Suppose that f X rightarrow Y and g Y rightarrow Z are surjections..pdf

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Suppose that f: X rightarrow Y and g: Y rightarrow Z are surjections. Prove that the composite g f: X rightarrow Z is a surjection. Solution Let z be in Z Since g is surjection there is a y in Y so that g(y)=z Since f is a surjection so there is x in X so that f(x)=y Hence, g(f(x))=g(y)=z ie (gof)(x)=z But z is arbitrary element in Z So for any z in Z there exists x in X so that (gof)(x)=z Hence, gof is a surjecxtion.

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Darwin\'s Theory had five main tenets (or components), what are they? ). These are not to be confused with the logic of Natural Selection, which is sometimes referred to as the tenets of Darwin’s theory Solution 5 major tenets of Darwins theory are : 1. The evolution stating that the world is directionally getting transformed continuously with tha passage of time, 2. The common descent which states that every group of the branching organism has decended from a common ancestor 3. The gradualism which states that during evolution, the transformation happens very gradually not at once, 4. The multiplication of species which states that the sympatric speciation leads to the diversification in species, and 5. The natural selection which states that the evolutionary change happens due to the survival of the fittest because the most fit would have a reproductive advantage, generation after generation..

Darwins Theory had five main tenets (or components), what are they.pdf

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Define: toxic shock syndrome (TSS) impetigo MRSA Solution Answer: 1. Toxic Shock Syndrome (TSS): It is a severe human disease caused by bacteria Staphylococcus aureus. The bacteria produced superantigens which act as a toxins (mainly Toxic Shock Syndrome Toxin-1) to cause this syndrome. Superantigens are bacterial proteins that stimulate the immune system much more extensively than do normal antigens. As they provoke such a drastic immune response thus, they are termed superantigens. Their action is to stimulate T cells to proliferate nonspecifically through simultaneous interaction with class II MHC proteins on antigen-presenting cells and variable regions on the chain of the T-cell receptor complex. It is most commonly seen in females who use superabsorbent tampons during menstruation. However, the toxin associated with this syndrome is also produced in men and in nonmenstruating women by S. aureus present at sites other than the genital area such as surgical wound infections. Symptoms: Low blood pressure, fever, diarrhea, an extensive skin rash (like sun burn), and ultimately shedding of the skin. 2. Impetigo: It is the most frequently diagnosed skin (superficial cutaneous) infection caused by bacteria Staphylococcuspyogenes and S. aureus. It is most commonly seen in children. Symptoms: The encrusted pustules, crusty lesions and vesicles surrounded by a red border, usually located on the face. 3. MRSA (Methicillin Resistant Staphylococcusaureus): Bacteria Staphylococcus aureus is a cause of considerable morbidity and mortality as a nosocomial or hospital-acquired pathogen since late 1950s and early 1960s. However, penicillinase-resistant, semisynthetic penicillins have proved to be successful antimicrobial agents in the treatment of staphylococcal infections. But it is unfortunate that the staphylococci become resistant through procuring of a chromosomal gene (mecA) that encodes an alternate target protein which is not inactivated by methicillin. The majority of the strains are resistant to several of the most commonly used antimicrobial agents, including macrolides, aminoglycosides, and the beta-lactam antibiotics, including the latest generation of cephalosporins. Thus, MSRA strains have recently emerged as a major, difficult to treat, hospital-acquired infection in humans, therefore considered as super-bug..

