Which phyla have megaphylls Bryophyta Monilophyta Coniferophyta .pdf

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Which phyla have megaphylls? Bryophyta Monilophyta Coniferophyta Anthophyta Which phyla have parenchyma cells? Bryophyta Monilophyta Coniferophyta Anthophyta Which phyla have phloem cells? Bryophyta Monilophyta Coniferophyta Anthophyta Solution Ans 1: Monilophyta (Phyla have megaphylls) Ans 2: Anthophyta (Phyla have parenchyma cells) Ans 3: Bryophyta (phyla have phloem cells).

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Help with implementing the four function the first is an example thanks def inorder(T,f): if not is_bst(T): return if not T: return inorder(T[1],f) f(T[0]) inorder(T[2],f) # programming project: provide implementations for the functions below # and add tests for these functions to the code in the block # that starts \"if __name__ == \'__main__\'\" def preorder(T,f): pass def postorder(T,f): pass def tree_height(T): pass def balance(T): # returns the height of the left subtree of T # minus the height of the right subtree of T pass Solution class Node: def _init_(self,val): self.value=val self.leftChild=None self.rightChild=None def preorder(self): if self: print(str(self.value)) if self.leftChild: self.leftChild. preorder() if self.rightChild: self. rightChild. preorder() def postorder(self): if self: if self.leftChild: self.leftChild. postorder () if self.rightChild: self. rightChild. postorder () print(str(self.value)) def inorder(self): if self: if self.leftChild: self.leftChild. inorder () print(str(self.value)) if self.rightChild: self. rightChild. inorder () def height(self,root): if root is None: reture 0 else: return max(self.height(root.left),self.height(root.right))+1 det balance(self,root): if root is None: return 0 else: leftHeight= height(root.left) rightHeight= height(root.right) if leftHeight- rightHeight<=1 : return 1 else: return 0 class Tree: def _init_(self): self.root=None def preorder(self): print(“Preorder)” self.root.preorder() def postorder(self): print(“Postorder)” self.root. postorder () def inorder(self): print(“Inorder)” self.root. inorder () bst=Tree() bst.preorder() bst.postorder() bst.inorder() print(\"Height of tree is \" (height (bst))) if (balance(bst): print (“Tree is balance”) else print (“Tree is not balance”).

Help with implementing the four function the first is an exampleth.pdf

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Encrypt the message \" HALT \" by translating the letters into numbers (via A=0,B=1,C=2,D=3,E=4,F=5,G=6,H=7,I=8, J=9,K=10,L=11,M=12,N=13,O=14,P=15,Q=16,R=17, S=18,T=19,U=20,V=21,W=22,X=23,Y=24,Z=25 ) and then applying the encryption function given, and then translating the numbers back into letters. (a) f(p)=(p+5) mod 26 _____________ (b) f(p)=(p+12) mod 26 _______________ (c) f(p)=(p+4) mod 26 ______________ Solution Encrypt message as : A=0,B=1,C=2,D=3,E=4,F=5,G=6,H=7,I=8, J=9,K=10,L=11,M=12,N=13,O=14,P=15,Q=16,R=17, S=18,T=19,U=20,V=21,W=22,X=23,Y=24,Z=25 and given encryption function :a) f(p)=(p+5)mod26 if p=Z=25 f(25)=(25+5)mod26 =30mod26=4 f(25)=E and f(p)=(p+5)mod26 if p=15 f(15)=(15+5)mod26 =20mod26=20 f(15)=U b) f(p)=(p+12)mod26 f(p)=(p+12)mod26 if p=Z=25 f(25)=(25+12)mod26 =37mod26=11 f(25)=L and for f(p)=(p+12)mod26 if p=15 f(15)=(15+12)mod26 =27mod26=1 f(15)=B c)f(p)=(p+4)mod26 f(p)=(p+4)mod26 if p=Z=25 f(25)=(25+4)mod26 =29mod26=3 f(25)=D and for f(p)=(p+4)mod26 if p=15 f(15)=(15+4)mod26 =19mod26=19 f(15)=T.

Encrypt the message HALT by translating the letters into numbe.pdf

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Each letter in the word Massachusetts is written on a card. The cards are placed in a basket. What is the probability of selecting two S\'s if the first card is replaced before selecting the second card? Solution It\'s a probability of 16 out of 169. There are 13 letters in MASSACHUSETTS. 4 of the 13 are the letter S, so there is a 4 in 13 chance to draw an S. Since the first card is replaced, there is still a 4 in 13 chance of drawing another S. So the combined probability is 4/13 x 4/13 = 16/169.

