------
---+-----+------+------+------
A C G T DNA Sequence
................................
T G C A Complement
---+-----+------+------+------.
Similar to Save Write a program to implement Binary search using recursive algo.pdf (20)
Top 20 java programming interview questions for sdet
Top 20 java programming interview questions for sdet
Workshop "Can my .NET application use less CPU / RAM?", Yevhen Tatarynov
Workshop "Can my .NET application use less CPU / RAM?", Yevhen Tatarynov
Average.javaimport java.util.Scanner; import java.util.Arra.pdf
Average.javaimport java.util.Scanner; import java.util.Arra.pdf
(JAVA NetBeans) Write a Java program able to perform selection sort-So.docx
(JAVA NetBeans) Write a Java program able to perform selection sort-So.docx
Use arrays to store data for analysis. Use functions to perform the .pdf
Use arrays to store data for analysis. Use functions to perform the .pdf
More from arihantmobileselepun
How are the epimysium and tendons relatedA. The epimysium surroun.pdf
How are the epimysium and tendons related?
A. The epimysium surrounds the tendon.
B. Tendons surround the epimysium.
C. Tendons and epimysium both surround the end of the muscle.
D. The tendon is a continuation of the epimysium.
Solution
Answer:
D. The tendon is a continuation of the epimysium.
Explanation:
At the ends of a muscle, the fibers of the Epimysium may form a cable-like tendon or a sheet-
like aponeurosis..
How does DNA differ from RNAA. RNA uses only purinesB. RNA uses.pdf
How does DNA differ from RNA?
A. RNA uses only purines
B. RNA uses a different five-carbon sugar
C. RNA has a contains different sized phosphate groups
D. RNA has multiple bases attached to the sugar
Solution
B. RNA uses a different five-carbon sugar. RNA has the ribose pentose; whereas in DNA, the
pentose is deoxyribose..
Gerard throws a discus distances of 10 meters, 14.5 meters, 14.8 met.pdf
Gerard throws a discus distances of 10 meters, 14.5 meters, 14.8 meters, 16.7 meters, and 15.4
meters. On his last attempt he faults, which counts as 0 meters. What is his average throwing
distance?
A. 11.9 meters
B. 12.8 meters
C. 13.78 meters
D. 14.28 meters
Solution
A. To find the average, add the numbers and divide by the number of numbers. In this problem
add 10, 14.5, 14.8, 16.7, 15.4, and 0 to get 71.4. Divide this by 6.
71.4/6 = 11.9.
For Printing and personal use ONLY. DO NOT Hand in this copy of the a.pdf
For Printing and personal use ONLY. DO NOT Hand in this copy of the assignment. Only
responses submitted to the corresponding D2L quiz will be graded. The D2L version is exactly
like this document. Membrane proteins are proteins that are suspended within the lipid bilayer of
a membrane. They are held in place because some R-groups of the protein interact with the
phosphate heads of the phospholipids of the membrane, while other R-groups of the protein
interact with the hydrocarbon tails of the phospholipids of the membrane. From the amino acids
shown below, which one likely has an R-group that interacts with the phopholipid tails? If a cell
has some salt inside it but lacks ion channels, and the cell is put into a beaker of pure water, what
will happen?
Solution
Leucine (green color) is most likely to be part of the protein that interact with the hydrocarbon
chain of phospholipid in the membrane (since it is hydrophobic amino acid).
Find and provide a link to a multi-year capital plan from any Illino.pdf
Find and provide a link to a multi-year capital plan from any Illinois municipality. Discuss the
presentation of the plan (i.e. - is there detail about specific projects, funding sources, etc) and
how you think it could be improved.
Solution
The Link can\'t be provided so you can find it in the document centre of the village of the round
lake beach and the multi year capital plan is for the years from 2015- 2024.
