Ques-1 Development of tube feet of CRINOIDEADevelopment of tube f.pdf
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Ques-1: Development of tube feet of CRINOIDEA Development of tube feet is from the larval stage in Crinoidea (echinoderms) and these tube foot useful for locomotion in adult species. Echinoderms are deuterostomes. Larval forms of echinoderms possess bilateral symmetry and a mouth arises from opposite end of the blastopore and alimentary canal forms connecting both during embryonic development (metamorphosis). Larval form is ciliated & from which this tube foot is developed in adults with pentaradial symmetry during metamorphosis. Adult echinoderms possess a water vascular system associated with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and peristaltic movement. Adult echinoderms possess a water vascular system associated with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and peristaltic movement. These are appeared in sea stars, these openings are smaller, and yellow button like warts located on the aboral surface of sea stars. This opening has other important parts, which perform meticulous function in water vascular system, and these parts are tube feet, stone canal, ring and radial canals. Ques-2: Spine structure of Ophiuroidea (echinoderm) are arranged in 5 rows i& connected to ligaments with mutable collagenous substance. These spines are composed of dermal calcareous ossicles or calcareous spicules Solution Ques-1: Development of tube feet of CRINOIDEA Development of tube feet is from the larval stage in Crinoidea (echinoderms) and these tube foot useful for locomotion in adult species. Echinoderms are deuterostomes. Larval forms of echinoderms possess bilateral symmetry and a mouth arises from opposite end of the blastopore and alimentary canal forms connecting both during embryonic development (metamorphosis). Larval form is ciliated & from which this tube foot is developed in adults with pentaradial symmetry during metamorphosis. Adult echinoderms possess a water vascular system associated with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and peristaltic movement. Adult echinoderms possess a water vascular system associated with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and peristaltic movement. These are appeared in sea stars, these openings are smaller, and yellow button like warts located on the aboral surface of sea stars. This opening has other important parts, which perform meticulous function in water vascular system, and these parts are tube feet, stone canal, ring and radial canals. Ques-2: Spi.
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using System;
using System.Collections.Generic;
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using System.Linq;
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}
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{
this.Validate();
this.cityBindingSource.EndEdit();
this.tableAdapterManager.UpdateAll(this.cityDataSet);
}
private void Form1_Load(object sender, EventArgs e)
{
// TODO: This line of code loads data into the \'cityDataSet.City\' table. You can move,
or remove it, as needed.
this.cityTableAdapter.Fill(this.cityDataSet.City);
}
private void btnAscPopulation_Click(object sender, EventArgs e)
{
// Sort the table by population in ascending order
this.cityTableAdapter.FillByAscPopulation(this.cityDataSet.City);
}
private void btnDescPopulation_Click(object sender, EventArgs e)
{
// Sort the table by population in descending order
this.cityTableAdapter.FillByDescPopulation(this.cityDataSet.City);
}
private void btnName_Click(object sender, EventArgs e)
{
this.cityTableAdapter.FillByCity(this.cityDataSet.City);
}
private void btnTotal_Click(object sender, EventArgs e)
{
// Declare variable to hold total population.
double totalPopulation;
// Get the total population.
totalPopulation = (double)this.cityTableAdapter.TotalPopulation();
// Display the total population.
MessageBox.Show(\"Total Population: \" + totalPopulation.ToString(\"n0\"));
}
private void btnAverage_Click(object sender, EventArgs e)
{
// Declare variable to hold average population.
double averagePopulation;
// Get the average population.
averagePopulation = (double)this.cityTableAdapter.AveragePopulation();
// Display the average population.
MessageBox.Show(\"Average Population: \" + averagePopulation.ToString(\"n0\"));
}
private void btnMaxPopulation_Click(object sender, EventArgs e)
{
// Declare variable to hold highest population.
double maxPopulation;
// Get the highest population.
maxPopulation = (double)this.cityTableAdapter.MaxPopulation();
// Display the highest population.
MessageBox.Show(\"Highest Population: \" + maxPopulation.ToString(\"n0\"));
}
private void btnMinPopulation_Click(object sender, EventArgs e)
{
// Declare variable to hold lowest population.
double minPopulation;
// Get the lowest population.
minPopulation = (double)this.cityTableAdapter.MinPopulation();
// Display the lowest population.