Define toxic shock syndrome (TSS) impetigo MRSASolutionAnswer.pdf

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Biochemistry homework: Focus concept ripening fruit and the activity tin methyl- The link between the production of methanol in wild type and transgenic tomato esterase, the enzyme responsible for methanol production, is examined in fruit. Prerequisite Enzyme kinetics and inhibition. Background major source ofthis methanol is likely Methanol is produced by fruit during the ripening process. The to be from the enzyme p methylesterase which acts upon pectin methyl esters to de-esterify them and produce methanol and other pectic substances One might assume that the production of methanol is correlated with PME activity, but this has not been firmly established. In this case, the investigators at the Plant Science Department at Rutgers University developed a transgenic tomato which was deficient in PME activity. By comparing the behavior of the transgenic tomato with the wild type tomato, the production of methanol in the ripening tomato fruit may be better understood. Questions breaker (Br), turning (Tru) and red ripe (RR) tomatoes. They also determined the amount of PME protein present by using a Western blot assay, which uses a specific antibody to detect the PME protein. The results are shown in Figure 12.1. What is your interpretation ofthese results? Transgenio ING MG Be Figure 12.1: PME protein levels (A) and activity (B) in wild-type and transgenic tomatoes. (Modified from Frenkel, et al. 1998.) Solution As given in A the protein level of PME is completely negligible in transgenic variety and present in wild type. In wild type variety the protein level in less in the IMG stage as compared to the subsequent stages this is because the expression of PME increases as the ripening of tomato increases. in fig. B : In the case of activity of the PME in the wild type variety increase is shown until it reaches the Tu stage and it starts decreasing in the fully ripening stage since the PME activity is not required to do the ripening. little activity in the transgenic variety is also shown because of the endogenous expression of PME because the PME is only altered not fully knockout.

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Which of the following statements is (are) true: I : Cash flow is about predicting when money will come in and will go out. II: Cash flow is not about making sure that there is enough money to pay your debts. Both statements are true Solution Answer: Both statements are false Explanation: Both the Statements are false because cash flow can be termed as total net amount of money being transfered into and out of business, specifically as effecting liquidity. So, Cash neither predicting nor making sure..

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What is the relationship of the epithelium of the epidermis to the connective tissues layers of the dermis? Describe the boundary between the epidermis and dermis and the various types of connective tissues that compose the dermis. Solution Epidermis is a type of epithelium tissue containing typically thin sheets of cells. Epidermis forms the outermost layer of the skin containing 4 to 5 layers of epithelial cells.The outermost part of the epidermis is comprises of flattened cells in stratified layers overlying on a basal layer (stratum basale) made up of columnar cells arranged perpendicularly. Beneath the epidermis lies the middle layer called dermis layer. It is composed of dense irregular connective tissue and areolar connective tissue such as a collagen with elastin arranged in bundled and woven pattern. This layer comprise of the papillary and reticular layers. Dermis contains connective tissues, vessels, glands, follicles, hair roots and sensory nerve endings. This superficial paillary layer of the dermis, made up of areolar connective tissue like collagen and elastin fibers in a loose mesh, projects itself into the stratum basale of the epidermis to form finger-like dermal papillae. The dermis layer also consists of blood and lymph vessels and nerves, which end in epidermal layer and also hair follicles and sweat glands which project out through epidermis. Thus dermis is the layer of the skin between the epidermic and the subcutaneous tissues, primarily composed of dense irregulat connective tissue. These connective tissue provides cushions to the body from stress.Its area adjoining the epidermis is called as papillary region and the deeper area is known as reticular dermis. Dermis is tightly connected to epidermis through a basement membrane. Papillary layercomprising of mainly collagen, intewines with the rete ridges of the epidermis. Reticular layer consists of irregular connective tissue fibers, densely packed collagen, elantin and reticular fibers. Structurally, dermis comprises of collagen, elastic fibers and fibrillar matrix. It also provides the sence of touch, heat etc..

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What is a linked list? What is a linked lists general syntax? Can you give me any tips on using linked lists or any extra general info I may need to know? Solution Linked Lists: It is a collection of data elements. A single linked list points to the next element of the linked list. A double linked list points to both the previous and next elements. For every node in the linked list, it has 2 parts - data and pointer Unlike arrays, linked lists are not stored at continguous location. As compared to arrays, linked lists offer dynamic size (we can add or delete elements from the linked list easily). A disadvantage as compared to arrays being, random access of the element is not possible in a linked list. General Syntax in C: //node struct node { int data; // data element struct node *next; // pointer to next element } Best way to learn linked lists is to implement the data structure along with some basic operations on it: Inserting an element (at the start, at the end, at the middle), Deleting an element (from the start, from the end, from the middle), Counting the number of elements etc..

