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The Program to find out the middle of a given linked list class LinkedList { Node head; class Node { int data; Node next; Node(int d) { data = d; next = null; } } void printMiddle() { Node slow_ptr = head; Node fast_ptr = head; if (head != null) { while (fast_ptr != null && fast_ptr.next != null) { fast_ptr = fast_ptr.next.next; slow_ptr = slow_ptr.next; } System.out.println(\"The middle element is [\" + slow_ptr.data + \"] \ \"); } } public void push(int new_data) { Node new_node = new Node(new_data); new_node.next = head; head = new_node; } public void printList() { Node tnode = head; while (tnode != null) { System.out.print(tnode.data+\"->\"); tnode = tnode.next; } System.out.println(\"NULL\"); } public static void main(String [] args) { LinkedList llist = new LinkedList(); for (int i=5; i>0; --i) { llist.push(i); llist.printList(); llist.printMiddle(); } } } Solution The Program to find out the middle of a given linked list class LinkedList { Node head; class Node { int data; Node next; Node(int d) { data = d; next = null; } } void printMiddle() { Node slow_ptr = head; Node fast_ptr = head; if (head != null) { while (fast_ptr != null && fast_ptr.next != null) { fast_ptr = fast_ptr.next.next; slow_ptr = slow_ptr.next; } System.out.println(\"The middle element is [\" + slow_ptr.data + \"] \ \"); } } public void push(int new_data) { Node new_node = new Node(new_data); new_node.next = head; head = new_node; } public void printList() { Node tnode = head; while (tnode != null) { System.out.print(tnode.data+\"->\"); tnode = tnode.next; } System.out.println(\"NULL\"); } public static void main(String [] args) { LinkedList llist = new LinkedList(); for (int i=5; i>0; --i) { llist.push(i); llist.printList(); llist.printMiddle(); } } }.

The Program to find out the middle of a given linked listclass Lin.pdf

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Question 1: Answer: b. The returned value, return address, parameters, and local variables Question 2: Answer: Pointers have to point to a certain type Question 3: Answer: True Question 4: Answer: True Solution Question 1: Answer: b. The returned value, return address, parameters, and local variables Question 2: Answer: Pointers have to point to a certain type Question 3: Answer: True Question 4: Answer: True.

Question 1Answer b. The returned value, return address, paramete.pdf

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Plasma membrane of human is made of phospholipids; these are fluids so fluidity is reflected to the viscosity of the membrane. The fluidity is reciprocal to viscosity of the lipids of membrane. High viscosity causes to low fluidity. Cholesterol plays an important role for the binding of phospholipids in the membrane. Cholesterol has hydrocarbon ring. They interact with the hydrophilic head of phospholipids through their polar hydroxyl groups. This binding helps to decrease mobility of membrane to outside; make more rigid. When the temperature downs from 37 0C (200C) the cholesterol plays such important role because it prevents the closeness of lipid hydrocarbon chains by binding with them. When the temperature rises (400C) cholesterol holds some methylene group of fatty acid chains and thus increases the melting temperature. Solution Plasma membrane of human is made of phospholipids; these are fluids so fluidity is reflected to the viscosity of the membrane. The fluidity is reciprocal to viscosity of the lipids of membrane. High viscosity causes to low fluidity. Cholesterol plays an important role for the binding of phospholipids in the membrane. Cholesterol has hydrocarbon ring. They interact with the hydrophilic head of phospholipids through their polar hydroxyl groups. This binding helps to decrease mobility of membrane to outside; make more rigid. When the temperature downs from 37 0C (200C) the cholesterol plays such important role because it prevents the closeness of lipid hydrocarbon chains by binding with them. When the temperature rises (400C) cholesterol holds some methylene group of fatty acid chains and thus increases the melting temperature..

Plasma membrane of human is made of phospholipids; these are fluids .pdf

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Please post the exact proplem in comments.SolutionPlease post .pdf

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P( both correct) = P( first correct)P( second correct) = 1212 = .pdf