The plan is presented in detail proving the names of the specific projects that would be
undertaken by the village and all the ways in which they will be collecting the funds needed. The
plan talks about the following specific projects:
The funding is alot dependent upon the taxes from the people inhabiting the place and a very
small amount has been estimated as allocations from outside. The plan is highly unrealistic as
there may be instances where the people residing the place may not choose to pay so much or
simply not buy the bonds they are expecting to issue..
A geneticist is working with a new bacteriophage called phage Y3 that.pdf
A geneticist is working with a new bacteriophage called phage Y3 that infects E. coli. He has
isolated eight mutant phages that fail to produce plaques when grown on E. coli strain K. To
determine whether these mutations occur at the same functional gene, he simultaneously infects
E. coli K with paired combination of the mutants and looks to see whether plaques are formed.
He obtains the following results. (A plus sign means that plaques were formed on E. coll K; a
minus sign means that no plaques were formed on E. coli K.) To how many functional genes
(cistrons) do these mutations belong? 5 1 8 2 7 3
Solution
Answer: Correct answer: 3
Reason: The geneticist has conducted the transformation experiment in which “trans” portion of
the cis–trans test to determine the complementation between the different phage mutants infected
with E. coli K cells. Finally, these cells were looked whether the plaques are generated in the
plates. It can be seen in the above results table, in which plus sign means that plaques were
formed due to complimentaion in 3 mutants (5, 6 & 8) have meticulously specific defects in the
same functional so that on E. coli K; a minus sign means that no plaques. Therefore, it can be
concluded that complementation can happen when the mutations of the different phages come
together in the sample as part of transformation.
Assume that the four living species in the figure below evolved from.pdf
Assume that the four living species in the figure below evolved from their common ancestor
without convergence or reversal.
1. Reconstruct their evolutionary relationships and label the character transitions. Show diagram.
2. Which evolved first: red eyes (living species 3) or fin stripes? Explain your answer.
***I have the answer but I don\'t know how we got it. I\'ve seen some incorrect/incomplete
answers here and that wasn\'t helpful. Please explain you answer for BOTH parts and show
diagram. If you\'re not sure don\'t answer! Thanks!*** Common ancestor Living species
Solution
a)the transition is as follows
common ancestor->Panderichthys,->Tiktaalik with the fins->tetrapods with legs and wings-
>amphibian
b)The eyes came first because the fish first has a body with eyes and then it has fins to swim.The
fins stripes increase with type into long wings and it can swim more and more.The eye is to see
the direction to swim.
Declining BalanceThe cost of the asset $10,000.00 The salvage v.pdf
Declining Balance
The cost of the asset: $10,000.00
The salvage value: $800.00
The life of the asset: 4 years
Calculate the Depreciation Expense to the Nearest One Dollar at End of Year 2.
Solution
Since No depreciation rate is given it is assumed theat depreciation is calclucated based on
double decling balance method
Hence depreciation Rate =1/4*2 =50%
Depreciation Exxpense at year 2 is $3750Carrying Value
at the beginningDepreciationCarrying value at the EndYear 11000025007500Year
2750037503750.
Create a java project that - Draw a circle with three random init.pdf
Create a java project that:
- Draw a circle with three random initial points on the circle.
- Link the points to form a triangle.
- Print the angles values in the triangle.
- Use the mouse to drag a point along the perimeter of the circle. As you drag it, the triangle and
angles are redisplayed dynamically.