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What is the compensation depth for phytoplankton? How does it differ from the critical depth? Solution Planktons are two types phytoplanktons and these are weakly swimming and free floating microalgae and photosynthetic whereas zooplanktons are fast swimming and feed on phytoplanktons. The compensation depth for phytoplankton is mainly useful to denote the growth of phytoplankton blooms and it is defined as the depth at which the net growth rate of phytoplankton is zero. The compensation depth of phytoplankton bloom is 0.1 to 1% (above this depth population grows) where solar radiation is going to reach adequately for their survival. The critical depth is defined as the critical depths is differ from it in which the phytoplankton biomass cannot grow adequately after a certain depth interval.

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what are structured data types ? give an example Solution Definition of structured data types: A structure is a user defined data type that groups logically related data items of different data types into a single unit.All the elements of a structure are stored at contiguous memory locations. A variable of structure type can store multiple data items of different data types under the one name.As the data of contact_information thst is firstname, lastname,homephone, mobilephone are structured data types Syntax: struct name_of_the_structure { type_1 field_name1; type_2 field_name2; ... type_N field_nameN; }; Example: struct contact_information { char firstname[FIRST_SIZE]; char lastname[LAST_SIZE]; unsigned int homephone; unsigned int mobilephone; };.

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VoIP uses which of the following because network congestion can be such a critical problem? A. Time-division multiplexing B. TCP protocol C. VLANs technology D. Isochronous design Solution D. VoIP is very time sensitive and, as such, should be based on an isochronous design. This means that the entire system must be engineered to deliver output with exactly the same timing as the input. FireWire is another example of a device that contains an isochronous interface. Answer A is incorrect because VoIP does not use time-division multiplexing. Answer B is incorrect because VoIP uses UDP for the voice portion of the call, not TCP. Some implementations of VoIP can use TCP for setup and call control. Answer C is incorrect because VLANs are not used for timing and delay problems, but are used to separate the VoIP from general traffic to make it more secure from sniffing..

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Under what assumptions would Multiple Regtression be used? Solution Firstly, multiple linear regression needs the relationship between the independent and dependent variables to be linear. Secondly, the multiple linear regression analysis requires all variables to be normal. Thirdly, multiple linear regression assumes that there is little or no multicollinearity in the data Fourthly, multiple linear regression analysis requires that there is little or no autocorrelation in the data. The last assumption the multiple linear regression analysis makes is homoscedasticity..

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Trying to figure out this sequencing quesiton. Please explain how you would approach this and what the sequence would be at the end. Thank you 2. Determine the sequence of the 14-mer peptide given the following data: a. Amino acid composition: (4S, 2L, F, G, I, K, M, T, W, Y) b. ment of peptide with 1-fluoro-2,4-dinitro benzene followed by hydrolysis DNP-S Solution (a) 4 serine, 2 leucine, phenyalanine, glycine, isoleucine, lysine, methoinine, thereonine, tryptophan, tyrosine.. (g) 2 serine, leucine, tryptophan, phenyalanine, glycine, isoleucine, lysine, leucine, homoserine, thereonine, tyrosine (e) Glycine, lysine, leucine, phenyalanine, isoleucine, tryptophan, thereonine, methoinine.