Each letter in the word Massachusetts is written on a card. The card.pdf

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Below is the question I need help with. It need to be done in Java. I need help making the priority queue part of the assignment, if you can add the GUI it will be helpful. You are to simulate a dispatcher using a priority queue system. New processes are to be entered using a GUI with priority included (numbering should be automatic). Processes are also to be terminated by GUI command. Context switches are to be by command with the cause of the switch being immaterial. Assume only one CPU. Priorities and numbers of processes can be kept small, just big enough to demonstrate the below listed functionality. You may pre-populate the queues initially from a data file. I am looking at the mechanism as you are NOT creating actual processes, just simulating them. Functionality to be provided by you: 1. Priority based Ready Queue(s). 2. Blocked list. 3. Output of complete system status after every context switch showing ready, blocked, and running processes. Solution import java.awt.Container; import java.awt.Dimension; import java.awt.FlowLayout; import java.awt.event.ActionEvent; import java.awt.event.ActionListener; import java.util.PriorityQueue; import java.util.Queue; import javax.swing.JButton; import javax.swing.JFrame; import javax.swing.JLabel; import javax.swing.JTextArea; import javax.swing.JTextField; class dispatcher extends JFrame implements ActionListener { static int processNo = 0; static Process currentlyRunning; static Queue readyQueue; static Queue blockedQueue; static JButton newProcess; static JButton contextSwitch; static JTextField priority; static JTextArea Queues; private static Process newProcess(int i, int i0, String running) { throw new UnsupportedOperationException(\"Not supported yet.\"); } dispatcher() { Container cp = getContentPane(); cp.setLayout(new FlowLayout()); Queues = new JTextArea(5, 15); Queues.setPreferredSize(new Dimension(500, 400)); cp.add(Queues); cp.add(new JLabel(\"Priority\")); priority = new JTextField(6); newProcess = new JButton(\"Create Process\"); newProcess.addActionListener(this); cp.add(priority); cp.add(newProcess); contextSwitch = new JButton(\"Context Switch\"); contextSwitch.addActionListener(this); cp.add(contextSwitch); setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); setTitle(\"Dispatcher\"); setSize(600, 500); setVisible(true); } public static void main(String args[]) { dispatcher d = new dispatcher(); readyQueue = new PriorityQueue<>(); blockedQueue = new PriorityQueue<>(); Process p1 = newProcess(processNo++, 2, \"Running\"); Process p2 = newProcess(processNo++, 1, \"Blocked\"); Process p3 = newProcess(processNo++, 4, \"Ready\"); Process p4 = newProcess(processNo++, 3, \"Ready\"); Process p5 = newProcess(processNo++, 5, \"Ready\"); Process p6 = newProcess(processNo++, 8, \"Blocked\"); readyQueue.add(p3); readyQueue.add(p4); readyQueue.add(p5); blockedQueue.add(p6); blockedQueue.add(p2); currentlyRunning = p1; contextSwitch(); } static void createProcess(int priority) { readyQueue.add.