Here is the formula to compute angles:
A = acos((a * a - b * b - c * c) / (-2 * b * c))
B = acos((b * b - a * a - c * c) / (-2 * a * c))
C = acos((c * c - b * b - a * a) / (-2 * a * b))
Solution
@Suppresswarnings(\"WeakerAccess\")
public class DragPoints extends Application {
@Override
public void start(Stage primaryStage) {
final PointPane pane = new PointPane(640, 480);
pane.setStyle(\"-fx-background-color: wheat;\");
Label label = new Label(\"Click and drag the points.\");
BorderPane borderPane = new BorderPane(pane);
BorderPane.setAlignment(label, Pos.CENTER);
label.setPadding(new Insets(5));
borderPane.setBottom(label);
Scene scene = new Scene(borderPane);
primaryStage.setTitle(\"Exercise15_21\");
primaryStage.setScene(scene);
primaryStage.show();
}
private class PointPane extends Pane {
final Circle circle = new Circle();
final Vertex[] v = new Vertex[3];
final int strokeWidth = 2;
final Color circleStroke = Color.GRAY, legStroke = Color.BLACK;
@SuppressWarnings(\"SameParameterValue\")
PointPane(double w, double h) {
this.setPrefSize(w, h);
this.setWidth(w);
this.setHeight(h);
circle.setStroke(circleStroke);
circle.setFill(Color.TRANSPARENT);
circle.setStrokeWidth(strokeWidth);
circle.radiusProperty().bind(this.heightProperty().multiply(0.4));
circle.centerXProperty().bind(this.widthProperty().divide(2));
circle.centerYProperty().bind(this.heightProperty().divide(2));
this.getChildren().add(circle);
for (int i = 0; i < v.length; i++) {
v[i] = new Vertex(circle, 2 * Math.PI / v.length * (i + Math.random()));
v[i].radiusProperty().bind(circle.radiusProperty().divide(10));
v[i].setPosition();
v[i].setStroke(new Color(i == 0 ? 1 : 0, i == 1 ? 1 : 0, i == 2 ? 1 : 0, 1));
v[i].setFill(Color.TRANSPARENT);
v[i].setStrokeWidth(strokeWidth);
this.getChildren().add(v[i]);
v[i].setOnMouseDragged(new EventHandler() {
@Override
public void handle(MouseEvent event) {
int i;
for (i = 0; i < v.length; i++)
if (v[i] == event.getSource())
break;
v[i].setAngle(event.getX(), event.getY());
moveUpdate((Vertex) event.getSource());
}
});
}
for (int i = 0; i < v.length; i++) {
int j = i + 1 < v.length ? i + 1 : 0;
int k = j + 1 < v.length ? j + 1 : 0;
v[i].bindLeg(v[j], v[k]);
v[i].leg.setStroke(legStroke);
v[i].leg.setStrokeWidth(strokeWidth);
this.getChildren().add(v[i].leg);
this.getChildren().add(v[i].text);
}
for(DoubleProperty p: new DoubleProperty[]
{circle.radiusProperty(), circle.centerXProperty(), circle.centerYProperty()})
p.addListener(new ResizeListener());
moveUpdate(v[0]);
}
void moveUpdate(Vertex vert) {
vert.setPosition();
double[] legLength = new double[3];
for (int i = 0; i < v.length; i++)
legLength[i] = v[i].getLegLength();
for (int i = .
aphase The spindle contracts and the sister chromatids are separated.pdf
aphase: The spindle contracts and the sister chromatids are separated at the centromere Once the
sister for each replicated chromosome have separated, each is now called in the cell an
unreplicated chromosome. It is at this point t chromosome num Telophase: Telophase reverse.
List the events below that happen in is essentially prophase in telophase SIMULATE CELL
DIVISION MITOSIS Using pop beads and magnetic centromeres, create 4 chromosomes as
follows: Red 6 pop beads on each side of the centromere Yellow: Same as Red. Blue 5 pop beads
on one sided of the centromere, 3 pop beads on the other side Green: Same as Blue. Work
through the phases of cell division, beginning at G1, with 4 unreplicated chromosomes
Remember that we can not see chromosomes during interphase, but this can not be modeled in
this lab. It is important to note what happens during the S phase when you do the simulation
Chromosome count Use the adjective \"unreplicated\" or \"replicated\" with your count G1: rom
DS ones S: osomes As best as you are able, and without looking at the descriptions above or
your text, describe what is happening in each phase of mitosis as you simulate the process. Avoid
using pronouns (it and they). Speak the language of genetics as you simulate mitosis. Help your
lab partner if she and provide feedback with the language. Chromosome counts and description
(replicated or unreplicated) in each phase of mitosis: Prophase: mm Somes Metaphase 1,-L
Chromosome Anaphase Telophase: What event occurs after Telophase to complete the process
of mitosis? What C4 Kenesis mitosis has been completed? is the chromosome count in each cell
after Summarizing Mitosis 1 parent cell produces 2 essentially identical daughter cells. Mitosis
separates sister chromatids.