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One of the functions of skin is to regulate the temperature. The skin under the influence of homeostatic control regulates the body temperature. Skin coordinates with the blood vessels present in it to regulate the temperature. When the outside environment is hot with high temperature, the blood vessels are dilated to initiate the heat loos. Simulatneously, the sweat glands in skin produce sweat or prespiration along with salts. The sweat with the blow of wind from outside cools the body immediately and allow the drop of internal temperature. When the external environment is cold with low temperatures, blood vessels constrict and pores and sweat glands in skin are tightened initiating shivering to retain or generate heat in the body and to resist the cold temperatures. The heat or cold temperatures are snesed by the thermorteceptors present in the skin and that carry the sensory information to temperature regulation center in the brain (pituitary gland). The signals received back by the receptors integrate the message with skin and regulates the temperature. Solution One of the functions of skin is to regulate the temperature. The skin under the influence of homeostatic control regulates the body temperature. Skin coordinates with the blood vessels present in it to regulate the temperature. When the outside environment is hot with high temperature, the blood vessels are dilated to initiate the heat loos. Simulatneously, the sweat glands in skin produce sweat or prespiration along with salts. The sweat with the blow of wind from outside cools the body immediately and allow the drop of internal temperature. When the external environment is cold with low temperatures, blood vessels constrict and pores and sweat glands in skin are tightened initiating shivering to retain or generate heat in the body and to resist the cold temperatures. The heat or cold temperatures are snesed by the thermorteceptors present in the skin and that carry the sensory information to temperature regulation center in the brain (pituitary gland). The signals received back by the receptors integrate the message with skin and regulates the temperature..

One of the functions of skin is to regulate the temperature. The ski.pdf

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None of the options.SolutionNone of the options..pdf

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Main class :- ------------------------- import java.awt.FlowLayout; import java.awt.Graphics; import java.awt.GridLayout; import java.awt.Insets; import java.awt.event.ActionEvent; import java.awt.event.ActionListener; import javax.swing.JButton; import javax.swing.JFrame; import javax.swing.JLabel; import javax.swing.JPanel; import javax.swing.JTextField; public class Main { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub final InfixExperssionEvaluator ie = new InfixExperssionEvaluator(); JFrame frame = new JFrame(\"FrameDemo\"); frame.setSize(600,200); frame.setLayout(new FlowLayout()); JPanel main = new JPanel(new GridLayout(3, 1)); JPanel panel1 = new JPanel(new GridLayout(1, 2)); panel1.setSize(500, 250); panel1.add(new JLabel(\"Enter Infix Expression\")); final JTextField field1 = new JTextField(); field1.setSize(150,100); panel1.add(field1); main.add(panel1); JPanel panel2 = new JPanel(new GridLayout(1, 1)); panel1.setSize(300, 150); JButton button = new JButton(); button.setSize(200, 100); button.setText(\"Submit\"); panel2.add(button); main.add(panel2); JPanel panel3 = new JPanel(new GridLayout(1, 2)); panel3.add(new JLabel(\"Value : \")); final JLabel lbl = new JLabel(); panel3.add(lbl); main.add(panel3); frame.getContentPane().add(main); frame.setVisible(true); button.addActionListener(new ActionListener() { @Override public void actionPerformed(ActionEvent e) { // TODO Auto-generated method stub if (\"Submit\".equals(e.getActionCommand())) { String expr = field1.getText(); lbl.setText(Double.toString(ie.infixToValue(expr))); } } }); button.setActionCommand(\"Submit\"); String expr = \"((2*5)+(6/2))\"; System.out.println(\"value is ::\"+ ie.infixToValue(expr)); } } Class for Evaluate methods :- ----------------------------------------- import java.util.ArrayDeque; import java.util.HashMap; import java.util.Map; import java.util.Stack; import java.util.StringTokenizer; public class InfixExperssionEvaluator { Stack expressionStack = new Stack(); Stack valueStack = new Stack(); //to hold only non-decimal values public static final String OPERATORS=\"+-/*()\"; //ignore the \"(\" as an operator //Map For Operator precedence Key(+,-,*,/) , value 0/1/2/3.. Map operator_levels = new HashMap(); public InfixExperssionEvaluator(){ operator_levels.put(\")\", 0); operator_levels.put(\"*\", 1); operator_levels.put(\"/\", 1); operator_levels.put(\"+\", 2); operator_levels.put(\"-\", 2); operator_levels.put(\"(\", 3); } double infixToValue(String inputExpression) { String eachNumber=\"\"; //to hold multiple digit values. lame idea, i know, but can\'t think of anything else at 2 AM int totalCharInInput = inputExpression.length(); inputExpression.replaceAll(\"\\\\s+\",\"\"); //Devide String to Tokens StringTokenizer tokens = new StringTokenizer(inputExpression, \"{}()*/+-\", true); //Loop through each Token while(tokens.hasMoreTokens()){ String eachChar = tokens.nextToken(); if (isValidOperator(eachCh.