MessageBox.Show(\"Lowest Population: \" + minPopulation.ToString(\"n0\"));
}
private void btnExit_Click(object sender, EventArgs e)
{
// Cl.
package chegg;import java.util.ArrayList; import java.util.List;.pdf
package chegg;
import java.util.ArrayList;
import java.util.List;
import java.util.ListIterator;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeMap;
public class Vowles {
public static void main(String args[]) {
/* creating instance variable of List */
List letters = new ArrayList();
/* vowels instance variable for holding vowels */
List vowels = new ArrayList();
/* creating HashMap for storing vowels and count */
TreeMap map = new TreeMap();
/* Inserting vowels into treemap */
map.put(\"a\", 0);
map.put(\"e\", 0);
map.put(\"i\", 0);
map.put(\"o\", 0);
map.put(\"u\", 0);
/*
* creating instance variable of Scanner class for taking input from
* console
*/
Scanner scanner = new Scanner(System.in);
/*
* prompt for user input
*/
System.out.println(\"Please enter the letters with comma seperated\");
String input = scanner.nextLine();
/* closing scanner object */
scanner.close();
String[] letter = input.split(\",\");
/* Inserting letters into List */
for (String values : letter) {
letters.add(values);
}
/* ListIterator for iterating list */
ListIterator itr = letters.listIterator();
while (itr.hasNext()) {
String val = itr.next();
/* checking if value is any vowels */
if (val.equals(\"a\") || val.equals(\"e\") || val.equals(\"i\") || val.equals(\"o\") ||
val.equals(\"u\")) {
/* If vowels already exist in tree map */
if (map.containsKey(val)) {
Integer value = map.get(val);
map.put(val, ++value);
vowels.add(val);
} else {
map.put(val, 1);
vowels.add(val);
}
}
}
/* Removing all element from letters list */
letters.removeAll(letters);
/* Getting size of vowels */
int listSize = vowels.size();
System.out.println(\"There are \" + listSize + \" vowels in the list:\");
/* Iterating treeMap */
Set keys = map.keySet();
for (String key : keys) {
System.out.print(key + \"- \" + map.get(key) + \" \");
}
}
}
/***************output******************/
Please enter the letters with comma seperated
a,f,v,q,e,a,o,g,y,h,e,q,u
There are 6 vowels in the list:
a- 2 e- 2 i- 0 o- 1 u- 1
please rate the answer as best answer.
Thanks
Solution
package chegg;
import java.util.ArrayList;
import java.util.List;
import java.util.ListIterator;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeMap;
public class Vowles {
public static void main(String args[]) {
/* creating instance variable of List */
List letters = new ArrayList();
/* vowels instance variable for holding vowels */
List vowels = new ArrayList();
/* creating HashMap for storing vowels and count */
TreeMap map = new TreeMap();
/* Inserting vowels into treemap */
map.put(\"a\", 0);
map.put(\"e\", 0);
map.put(\"i\", 0);
map.put(\"o\", 0);
map.put(\"u\", 0);
/*
* creating instance variable of Scanner class for taking input from
* console
*/
Scanner scanner = new Scanner(System.in);
/*
* prompt for user input
*/
System.out.println(\"Please enter the letters with comma seperated\");
String input = scanner.nextLine();
/* closing scanner object */
scanner.close();
String[] letter = in.
L{e^{-t}} = int _0 to infinty [e^{st}e^{2-t} dt = int _0 to infinty.pdf
L{e^{-t}} = int _0 to infinty [e^{st}*e^{2-t} dt = int _0 to infinty [e^{st}*e^{2} *e^{-t}dt =
e^{2} int _0 to infinty [e^{(s-1)t} dt
L{e^{-t}} = e^2/(s+1)
Solution
L{e^{-t}} = int _0 to infinty [e^{st}*e^{2-t} dt = int _0 to infinty [e^{st}*e^{2} *e^{-t}dt =
e^{2} int _0 to infinty [e^{(s-1)t} dt
L{e^{-t}} = e^2/(s+1).
Issue of Equity or Debt are commonly used methods for business finan.pdf
Issue of Equity or Debt are commonly used methods for business financing.