Solution
G1: 4 unreplicated chromosomes (singlet chromosomes)
S: 4 replicated chromosomes (doublet chromosomes)
G2: 4 replicated chromosomes (doublet chromosomes)
Prophase: 4 replicated chromosomes
Metaphase: 4 replicated chromosomes arranged on metaphasic plate.
Anaphase: 8 unreplicated (singlet) chromosomes (4 singlet chromosomes move toward one end,
while the other 4 singlet (unreplicated) chromosomes move to the opposite end.)
Telophase: 8 unreplicated chromosomes (4 unreplicated chromosomes in each of the two
dividing daughter cells)
cytokinesis occurs after telophase that completes mitosis
after mitosis, 2 daughter cells are formed. each daughter cell contains 4 unreplicated
chromosomes. (similar to parent cell at G1 phase).
A population of wild deer mice includes individuals with long or sho.pdf
A population of wild deer mice includes individuals with long or short hair. When two short
haird indviduals breed, one-third of the pups have long hair. How do you explain the unusual
phenotype ratio?
Solution
it can be autosomal inheritance where homozygous dominant is lethal.
consider short hair to be dominant S, long hair to be recessive s
so cross between two short haired can be
Ss X Ss
S s
S SS Ss
s Ss ss
1:2:1 ratio for SS:Ss:ss
but we considered homozygous dominant to be lethal, so SS are dead, therefore Ss:ss is 2:1 and
1/3rd population have long hair..
Who is it important that the lymphatic system operate separately f.pdf
Who is it important that the lymphatic system operate separately from the cardiovascular
system? Explain
Describe the relationship between the lymphatic vessels and blood vessels.
Who is it important that the lymphatic system operate separately from the cardiovascular
system? Explain
Describe the relationship between the lymphatic vessels and blood vessels.
Describe the relationship between the lymphatic vessels and blood vessels.
Solution
It is important that the lymphatic system operate separately from the cardiovascular system
becuase lymphatic system is a closed circulatory system that consists of lymph capillaries, lymph
vessels, and lymph nodes different to blood vessels. The lymph present in capillaries flows into
lymphatic vessels, which are presented deep or superficial but blood vessels are deep innervating
various parts of body to maintain pheripheral resistance & mean arterial blood pressure. Blood
flows through three types of blood vessels (arteries, capillaries and veins) distributed throughout
the body. Capillaries are the fine endings of blood vessels. These capillaries are responsible for
the exchange of respiratory gases, nutrients and waste products. The molecules move across the
capillaries through a process called “diffusion.” These lymph vessels combine to form lymphatic
trunks that empty into the large collecting vessels of lymphatic system called “thoracic duct and
right lymphatic duct.”
The lymphoid tissue is mostly composed of a reticular connective tissue (a type of loose
connective tissue). The fibres of the reticular network contain macrophages. A number of
lymphocytes from the postcapillary venule walls occupy this network and leave to the body
again. This ensures the lymphocytes to identify the infected sites at a faster rate.
Lymphatic system is a closed circulatory system that consists of lymph capillaries, lymph
vessels, and lymph nodes. Lymph is a tissue fluid, which is formed by the diffusion of
substances from blood capillaries into tissue spaces. The lymph present in capillaries flows into
lymphatic vessels, which are presented deep or superficial. These lymph vessels combine to form
lymphatic trunks that empty into the large collecting vessels of lymphatic system called “thoracic
duct and right lymphatic duct.”