Below is the question I need help with. It need to be done in Java. .pdf

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Css code for a awoko newspaper html website which provides business and finance, editorial, entertainment.,features and sport. Css code for a awoko newspaper html website which provides business and finance, editorial, entertainment.,features and sport. Solution //Name this css file as style.css @charset \"utf-8\"; /* General */ body { margin:0px; font-size:12px; background-color:#ffffea; } body * { outline:none; } .float { float:left; } body form { margin:0px; } #page { width:100%; background-image:url(images/body_bg2.png); background-repeat:repeat-x; } #wrapper { width:966px; margin:0px auto 0px auto; } #left { width:12px; height:219px; background-image:url(images/left_effect.png); background-repeat:no-repeat; background-position:0px 157px; } #center { width:942px; } #right { width:12px; height:219px; background-image:url(images/right_effect.png); background-repeat:no-repeat; background-position:0px 157px; } /* Header */ #header { padding:2px 0px 2px 0px; } #search { width:50%; float:left; } .search_field { height:18px; border:1px solid #e3c6c4; } #top_links { width:40%; float:right; margin:4px 0px 4px 0px; font-family:Georgia, Arial, Helvetica, sans-serif; } #top_links * a { color:#FFF; text-decoration:none; } #top_links ul { padding:0px; margin:0px; float:right; } #top_links ul li { list-style:none; margin-left:15px; float:left; } #logo { width:222px; margin-right:16px; margin-top:15px; font-family:Georgia, \"Times New Roman\", Times, serif; color:#000; line-height:25px; text-align:center; } #logo a { background-image:url(images/logo.png); background-repeat:no-repeat; width:202px; height:72px; margin-right:auto; margin-left:auto; display:block; } #top_ad { width:374px; margin-right:21px; margin-top:15px; height:90px; } #today_quote { width:309px; margin-top:15px; } #quote { width:219px; text-align:right; margin-right:5px; padding:5px; font-size:14px; } /*Middle*/ #middle { border-top:1px solid #df202d; border-left:1px solid #df202d; border-right:1px solid #df202d; } .scroller { background-color:#fef407; height:22px; width:auto; color:#959595; font-family:Georgia, \'Times New Roman\', Times, serif; line-height:22px; padding:0px 5px 0px 5px; } #main_menu { width:100%; font-family:Georgia, Arial, Helvetica, sans-serif; line-height:29px; font-size:12px; text-transform:uppercase; background-image:url(images/body_bg2.png); background-repeat:repeat-x; background-position:top; background-color:#c0010e; height:30px; } #main_menu * a { color:#FFF; text-decoration:none; } #main_menu ul { padding:0px; margin:0px; width:100%; text-align:center; } #main_menu ul li { list-style:none; margin:0px 10px; display:inline-block; } /*Three Columns*/ /* column container */ .colmask { position:relative; /* This fixes the IE7 overflow hidden bug */ clear:both; float:left; width:100%; /* width of whole page */ overflow:hidden; /* This chops off any overhanging divs */ } /* common column settings */ .colright, .colmid, .colleft { float:left; width:100%; /* width of .

Css code for a awoko newspaper html website which provides busines.pdf

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Case study: Job Satisfaction in the Banking Industry: Or the logistical nightmare of conducting large scale quantitative research Job characteristics are believed to have an impact on stress and well-being at work (Karasek& Theorell, 1990). The demands of the job on the one hand and theextentto which you have control over your own activities (decision latitude) on the other, are two factors which together define how stressful a job is. Those jobs which are high demand, but offer limited control, are considered to be high-strain and carry an increased risk of job dissatisfaction, stress and burnout. Based on this theoretical framework, the Union of Belgian Banks sent out a research call to several institutions, with a bidding process based on criteria such as quality of the proposal, timing, and – above all – budget. The aim of the research was to carry out quantitative research to measure the relationship between job characteristics and job satisfaction in all Belgian banks (seeCambréet al., forthcoming). But in order to do this effectively, several methodological issues needed to be resolved during the research process. First of all, aresearch consortiumwas selected to conduct the research, or more precisely, the two highest ranked bidders were asked to jointly undertake the research. This was the outcome of a political decision by the banks (see also p. 142, ‘Affiliation and conflicts of interest’),since the employers preferred one partner and the unions (employee representatives) preferred the other. The two competing research institutes, a private company specialising in stress at work and theKatholiekeUniversiteitLeuven (Belgium), were required to co-operate and develop a level of trust in order to conduct the research. For example, both research institutes had different ideas as to which scale should be used in the questionnaire. They could not just combine the scales or include both scales, simply because they are supposed to measure the same concept. Furthermore, this would also make the questionnaire too complex. Therefore, the research institutes had to combine their knowledge, look for compromises and jointly work on a shared vision, which is, to say the least, rather time consuming. A second obstacle that needed to be overcome was thesample(see Chapter 7). Intotal, 69,000 employees work for Belgian banks and it was decided that questioning all employees would be too complicated and too expensive. Therefore the research committee, consisting of representatives of the banks, the unions and the research consortium, opted for a cross-sectional design (p. 45) with a fixed sample of 15,000 employees (roughly 21%; see p. 187 ‘Absolute and relative sample size’). In this sample, the small banks were over-represented in order to be able to make conclusions at the level of each bank. Within each bank, the respondents were selected at random with no particular quota for gender, age or employee level. In the postal survey (see p.231 ‘Self- co.