Main class --------------------------import java.awt.FlowLayout.pdf

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It is a test for iron(III) ion present in the solution , so it is a .pdf

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C. 12Solution C. 12.pdf

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I think that y=21SolutionI think that y=21.pdf

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Connects corresponding regions of left and right cerebral hemispheres - Corpus callosum Connects cerebral cortex to other brain regions - Projection track Connects different parts of the same hemisphere - Association track Conscious planning, control of voluntary movement- Frontal lobe Control center for homeostasis and circadian rhythm- Hypothalamus Deep gray matter involved in arousal and memory- Basal forebrain nuclei Process sensory information and relay it to cerebral cortex - Thalamus Secretes melatonin to coordinate circadian functions - Epithalamus Senses of smell and hearing - Temporal lobe Somatosensory cortex, integration of sensory information - Parietal lobe Specialized for visual processing - Occipital lobe Unconscious control of repetitive movements- basal ganglia Visceral senses, taste, integration of sensory information - Insula Solution Connects corresponding regions of left and right cerebral hemispheres - Corpus callosum Connects cerebral cortex to other brain regions - Projection track Connects different parts of the same hemisphere - Association track Conscious planning, control of voluntary movement- Frontal lobe Control center for homeostasis and circadian rhythm- Hypothalamus Deep gray matter involved in arousal and memory- Basal forebrain nuclei Process sensory information and relay it to cerebral cortex - Thalamus Secretes melatonin to coordinate circadian functions - Epithalamus Senses of smell and hearing - Temporal lobe Somatosensory cortex, integration of sensory information - Parietal lobe Specialized for visual processing - Occipital lobe Unconscious control of repetitive movements- basal ganglia Visceral senses, taste, integration of sensory information - Insula.

Connects corresponding regions of left and right cerebral hemisphere.pdf

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Comments added... // method to check divisble by particular number.... public boolean isDivisibleBy(Node firstNode, int divisor){ // current node and previous node declaring for iterating purpose Node curr = firstNode; Node prev = null; // boolean value holds divisble or not boolean isDivible = true; while(curr != null) // iterating until list ends reach { if(curr.data % divisor != 0){ // if not divisble then exit and change flag to false isDivible = false; break; } // iterate to next node prev = curr; curr = curr.next; } // return result return isDivible; } Solution Comments added... // method to check divisble by particular number.... public boolean isDivisibleBy(Node firstNode, int divisor){ // current node and previous node declaring for iterating purpose Node curr = firstNode; Node prev = null; // boolean value holds divisble or not boolean isDivible = true; while(curr != null) // iterating until list ends reach { if(curr.data % divisor != 0){ // if not divisble then exit and change flag to false isDivible = false; break; } // iterate to next node prev = curr; curr = curr.next; } // return result return isDivible; }.

Comments added... method to check divisble by particular number..pdf

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c) Li is oxidized F is reduced as li goes to +1 oxidation state an.pdf

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Balance sheet of Doctor\'s Partnership Notes: Partners capital contribution is computed as follows:Current assetsAmountCash (A)$37,737.00Supplies (B)$2,513.00Accounts receivable (C )$35,661.00$4,227.00$31,434.00Total Current assets D= A+B+C$71,684.00Fixed assets (E )Land (F)$17,589.00Buildings (G)$75,344.00Equipment (H)$47,114.00Total Fixed Assets I =F+G+H$140,047.00Total Assets J =D+ I$211,731.00Liabilities and Owner\'s EquityCurrent LiabilitiesMortagage loan (K)$20,134.00Partners equity:Terry (L)$55,070.00Nick (M)$72,799.00Frank (N)$63,728.00Total Partner\'s equity O=L+M+N$191,597.00Total Liabilities and Owner\'s Equity P=K+O$211,731.00 Solution Balance sheet of Doctor\'s Partnership Notes: Partners capital contribution is computed as follows:Current assetsAmountCash (A)$37,737.00Supplies (B)$2,513.00Accounts receivable (C )$35,661.00$4,227.00$31,434.00Total Current assets D= A+B+C$71,684.00Fixed assets (E )Land (F)$17,589.00Buildings (G)$75,344.00Equipment (H)$47,114.00Total Fixed Assets I =F+G+H$140,047.00Total Assets J =D+ I$211,731.00Liabilities and Owner\'s EquityCurrent LiabilitiesMortagage loan (K)$20,134.00Partners equity:Terry (L)$55,070.00Nick (M)$72,799.00Frank (N)$63,728.00Total Partner\'s equity O=L+M+N$191,597.00Total Liabilities and Owner\'s Equity P=K+O$211,731.00.