Buyer\'s of the Comany\'s Debt instruments have no claim to ownership in the business. This
prevents dilution of the Company\'s ownership. The holder of the debt instrument is akin to a
lender being entitled to principal / face value / base value and interest payments on the said
amount at the specified rate. The holder of the debt instrument in the Company has no right to
the future profits of the Company. However, due to the fixed nature of the returns on these
instruments, risk borne by the holder is low. Hence, the Cost of Debt Capital is also lower than
Equity Capital.
Issue of New Equity dilutes the existing ownership interest of the Company. In most cases it also
increases the shareholder base (number of shareholders) of the Company. The major
disadvantage is the possibility of future loss of control and decision making power (if issued to
large investors). However, the Company is not obligated to make payments of interest regularly
as in the case of Debt Instruments. This in turn preserves the cash flow of the Company and
makes profits available for reinvestment into the business. Equity holders take on more risk; due
to the uncertainty of business success and growth leading to uncertainty of cash inflows. This is
further magnified under a complex capital structure where the Company may have several other
obligations to meet in the pecking order first, such as secured debt, convertible hybrid
instruments, etc. The Company may also have obligations to Government Departments. These
factors result in Equity holders demanding a higher rate of return, for absorbing this risk. Thus,
the Cost of Equity is higher than cost of Debt.
When new debt is issued at high finance costs, i.e. high rates of interest, the Company may face
additional turmoil during uncertain or difficult financial situations as it increase the risk of
insolvency. Where the Company is already highly leveraged (high debt to equity ratio) and new
debt is issued at a high rate, the Company may face slower growth rates because of the obligation
to service debt before pumping resources into high growth projects and endeavours. Further, the
cost of debt also determines the cash flow to equity.
However, the one of the biggest advantages of debt over equity is that interest cost is allowed as
a deduction for the computation of tax. This in turn lowers the tax liability. Thus, the actual cost
of debt is lower than stated cost by the amount of the tax saving benefit.
The management of the Company is required to provide regular information for the investors /
shareholders to monitor. Investors / Shareholders regularly seek comprehensive updates about
the business, financial and performance information, guidace for the future, etc. This is
especially true for newer investors that are still gaining comfort about the Company This is
largely avoided when Debt instruments are issued.
Raising equity finance is very d.
In probability we have two types of eventsIndependent Events who.pdf
In probability we have two types of events:
Independent: Events who\'s probability is not affected by the previous event(s).
Dependent: Events who\'s probability can be affected by preious event(s).
What we have in this case is a dependent event because when we don\'t replace the marble, it
changes the amount of total marbles inside the jar and thus the chances of drawing another
marble changes.
Because one event relys on the other we must find their probabilities separately at first. So what
we do is figure out the probability of pulling out a blue marble and then the probability of pulling
out another after not replacing the first. We will call these event 1 and 2.
Event 1: To find the probability of pulling out a blue marble we just divide the amount of blue
marbles we have by the total marbles in the jar:
18/45. We will keep it in this form because this is the standard way of representing chances in
stats classes.
Event 2: Now what we have left are 17 blue marbles and a total of 44 marbles so the chances of
drawing a blue marble now become:
17/45.
To combine these two chances we now multiply them together:
18/45*17/44. We will of course have to find a least common denominator but that\'s not hard and
I\'m sure you could do that part on your own.
Solution
In probability we have two types of events:
Independent: Events who\'s probability is not affected by the previous event(s).
Dependent: Events who\'s probability can be affected by preious event(s).
What we have in this case is a dependent event because when we don\'t replace the marble, it
changes the amount of total marbles inside the jar and thus the chances of drawing another
marble changes.
Because one event relys on the other we must find their probabilities separately at first. So what
we do is figure out the probability of pulling out a blue marble and then the probability of pulling
out another after not replacing the first. We will call these event 1 and 2.
Event 1: To find the probability of pulling out a blue marble we just divide the amount of blue
marbles we have by the total marbles in the jar:
18/45. We will keep it in this form because this is the standard way of representing chances in
stats classes.
Event 2: Now what we have left are 17 blue marbles and a total of 44 marbles so the chances of
drawing a blue marble now become:
17/45.