Case study Job Satisfaction in the Banking Industry Or the logisti.pdf

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You are constructing working models of the heart for extra credit Your first attempt failed because the cells separated from each other during beating. What structures did you omit, and why did this allow the heart to fall apart? You omitted the gap junctions in your second attempt How would a real heart fail to function normally without gap junctions? Solution Ans. In first attempet the page maker is missing. In the heart, crevice intersections assume a vital part in motivation conduction. Concentrates in the course of the most recent decade have uncovered that hole intersections are encoded by a multigene family known as the connexins. Crevice intersection correspondence (GJC) interceded by connexins is basic for heart function.Heart will get to obtaine these capacity and quit working..

You are constructing working models of the heart for extra credit You.pdf

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why does countercurrent flow yields greater oxygen uptake than concurrent flow? Mention diffusion. Solution Answer: The orientation of medium is perpendicular to the flow of blood in countercurrent flow so that it is leading to higher efficiency of oxygen diffusion to medium flow at differential points whereas in concurrent flow, medium is synchronous and parallel to the flow of blood so that the rate of diffusion from the alveoli to blood capillaries are lower. The both bloodstreams are going to move in a similar or simultaneous pathway. The concurrent flow yields intermediate in transfer efficiency with multivillous oxygen diffusion main found in humans and sheep.

why does countercurrent flow yields greater oxygen uptake than concu.pdf

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Wind Shear can be defined in the horizontal plane as well as the vertical plane (True/False) Solution True Wind shear, sometimes referred to as wind gradient, is a difference in wind speed and/or direction over a relatively short distance in the atmosphere. Atmospheric wind shear is normally described as either vertical or horizontal wind shear. Horizontal Plane :Horizontal wind shear is a change in wind speed with change in lateral position for a given altitude. Vertical Plane :Vertical wind shear is a change in wind speed or direction with change in altitude..

Wind Shear can be defined in the horizontal plane as well as the vert.pdf

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While hiking on some small islands in Lake Superior, you stumble across a small frog. Thinking about your respective evolutionary histories, what is a likely homology between you and this animal: Number of limbs External fertillization Amniotic egg Four-chambered heart Live birth Solution ANS: A. Number of limbs. Explanation: According to the evolution, comparative anatomy of limb bone structure (pentadactyl ,5 digit) is same in most vertebrates. It is called homology. Homologies clearly indicate the phenomenon of common ancestry. Like humans frogs also contains same bone structure. They are…. Arms: Like Human arms, frog front leg bones are called as humerus, radius and ulna (radius and ulna are fused into one bone). Legs: Like Human legs, frog femur supports its upper leg and lower leg bone (tibia and fibula, are fused). And also frog contains two scapulae (shoulder blades), and clavicles (collarbones) like human..

While hiking on some small islands in Lake Superior, you stumble acr.pdf

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Where do endocyotic vesicles come from? A. Plasma membrance B. Nuclear membrance C. Smooth ER D. Lysosome Solution Answer: A. plasma membrane Reason: Cells are going to ingest fluid, molecules, and waste particles Clathrin mediated endocytosis result in internalization of the receptor further leading these receptors to be degraded by lysosomal endosome formation when there is not ligand to bound. This endocytosis process is going to produce endocytic vesicles from localized regions of the plasma membrane through invagination.

Where do endocyotic vesicles come fromA. Plasma membranceB. Nuc.pdf

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When considering the AFN equation, we exclude notes payables from our calculation of Spontaneous increases in Accounts Payables and Accruals because Notes Payable is a source of external capital. True or false? Solution True, spontaneous funds are funds arising out of normal operations from suppliers, employees, & government, including immediate increase in accounts payable and accruals. But they dont include notes payable because it is a source of external capital and not internally generated through normal business operations..