It is important that the lymphatic system operate separately from the cardiovascular system due
to the follwoing reasons and exerts differentail effects of innate immune system. The lymph is
generated to produce defensive mechanisms via innate immunity include physical barriers,
phagocytes, immune surveillance,interferons, complement system, inflammation and fever.
1. Fluid recovery: Filtration of the fluid into the tissue spaces from the blood capillaries will
occur continuously. 85% of it will be absorbed by the blood capillaries. About 15% will be
remained unabsorbed. This unabsorbed portion of the fluid is about 2-4 litres of water and 1/4th
to 1/2th of plasma protein. If this water protein an.
Provide a full explanation to the below question.1. Summarize the .pdf
Provide a full explanation to the below question.
1. Summarize the 802.11 standard and describe the various flavors of the standard, its
architecture and how the standard is contributing to easing of congestion in cellular networks.
Solution
IEEE 802.11 is a standard that defines the physical and MAC layers of Wireless Local Area
Network (WLAN). Under this standard, the Mobile terminals (MTs) can communicate with a
Access Point (AP) in two modes. First mode is the Infrastructure Mode, in which the MTs
communicate with the APs which forward their data to the WLAN. Second mode is the Adhoc
mode, in which the MTs communicate directly with each other without the use of a AP. The
IEEE 802.11 has a very robust Medium Access Control (MAC) mechanism that helps in
alleviating te congestion in the LANs. The inherent working of MAC protocol employed by
802.11 standard is same as that of Carrier Sense Multiple Access/ Collision Aviodance
(CSMA/CA). However, in IEEE 802.11 the protocol is implemented in two different ways. First
way is called as the Distributed Coordination Function based Wireless MAC (DCFWMAC), and
the second way is Point Coordination Function based Wireless MAC (PCFWMAC).
In the DCFWMAC, every node tries to access the medium based on some fixed duration of time
defined in the standard. Normally there are three such time intervals defined based on which a
node attempts to access the channel and transmit its packets. The first interval is the Short Inter
frame Spacing (SIFS). It is the smallest duration of time between two frames and gives the node
higher priority for sending its data. Such a interval is employed only when the node has correctly
sent its data through the channel and the receiver node needs to send a Acknowlegement (ACK)
after waiting for SIFS period of time. The second most priority interval is PCF Inter Frame
Spacing (PIFS) used during Polling mechanism by the AP. Its duration is between SIFS and
DCF Inter frame Spacing (DIFS). The DIFS is the longest time duration and hence of least
priority, in which a AP has to wait between two successive channel accesses for the given
duration. Once the channel is sensed idle, the AP waits for DIFS period of time to transmit data
through the channel. If the channel is sensed busy, then it backs off for a period of time based on
the minimum and maximum value of a contention window. During the back off period, if the
node senses the channel as busy, then it freezes the backoff counter and starts the backoff(with
the remaining time left) once the channel is idle. In this manner, the nodes which have waited
longer get higher priority over others which accessing the channel. Once the node sends the data
after accessing the channel, the receiver waits for SIFS duration of time to finally send the
acknowledgement to the sender. In order to tackle the hidden terminal problem, the DCFWMAC
employs an additional mechanism called as the RTS-CTS mechanism, in which any node that
wants to transmit.
Write a program to generate the entire calendar for one year. The pr.pdf
Write a program to generate the entire calendar for one year. The program must get two values
from the user: (1) the year and (2) the day of the week for January 1st of that year. The year,
which should be positive, is needed to check for and handle leap years1. The day of the week for
January 1st is needed so that you know where to start the calendar. The user should enter 0 for
Sunday, 1 for Monday, ... or 6 for Saturday. As always, you need to validate the user\'s input. To
actually print the calendar, you must use a single method that prints out the calendar for one
month and then call this function 12 times from main(), once for each month in the year. To
check for a leap year you will need to write another method that takes the year as a parameter
and returns true if it’s a leap year, or false otherwise. Stubs (i.e. method signatures without any
code) for both of these methods have been provided for you.