To combine these two chances we now multiply them together:
18/45*17/44. We will of course have to find a least common denominator but that\'s not hard and
I\'m sure you could do that part on your own..
i have written ths code as per your requirements with clear comments.pdf
i have written ths code as per your requirements with clear comments as shown below with
output
package com.sanfoundry.hardgraph;
import java.util.Scanner;
import java.util.Vector;
class Node1
{
int name; // node ID, started from 0 to n-1
Vector preds; // predecessors (String)
Vector neibs; // neighbors (String)
Vector backs; // backward edges -node is end vertex (Integer)
Vector fors; // forward edges -node is start vertex (Integer)
int pNode; // previous node on the augmenting path
int pEdge; // from which edge this node comes on the augmenting
// path
public Node1(int id)
{
name = id;
backs = new Vector();
fors = new Vector();
pNode1 = -1;
pEdge = -1;
}
}
class Edge
{
int name; // edge ID, started from 0 to n-1
int start; // start vertex of this edge
int end; // end vertex of this edge
int direct; // forwards (+1) or backwards (-1) on augmenting path
// if 0 then not part of augmenting path
int capacity; // capacity
int flow; // current flow
public Edge(int id)
{
name = id;
start = -1;
end = -1;
direct = 0; // default is neither
capacity = 0;
flow = 0;
}
public String toString()
{
return name + \": s=\" + start + \" e=\" + end + \" d=\" + direct;
}
}
public class LongestPathinDAG
{
int n; // number of nodes
int target; // destination node
int minLength; // the minimal length of each path
Node1[] v; // used to store Nodes
Edge[] e; // used to store Edges
int[] path; // used to store temporary path
int length = 0; // length of the path
int distance = 0; // distance of the path
int[] bestPath; // used to store temporary path
int bestLength = 0; // length of the longest path
int bestDistance = -1000000; // distance of the longest path
int[] visited; // used to mark a node as visited if set as
// 1
public LongestPathinDAG()
{
Scanner sc = new Scanner(System.in);
System.out.println(\"Enter the number of vertices: \");
n = sc.nextInt();
System.out.println(\"Enter the number of edges: \");
int m = sc.nextInt();
v = new Node1[n];
e = new Edge[m];
System.out.println(n + \" nodes and \" + m + \" edges.\");
for (int i = 0; i < n; i++)
v[i] = new Node1(i);
int i = 0;
while (i < e.length)
{
Edge edge = new Edge(i);
int sVal = sc.nextInt();
edge.start = sVal;// sc.nextInt();
int eVal = sc.nextInt();
edge.end = eVal;// sc.nextInt();
edge.capacity = sc.nextInt();
System.out.println(\" edge: \" + edge.start + \" - \" + edge.end
+ \" : \" + edge.capacity);
edge.flow = 0;
e[i] = edge;
v[sVal].fors.add(i);
v[eVal].backs.add(i);
i++;
if (i == m)
break;
}
visited = new int[v.length];
path = new int[v.length];
bestPath = new int[v.length];
sc.close();
}
/*
* this function looks for a longest path starting from being to end,
* using the backtrack depth-first search.
*/
public boolean findLongestPath(int begin, int end, int minLen)
{
/*
* compute a longest path from begin to end
*/
target = end;
bestDistance = -100000000;
minLength = minLen;
dfsLongestPath(begin);
if (bestDistance == -100000000)
return false;
else
return true;
}
private void dfsLongestPath(int curr.
Hi,Please fidn the Answer.Sorting,h is Header .pdf
Hi,
Please fidn the Answer.
//////////////////Sorting,h is Header file which contians all Function decalartion
#include
#include // std::cout
#include // std::shuffle
#include // std::array
#include // std::default_random_engine
#include
using namespace std;
class Sorting
{
private:
int num;
public:
Sorting::Sorting() {
};
///Buble Sort Input array and n Number of Elemnets
void BubbleSorting(int array[], int n);
void SelectionSorting(int array[], int n);
void ShuffleArray(int array[], int n);
void DisplayArray(int array[], int n);
void displayMenu();
int choice;
};
//////////////////////////Sorting.cpp is class fuction Which Implemnets the Function
#include \"Sorting.h\"
#include
using namespace std;
///Sorting the Array Using Bubble Sort...