Balance sheet of Doctors PartnershipNotesPartners capital con.pdf

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As per the above scenario, please find the short report below. The art of sending the message that is hidden in the grapical images is mainly called as \"STEGANOGRAPHY\". It is mainly followed , to protect the message from the unknown source. were other they cannot see the message that is hidded. Only the reciever and sender alone the way to encrypt or decrypt the hidden message from the images that are sent. 1) To make the message hidden in the image, first choose the message. There are so many softwares that are available, were the images if that are stored in the local drive, they can choosen. 2) Then type the information that you like to transfer , in terms of message. It may ask you for the password. Then you can create the password and enter it. Its mainly for the purporse of security. 3) After that if you , click hide message button in the available software, the message will be hidden. It will take few minutes to create the image with message. ONce it is created we will not able to crop the image of resize the image. Once the encryption is done, sender will send the message to intended reciepent. After recieving the message, the reciever need to decrypt the message. Following steps need to be followed to see the hidden messages. 1) Select the image needed for the decryption. 2) Enter the password that was sent by the sender, once the password is entered the password check will be done and reveal your message option need to be clicked. 3) Atlast the hidden message will be shown. It can be stored in the local drive and used. Solution As per the above scenario, please find the short report below. The art of sending the message that is hidden in the grapical images is mainly called as \"STEGANOGRAPHY\". It is mainly followed , to protect the message from the unknown source. were other they cannot see the message that is hidded. Only the reciever and sender alone the way to encrypt or decrypt the hidden message from the images that are sent. 1) To make the message hidden in the image, first choose the message. There are so many softwares that are available, were the images if that are stored in the local drive, they can choosen. 2) Then type the information that you like to transfer , in terms of message. It may ask you for the password. Then you can create the password and enter it. Its mainly for the purporse of security. 3) After that if you , click hide message button in the available software, the message will be hidden. It will take few minutes to create the image with message. ONce it is created we will not able to crop the image of resize the image. Once the encryption is done, sender will send the message to intended reciepent. After recieving the message, the reciever need to decrypt the message. Following steps need to be followed to see the hidden messages. 1) Select the image needed for the decryption. 2) Enter the password that was sent by the sender, once the password is entered the password check will be done and reveal you.

As per the above scenario, please find the short report below.The .pdf

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Answer B.nitrile hydrolysis reactionNitrilase which catalyzes the.pdf

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Answer: Question no 37 and 38 are not clear. 37) what molecules absorbed? 38) which enter the bloodstream or lymphatic system? 39) Large intestine is also called as colon and it is last part of gastrointestinal tract (of the digestive system of vertebrates) where the final stages of digestion occur. The major function of the large intestine is to absorb water from the remaining indigestible food matter and remaining waste material is stored as feces before elimination from the body. 40) Rectum is a component of the lower gastrointestinal tract. It is a straight portion which is continuation of the sigmoid colon, and connects to the anus. The function of rectum is the temporary storage of feces. When the rectum becomes full, the increase in intrarectal pressure forces the walls of the anal canal apart, allowing the fecal matter to enter the canal. The rectum shortens as material is forced into the anal canal. 41) Orifice is the external opening of the rectum, controls the expulsion of feces. Its function is to push the feces out of the anal canal by the help of internal and external anal sphincters (two muscular rings which control defeaecation). 42) Pancreas is an endocrine gland and a digestive organ as well, which secrete: Solution Answer: Question no 37 and 38 are not clear. 37) what molecules absorbed? 38) which enter the bloodstream or lymphatic system? 39) Large intestine is also called as colon and it is last part of gastrointestinal tract (of the digestive system of vertebrates) where the final stages of digestion occur. The major function of the large intestine is to absorb water from the remaining indigestible food matter and remaining waste material is stored as feces before elimination from the body. 40) Rectum is a component of the lower gastrointestinal tract. It is a straight portion which is continuation of the sigmoid colon, and connects to the anus. The function of rectum is the temporary storage of feces. When the rectum becomes full, the increase in intrarectal pressure forces the walls of the anal canal apart, allowing the fecal matter to enter the canal. The rectum shortens as material is forced into the anal canal. 41) Orifice is the external opening of the rectum, controls the expulsion of feces. Its function is to push the feces out of the anal canal by the help of internal and external anal sphincters (two muscular rings which control defeaecation). 42) Pancreas is an endocrine gland and a digestive organ as well, which secrete:.

AnswerQuestion no 37 and 38 are not clear. 37) what molecules abs.pdf

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Allopatric speciation occurs due to change of environment, like jumping of gorilla louse to hominins and adapted to develop into hominin pubic louse. A sympatric speciation occurs in the same environment often due to polyploidy.ie., the fusion of 2 different species to form a new species. Louse speciation is not a sympatric speciation. Solution Allopatric speciation occurs due to change of environment, like jumping of gorilla louse to hominins and adapted to develop into hominin pubic louse. A sympatric speciation occurs in the same environment often due to polyploidy.ie., the fusion of 2 different species to form a new species. Louse speciation is not a sympatric speciation..

Allopatric speciation occurs due to change of environment, like jump.pdf

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An adaptation of the epithelial cells of the proximal convoluted tub.pdf

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