The calendar should be printed in the form below. In this example, January starts on a Saturday
(day 6). Note that February starts on a Tuesday in this example because January ended on a
Monday.
January
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 2829
30 31
February
1 2 3 4 5
6 7 8 9 10 11 12 ...
The format he gave us is as follows:
public class PA6a {
/**
* Error to output if year is not positive
*/
static final String E_YEAR = \"The year must be positive!\";
/**
* Error to output if the day is not between 0 and 6
*/
static final String E_DAY = \"The day of January 1st must be between 0 and 6!\";
/**
* Determines if an input is a leap year
*
* @param year year in question
* @ereturn true if a leap year
*/
public static boolean isLeapYear(int year) {
return false; // TODO: replace with your code
}
/**
* Outputs a month to the console
*
* @param month title
* @param startDay 0=Sunday ... 6=Saturday
* @param numDays number of days in the month
* @return day of the week of the last day of the month
*/
public static int printMonth(String month, int startDay, int numDays) {
return 0; // TODO: replace with your code
}
/**
* Program execution point:
* input year, day of the week (0-6) of january 1
* output calendar for that year
*
* @param args command-line arguments (ignored)
*/
public static void main(String[] args) {
// TODO: write your code here
}
I am confused especially with the print month method, help is greatly appreciated.
Solution
package org.students;
import java.util.Scanner;
import com.sun.org.apache.bcel.internal.generic.IUSHR;
public class Calander {
//declaring static variables
static final String E_YEAR = \"The year must be positive!\";
static final String E_DAY = \"The day of January 1st must be between 0 and 6!\";
public static void main(String[] args) {
//Declaring variables
int getFirstDay;
int year;
Scanner scanner = new Scanner(System.in);
//This while loop continue to execute until user enters valid year
while (true) {
//Getting the year entered by the user
System.out.print(\"Enter the year:.
Write the interval notation for the set of numbers graphed. Solu.pdf
Write the interval notation for the set of numbers graphed.
Solution
The open circle represents that the value is not included in the interval while a solid (filled) circle
represents that the value is included in the interval.
So, we can see that there is open circle for (-2) and closed for (5).
Hence the required interval is (-2,5]..
Hyenas are diploid organisms, and their gametes contain 20 chromosome.pdf
Hyenas are diploid organisms, and their gametes contain 20 chromosomes. Fill in the blank in
each sentences using the terms below. Each term may be used more than once or not at all.
When the somatic cells of a hyena are in interphase, the ploidy level is denoted as The ploidy
level of the hyena\'s gametes is denoted as The 2n number of a hyena is 40 20 1n 80 2n
Solution
When the somatic cells of hyena in interphase = diploid [2n] because they are not divided and
tend to divide by preparing enzymes and multiplying DNA.
The ploidy level of hyena is haploid or 1n because they are developed through meiosis process.
The 2n number of chromosomes of hyena is 40 because gametes are haploid so n=20 [as given]
and the 2n = 2x 20 = 40.
Why doesnt hemoglobin have any intermediate conformations between .pdf
Why doesn\'t hemoglobin have any intermediate conformations between R-state and T-state?
Solution
Answer:
X-ray crystallogarphy has determined two major conformations of hemoglobin- the T state
which is stabilized in the absence of oxygen and the R state which is relatively more stable in the
presence of oxygen.
The pair of identical dimers of Hemoglobin are linked by several ion pairs at the interface that
stabilize it in its deoxy-state. Many of these interactions are disrupted upon binding to oxygen
and new ones are formed.
There are intermediate conformations between T and R states, but the intermediate
conformations have intermediate binding affinities. Also, these transition states are
thermodynamically unfavourable and short lived. These states have been utilized in studying the
allosteric pathways of hemoglobin. The binding of the ligand destabilizes the T state and makes
it more likely to switch to the high-affinity oxygenated (oxy)-R state..