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To create a node for a binary search tree (BTNode, ) and to create a binary search tree of comparable objects (Tree BTNode BTNode -data:E -left: BTNode -right: BTNode +BTNode (newData:E, newLeft: BTNode, newRight: BTNode) +getData():E +getLeft():BTNode +getRight():BTNode +getRightMostData():E +inOrderPrint():void +removeRightmost(): BTNode +setData(newData:E):void +setLeft(newleft: BTNode):void +setRight(newRight: BTNode):void needed for remove needed for remove Tree for Lab 7 Tree -root: 8TNode - numltems:int +Tree() +add(element:E):void +remove(element:E):boolean + remove +size():int + printTree():void Solution //BTNode.java public class BTNode { private E data; private BTNode left; private BTNode right; public BTNode(E newData, BTNode newLeft, BTNode newRight) { data = newData; left = newLeft; right = newRight; }//end constructor /** * Returns the data * @return The data */ public E getData() { return data; }//end getData /** * Returns the left link * @return the left link */ public BTNode getLeft() { return left; }//end getLeft /** * Returns the right link * @return the right link */ public BTNode getRight() { return right; }//end getRight /** * Prints the tree in order from left to right. */ public void inOrderPrint() { if (left != null){ left.inOrderPrint(); }//end left System.out.println(data); if(right != null){ right.inOrderPrint(); }//end right }//end inOrderPrint /** * Sets the data in the node * @param newData The new data to be passed */ public void setData(E newData) { data = newData; }//end setData /** * Sets a new link to left * @param newLeft the new link to go left */ public void setLeft(BTNode newLeft) { left = newLeft; }//end setLeft /** * Sets a new link to the right * @param newRight the new link to the right */ public void setRight(BTNode newRight) { right = newRight; }//end setRight public E getRightMostData() { if(right == null){ return data; } else { return right.getRightMostData(); }//end right == null if-else }//end getRightMostData public BTNode removeRightmost() { if (right == null){ return left; } else { right = right.removeRightmost(); return this; }//end if right == null if=-else }//end removeRightMost }//end BTNode ===================================================================== == //Tree.java public class Tree> { private BTNode root; private int numItems; /** * No-arg constructor */ public Tree() { root = null; numItems = 0; }//end constructor public void add(E element) { if (root == null) { root = new BTNode(element, null, null); }else{ BTNode cursor = root; boolean done = false; while(!done){ if (element.compareTo(cursor.getData()) <= 0){//cursor.getData().compareTo(element) > 0){ if (cursor.getLeft() == null) { cursor.setLeft(new BTNode(element,null,null)); done = true; } else { cursor = cursor.getLeft(); } //end left is null if-else }else{ if(cursor.getRight() == null){ cursor.setRight(new BTNode(element, null, null)); done = true; }else{ cursor = cursor.getRight(); }//end cursor.getRight if-else }/.

To create a node for a binary search tree (BTNode, ) and to create a .pdf

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The Morse code (see Table 6.10 in book) is a common code that is used to encode messages consisted of letters and digits. Each letter consists of a series of dots and dashes; for example, the code for the letter a is *- and the code for the letter b is -***. Store each letter of the alphabet in a node of a binary tree of level 5. The root node is at level 1 and stores no letter. The left node at level 2 stores the letter e (code is *), and the right node stores letter t (code is -). The 4 nodes at level 3 store the letters with codes (**, *-, -*, --). To build the tree (see Figure 6.38 in book), read a file in which each line consists of a letter followed by its code. The letters should be ordered by tree level. To find the position for a letter in the tree, scan the code and branch left for a dot and branch right for a dash. Encode a message by replacing each letter by its code symbol. Then decode the message using the Morse code tree. Make sure you use a delimiter symbol between coded letters. Implementation Notes: You will need to complete the generic implementation of the BinaryTree data structure first (see slides). Once it is complete, solve the problem above (note that the book uses dots and dashes, we will use the asterisk (*) as the dot, hyphen (-) as a dash, and space ( ) as the delimiter symbol between coded letters). Provide a menu program that asks the user to 1) test output for all morse code letters with their respective translated alphabet letters (make sure you are using the binary tree to do the actual translation, output as a nicely formatted table), 2) enter an input file name to decode morse code and output the translated text to the screen, or 3) enter in a line of morse code through the console to decode morse code and output the translated text to the screen relevenat picture of both the tree diagram and the table of morse code Solution JAVA CODE: import java.io.*; import java.util.*; public class MorseCoder implements MorseCodeInterface { private MorseNode root; /** * constructor to build the tree */ public MorseCoder() { root = new MorseNode(); readTreeInfo(); } /** * reads in the tree info from the text file (helper method) */ private void readTreeInfo() { Scanner input = null; try { input = new Scanner(new File(\"encodings.txt\")); } catch (FileNotFoundException exception) { System.out.println(\"File not found!\"); } while (input.hasNextLine()) { String data = input.nextLine().trim(); if (data.length() > 0) { add(data.substring(1).trim(), data.charAt(0)); } } input.close(); } /** * adds the letter to the tree based on the mcode string (helper method) * @param mcode the string being fed in * @param ltr the letter being added at the node */ private void add(String mcode, char ltr) { MorseNode current = root; String signal = \" \"; for (int i = 0; i < mcode.length(); i++) { signal = mcode.substring(i, i + 1); if (signal.equals(\".\")) { if (current.getLeft() != null) { current = current.getLeft(); } else { current.set.

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