When considering the AFN equation, we exclude notes payables from ou.pdf

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What lessons can we infer from history around the efforts to enact National Health Insurance and/or Health Reform Solution In the US, “history of health care reform” is one of the most important topics in the political debate since 20th Century. Candidates try to increase their vote bank by proposing alternative reform proposals in the presidential election. Several lessons can be learned from the history around the efforts to enact National Health Insurance and/or Health Reform. The national health care reform in the Country is majorly influenced by the power of president, lobbying strength, costs, and complexity of the issues. President Truman recommended for universal health insurance coverage that would be administered by the National Health insurance board. However, his proposal was rejected due to huge oppositions from opponents as well as AMA. However, Mr. Truman’s efforts to increase the number of hospitals and modernizing them were achieved. Later on, President Kennedy supported the King-Anderson bill that was designated to give insurance coverage to the people having the age of 65 or more. However, this bill was also rejected due to the opposition of AMA and other powerful people. From President Johnson’s efforts, Social Security Amendments of 1965 was passed that provides insurance coverage to people having age of 65 or more. Besides this it also gives coverage to the poor, blind and disabled people. President Nixon has played a crucial role in passing of the Health Maintenance Organization Act of 1973. President Ford has experienced escalating healthcare costs due to medical inflation. To cope with the problems, the National Health Planning and Resource Development Act of 1974 were passed that helped to remove duplication of healthcare facilities and services. President Jimmy Carter put lot of efforts to calling for the National health insurance having universal coverage, although he did not received much success. President Reagan has enacted several laws to reduce the federal expenses on health care. However, his administration passed the Medicare Catastrophic Coverage Act of 1988. President George HW Bush faced a huge disappointment, resentment and revolt due to the Medicare Catastrophic Coverage Act of 1988, because the Act forced the people to pay higher premiums and taxes. Most of the Act was repealed after 16-17 months. President Bush has helped to reform major healthcare legislations. After that, President Clinton was elected as the president. He proposed “American Health Security Act of 1993 to Congress. The Act was rejected due to huge opposition from most of the healthcare and health insurance industry. This Act proposed that the health insurance coverage should be provided by the private insurers and coordinated by the regional health alliances. The government passed one of the important Act, i.e. The Health Insurance Portability and Accountability Act. President George W. Bush passed the Medicare Drug Improvement and M.

What lessons can we infer from history around the efforts to enact N.pdf

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What fungal structure functions in absorbing nutrients from the environment Solution Fungi are unicellular or multi-cellular thick-cell-walled heterotrophy decomposers that eat decaying matter Hyphae absorb nutrients from the environment and transport them to other parts of the thallus or fungus body. Hyphae are long tubular structures. They have rigid cell walls that may be reinforced by perforated cross-walls called septa. Hyphae perform a variety of functions in fungi. They contain the cytoplasm or cell sap, including the nuclei containing genetic material.

What fungal structure functions in absorbing nutrients from the envi.pdf

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What are the main dorsalizing and neuralizing factors in Xenopus? What are the main inducing factors controlling anteroposterior pattern? Solution Ques-12: Answer: The Xenopus dorsalizing factors such as “noggin” are responsible for formation of Spermann organizer in Xenopus embryos finally promote “dorsal & ventral cell fate inside the mesoderm”. The other cereberus like –neutralizing factors are going tot promote formation of neural development via Wnt signaling.

What are the main dorsalizing and neuralizing factors in Xenopus Wh.pdf

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We have a cross between a heterzygous A/a, B/b with a homozygous recessive a/a, b/b. What is the ratio of parental type to recombinant type of offspring assuming that the two loci are on separate chromosomes? What is the ratio assuming that the two loci are closely linked on the same chromosome (so that no recombination or crossing over between the two loci is observed)? Solution So, when genes are on different chromosomes, 50% of the gametes produced by a this individual would be recombinant, when compared to the gametes produced by its parents. The other 50% are parental. If two genes occur on the same chromosome, they may or may not assort independently depends on linkage distance. Now when Linkage is present then fewer than 50% of the gametes produced by a double heterozygote are recombinant and this percentage will decrease or increase with the gene on same chromosome.ababababABAbBbAbBbAbBbAbBbAbAabbAabbAabbAabbaBaaBbaaBbaaBbaaB babaabbaabbaabbaabb.

We have a cross between a heterzygous Aa, Bb with a homozygous rec.pdf

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Using Python: Write a program called ExceptionHandling.py that demonstrates exception handling. In this program write code using your own data that generates the following errors – ZeroDivisionError, ValueError, TypeError, and IndexError. Catch all the errors in the main function. If the error doesn’t fall into any of these types include general Exception to handle other error types. Your data/code should generate the errors. Do not simulate the errors, the way we did in the ExceptionChaining program. Generate errors in different functions. Solution def div(a,b): return a/b def getElement(): list_ = [1] return list_[1] def addTwoElement(a,b): return a+b def getIntInput(): return int(raw_input(\'Enter a number : \')) def main(): try: res = div(a,0) except ZeroDivisionError: print \'Cannot divide by zero.\' except: print \'Unknown error.\' try: res = addTwoElement(1,\'str\') except TypeError: print \'Please provide correct types while performing addition.\' except: print \'Unknown error.\' try: res = getIntInput() except ValueError: print \'Cannot convert from string to int.\' except: print \'Unknown error.\' try: res = getElement() except IndexError: print \'Accessing an element which is out of bound.\' except: print \'Unknown error.\' main() OUTPUT:.