void Sorting::BubbleSorting(int Array[],int n)
{
int temp = 0;
cout << \"\ ------------ BUBBLE SORT ------------ \ \ \";
for (int i = 1; iArray[j + 1])
{
temp = Array[j];
Array[j] = Array[j + 1];
Array[j + 1] = temp;
}
}
cout << \"\ ------------ RESULTS BUBBLE SORT ------------ \ \ \";
DisplayArray(Array, n);
}
//////SelectionSorting input Array and n is number of Elemnets//////////////////////
void Sorting::SelectionSorting(int Array[], int n)
{
int i, n, p, k, min, loc, temp;
for (p = 1; p <= n - 1; p++) // Loop for Pass
{
min = Array[p]; // Element Selection
loc = p;
for (k = p + 1; k <= n; k++) // Finding Min Value
{
if (min > Array[k])
{
min = Array[k];
loc = k;
}
}
temp = Array[p]; // Swap Selected Element and Min Value
Array[p] = Array[loc];
Array[loc] = temp;
}
cout << \"\ ------------ RESULTS SELECTION SORT ------------ \ \ \";
DisplayArray(Array, n);
}
//////SelectionSorting input Array and n is number of Elemnets//////////////////////
///exchange each element with a randomly chosen element. It\'s possible that an element will be
exchanged with itself, but there is no problem with that
void Sorting::ShuffleArray(int Array[], int n)
{
cout << \"\ ------------ ShuffleArray SORT ------------ \ \ \";
for (int i = 0; i
#include \"Sorting.h\"
using namespace std;
int main()
{
bool SelectionOn = true;
Sorting Sorting;
int choice;
int n = 10;
int *Arry;
while (SelectionOn != false) {
cout << \"*******************************\ \";
cout << \" 1 - Enter User Input Array of Elements.\ \";
cout << \" 2 - Display User Input.\ \";
cout << \" 3 - Shuffle User Input the Elements i.\ \";
cout << \" 4 - Bubble Sorting.\ \";
cout << \" 5 - Selection Sorting.\ \";
cout << \" 6 - Exit.\ \";
cout << \" 7 - Display Menu Again. \ \";
cout << \" Enter your choice and press return: \";
cin >> choice;
switch (choice)
{
case 1:
cout << \"Enter User Input Array of Elements\ \";
cout << \"Enter No. of Elements=\";
cin >> n;
Arry = new int[n];
cout << \"\ Enter Elements=\ \";
for (int i = 1; i <= n; i++)
{
cin >> Arry[i];
}
break;
case 2:
cout << \"DISPLAY OF ARRAY\ \";
if (Arry != NULL)
{
Sorting.DisplayArray(Arry, n);
}
else
{
cout << \"ENTER ELEMENTS TO ARRAY BY SELCTING CHOICE 1\ \";
}
break;
case 3:
cout << \"SHUFFLE OF A.
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Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
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Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
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- 1. Ques-1: Development of tube feet of CRINOIDEA Development of tube feet is from the larval stage in Crinoidea (echinoderms) and these tube foot useful for locomotion in adult species. Echinoderms are deuterostomes. Larval forms of echinoderms possess bilateral symmetry and a mouth arises from opposite end of the blastopore and alimentary canal forms connecting both during embryonic development (metamorphosis). Larval form is ciliated & from which this tube foot is developed in adults with pentaradial symmetry during metamorphosis. Adult echinoderms possess a water vascular system associated with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and peristaltic movement. Adult echinoderms possess a water vascular system associated with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and peristaltic movement. These are appeared in sea stars, these openings are smaller, and yellow button like warts located on the aboral surface of sea stars. This opening has other important parts, which perform meticulous function in water vascular system, and these parts are tube feet, stone canal, ring and radial canals. Ques-2: Spine structure of Ophiuroidea (echinoderm) are arranged in 5 rows i& connected to ligaments with mutable collagenous substance. These spines are composed of dermal calcareous ossicles or calcareous spicules Solution Ques-1: Development of tube feet of CRINOIDEA Development of tube feet is from the larval stage in Crinoidea (echinoderms) and these tube foot useful for locomotion in adult species. Echinoderms are deuterostomes. Larval forms of echinoderms possess bilateral symmetry and a mouth arises from opposite end of the blastopore and alimentary canal forms connecting both during embryonic development (metamorphosis). Larval form is ciliated & from which this tube foot is developed in adults with pentaradial symmetry during metamorphosis. Adult echinoderms possess a water vascular system associated with external tube feet as well as a calcareous endoskeleton along with ossicles interconnected collagen fibres. Sea urchins often use tube feet along with spines for jerking movement and peristaltic movement. Adult echinoderms possess a water vascular system associated with external tube feet as well as
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