4. What is gyanandromorphy How does this happenSolutionAnswe.pdf
4. What is gyanandromorphy? How does this happen?
Solution
Answer : Gynandromorphy is a phenamenan in which there is an occurence of both male and
female charectorstics in a single individual (part of the body expresses as male charectors, where
as the other part express female charectors. Individuals with this phenotype are called
gynandromorphs.
It results due to the loss of an X chromosome in a particular cell during development through
mitosis.For instance, a cell with XY through mitosis produces XXYY upon duplication which in
turn divides to form XY cells. But in gynandromorphs, this cell divides into an X cell and an
XYY cells. Due to this the organism develops with both male and female charectors..
Which of the following requires the mitochondria to create contact po.pdf
Which of the following requires the mitochondria to create contact points with the ER? mt
fission mt fusion mt biogenesis of more mitochondria autophagy of mitochondria (Mitophagy)
apoptosis
Solution
Answer= The mitochondria requires the following to create a contact with ER is
mt fission - mitochondrial fission
Explaination= The mitochondria & RE are tubular organelles with a characteristic network
structure that facilitates the formation of interorganeller connections
The mito.& ER joins together at multiple contact sites to form specific domains called as
mitochondria ER associated membranes.
why nitrogen fixation is so importantSolutionNitrogen fixation.pdf
why nitrogen fixation is so important
Solution
Nitrogen fixation
It is characteristic and manufactured, is basic for all types of life since nitrogen is required to
biosynthesize fundamental building pieces of plants, creatures and other living things, e.g.,
nucleotides for DNA and RNA, the coenzyme nicotinamide adenine dinucleotide for its part in
digestion system (exchanging electrons amongst atoms), and amino acids for proteins. Along
these lines, as a major aspect of the nitrogen cycle, it is fundamental for horticulture and the
fabricate of compost. It is additionally, in a roundabout way, important to the make of every
concoction intensify that contain nitrogen, which incorporates explosives, most pharmaceuticals,
colors, and so on.
Nitrogen fixation : it happens normally in the dirt by nitrogen settling microscopic organisms
(for instance, Azotobacter partnered with a few plants, particularly vegetables). Some nitrogen-
settling microbes have cozy associations with plants, alluded to as harmonious nitrogen
obsession. Looser connections between nitrogen-settling microorganisms and plants are regularly
alluded to as cooperative or non-advantageous, as found in nitrogen obsession happening on rice
roots. It additionally happens actually noticeable all around by method for lightning.
All natural nitrogen fixation is finished by method for metalloenzymes called nitrogenases.
These catalysts contain press, frequently with a moment metal, generally molybdenum yet here
and there vanadium. Microorganisms that can settle nitrogen are prokaryotes (both microscopic
organisms and archaea, conveyed all through their particular Domains) called diazotrophs. Some
higher plants, and a few creatures (termites), have shaped affiliations (advantageous interaction)
with diazotrophs..
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- 1. Save Write a program to implement Binary search using recursive algorithm. Submit a screenshot of your Java source codes and IDE/Comment line results in PDF file. Solution BinarySearchTest.java import java.util.Scanner; public class BinarySearchTest { public static void main(String[] args) { int[] arr1 = { 1,2,3,4,5,6,7,8,9,10}; Scanner scan = new Scanner(System.in); System.out.println("Enter the key: "); int key = scan.nextInt(); int index = binarySearch(arr1,0,arr1.length,key); System.out.println("Index is "+index); } public static int binarySearch(int[] array, int a, int e, int key) { if (a < e) { int mid = a + (e - a) / 2; if (key < array[mid]) { return binarySearch(array, a, mid, key); } else if (key > array[mid]) { return binarySearch(array, mid+1, e , key); } else { return mid; } } return -1; } } Output: Enter the key: 4
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