Using Python Write a program called ExceptionHandling.py that demon.pdf

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This lab involves either AVL trees or Red-Black trees. Overview: The game involves asking a player how many letters they want, then generating a list of that many random letters. The user must then generate as many random words as s/he can using only those letters. When the user is done, the user should enter a dummy value (I used “- 1”). The list of words entered is then checked with the dictionary. For every word entered that is in the dictionary, the user gets 3 points. For every word entered that is not in the dictionary, the user loses 6 points. That’s basically it. A couple of notes: When generating the list of x random letters, the game is pretty dull if no vowels are included. So I generated a random number y between 1 and x, and then got that many random vowels. I then got x-y consonants. letters can be repeated. The AVL tree should be designed so that no duplicates are allowed in the tree. To test to see whether you’ve loaded the file and created the AVL tree properly, you should write an in-order printTree method as part of your AVL class and call it. If the words print out in order alphabetically, your AVL tree is most likely being created correctly. The dictionary I used was a scrabble dictionary, which means that most if not all of the words are 8 characters or less. You may not want to pick a number much bigger than 8, or, as an alternative, you may want to find your own dictionary. You can use .size() to get the number of characters in a string. You must include . The text file I am giving you is called dict.txt, and has over 80000 words in it. Every time I ran the program, it took my computer about a minute to load the entire dictionary into the AVL tree (and my computer is relatively fast – feel free to one-up me with your faster computer!). Thus for testing purposes, I created a much smaller dictionary file so I didn’t have to wait that whole minute while I was debugging. I’ve included that as well. Equally, for testing my tree, I created a much smaller file. I’ve included it, along with the pre-order, in-order, and post-order printout. I included my output so you can test your tree from the test file. Outline: I am including a general outline, which you may or may not choose to use for this lab: Files created include: BSTB.hpp // the Binary Search Tree class declarations BSTB.cpp // the Binary Search Tree class method definitions (using AVL balanced tree methods) /* note: you can start writing this as a regular binary search tree, then modify it to be an AVL tree later */ NodeTB.hpp // the NodeTree class declaration NodeTB.cpp // the NodeTree class method definitions (this is for the tree) Game.hpp // the class declarations for the Boggle-ish game Game.cpp // the class definitions for the Boggle-ish game Main.cpp // the main file, which creates a game object and then runs the game’s startGame method More Details: My BSTB.hpp: (not including include files) class BSTB { NodeTB *root; int score; //for wordlist when calculatin.

This lab involves either AVL trees or Red-Black trees.Overview Th.pdf

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This is a 16 part assignment. (PART 1 of 16) Back in the 1970s, there was a fad called “biorhythms”. People believed that their physical, emotional, and intellectual abilities could be modeled by the following equations: (PART 1 of 16) Find the expressions that would identify all the days a person would experience a critical day.(hint: there will be two expressions for each ability): a. Physically b. Emotionally c. Intellectually Physical: sin cal: si23 Emotional : sin Emotional: S1n 28 Intellectual: sin Solution physical :sin(2pi*t/23) emotionally : sin(2pit/28) intellectualy : sin(2pit/33) Expression = phsical +emotional +intellectual = sin(2pit/23) +sin(2pit/28) +sin(2pit/33).

This is a 16 part assignment. (PART 1 of 16)Back in the 1970s, the.pdf

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The members of a board of directors have the following roles president (P), we president (V), secretary (S), treasury (T), and fundraiser (F) Consider these board members to be a population of interest List the 10 possible samples (without replacement) of size two from this population of five board members. Solution no.of possible samples = 5C2 = 10 To find the possible samples first let us start with P, So it\'s - PV, PS, PT, PF with V it is - VS, VT, VF [ since VP = PV is already there in starting with P list] similarly with S it is - ST, SF [ Since SP,SV are already listed above] The 10 possible samples are PV, PS, PT, PF,VS, VT, VF,ST, SF Option A is the answer.

The members of a board of directors have the following roles presiden.pdf

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