Let X and Y have joint probability density function f(x, y) = 2 I(0, .pdf

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Let X and Y have joint probability density function f(x, y) = 2 I(0, 1)(x)I (0, 1)(y)I(0, 1)(x + y) (a) Compute the marginal densities fx(x) and fy(y). Are X and Y independent? Solution f(x)= integration from (0 to 1-x) [ 2dy] = [2(1-x)].... f(y)= integration from (0 to 1-y) [ 2dx] = [2(1-y)].... f(x)*f(y) = [2(1-x)]* [2(1-y)] which is not equal to f(x,y)...so, they ar enot independent....

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Consider the following C code snippet: // C code void setArray(int num) { int i; int array[10]; for (i=0; i < 10; i=i+1) { array[i]=compare(num, i); } } int compare(int a, int b) { if (sub(a, b) >= 0) return 1; else return 0; } int sub(int a, int b) { return ab; } Implement the C code snippet above in MIPS assembly language. Use $s0 to hold the variable i. Be sure to handle the stack pointer appropriately. The array is stored on the stack of the setArray function (i.e. when setArray function is called it allocates space for the array on the stack.). Solution I have used the Test function i.e ain function to get this module of the fuction definition working. Thw below is the MIPS code for the complete program:- .file 1 \"\" .section .mdebug.abi32 .previous .gnu_attribute 4, 1 .abicalls .text .align 2 .globl _Z3subii $LFB0 = . .set nomips16 .ent _Z3subii .type _Z3subii, @function _Z3subii: .frame $fp,8,$31 # vars= 0, regs= 1/0, args= 0, gp= 0 .mask 0x40000000,-4 .fmask 0x00000000,0 .set noreorder .set nomacro addiu $sp,$sp,-8 $LCFI0: sw $fp,4($sp) $LCFI1: move $fp,$sp movz $31,$31,$0 $LCFI2: sw $4,8($fp) sw $5,12($fp) lw $3,8($fp) lw $2,12($fp) nop subu $2,$3,$2 move $sp,$fp lw $fp,4($sp) addiu $sp,$sp,8 j $31 nop .set macro .set reorder .end _Z3subii $LFE0: .size _Z3subii, .-_Z3subii .align 2 .globl _Z7compareii $LFB1 = . .set nomips16 .ent _Z7compareii .type _Z7compareii, @function _Z7compareii: .frame $fp,32,$31 # vars= 0, regs= 2/0, args= 16, gp= 8 .mask 0xc0000000,-4 .fmask 0x00000000,0 .set noreorder .cpload $25 .set nomacro addiu $sp,$sp,-32 $LCFI3: sw $31,28($sp) $LCFI4: sw $fp,24($sp) movz $31,$31,$0 $LCFI5: move $fp,$sp $LCFI6: .cprestore 16 sw $4,32($fp) sw $5,36($fp) lw $4,32($fp) lw $5,36($fp) lw $2,%got(_Z3subii)($28) nop move $25,$2 jalr $25 nop lw $28,16($fp) nor $2,$0,$2 srl $2,$2,31 andi $2,$2,0x00ff beq $2,$0,$L4 nop li $2,1 # 0x1 b $L5 nop $L4: move $2,$0 $L5: move $sp,$fp lw $31,28($sp) lw $fp,24($sp) addiu $sp,$sp,32 j $31 nop .set macro .set reorder .end _Z7compareii $LFE1: .size _Z7compareii, .-_Z7compareii .align 2 .globl _Z8setArrayi $LFB2 = . .set nomips16 .ent _Z8setArrayi .type _Z8setArrayi, @function _Z8setArrayi: .frame $fp,88,$31 # vars= 48, regs= 3/0, args= 16, gp= 8 .mask 0xc0010000,-4 .fmask 0x00000000,0 .set noreorder .cpload $25 .set nomacro addiu $sp,$sp,-88 $LCFI7: sw $31,84($sp) $LCFI8: sw $fp,80($sp) movz $31,$31,$0 $LCFI9: sw $16,76($sp) $LCFI10: move $fp,$sp $LCFI11: .cprestore 16 sw $4,88($fp) sw $0,24($fp) b $L8 nop $L9: lw $16,24($fp) lw $4,88($fp) lw $5,24($fp) lw $2,%got(_Z7compareii)($28) nop move $25,$2 jalr $25 nop lw $28,16($fp) move $3,$2 sll $2,$16,2 addiu $4,$fp,24 addu $2,$4,$2 sw $3,4($2) lw $2,24($fp) nop addiu $2,$2,1 sw $2,24($fp) $L8: lw $2,24($fp) nop slt $2,$2,10 andi $2,$2,0x00ff bne $2,$0,$L9 nop move $sp,$fp lw $31,84($sp) lw $fp,80($sp) lw $16,76($sp) addiu $sp,$sp,88 j $31 nop .set macro .set reorder .end _Z8setArrayi $LFE2: .size _Z8setArrayi, .-_Z8setArrayi .align 2 .globl ma.

Consider the following C code snippet C codevoid setArray(int.pdf

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Describe a mechanism that the body uses to prevent a mutation from becoming an abnormal cell. Solution The body has several mechanisms in place to ensure that mutations do not turn lethal. The first of those steps is the proof reading activity of the DNA polymerase which reads the DNA strand and corrects the sequence as it synthesizes the new DNA fragment. Thus, the daughter cells contain the original non-mutated sequence. Secondly, organisms are mostly diploid i.e. they contain two sets of chromosomes which includes two forms of the same gene. The mutation would be lethal or cause loss of function only when the other copy of the gene is also mutated (recessive mutation). Also, mutant cells are usually cleared by the immune system which can detect and produce specific antibodies against these cells..

Describe a mechanism that the body uses to prevent a mutation from be.pdf

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classify domian of life differences Solution Classify the domain of life differences: the domain of life is mainly classified based on the following differences 1. Ribosomal RNAs (rRNA) inside the cell 2. The structure of cell membrane with lipid structure 3. Cell sensitivity to antibiotics Three domains of life such as \"prokaryota\", \"archae\" and \"eukaryota (kingdom plantae, kingdom Animalia)\" Explanation: The mitochondria, histone (nucleosomes) are not evolved in archae domain & prokaryote domain but evolved in the “kingdom of eukaryotes”, as an evolutionary adaptation difference. This is a major differential aspect among the three domains of life in their classification according to endo-symbiotic theory. The aspects of cell wall are constructed differently in all three domains of life such as bacteria, archaic species, and eukaryotes. The archae has S -layer in their cell wall but do not possess peptidoglygan layer as bacteria do. Lipopolysaccharide, peptidoglycan layers are absent in eukaryotic cell walls and their mechanism of synthesis is absent in eukaryotes. Plant cells are eukaryotic cells and contain cell wall but animal cells do not contain cell wall. Bacteria are prokaryotes & do not possess histone proteins but eukaryotes contain histone proteins associated with DNA. Any morphological characteristics of group of species are completely different to that of three domains of life such as \"prokaryota\", \"archae\" and \"eukaryota (kingdom plantae, kingdom Animalia)\". These morphological features are completely different to the ancestral species and leading to new phylum evolvement. Internal structural features, embryological development, nucleus structure, genome nature (RNA or DNA), type of cellular respiration & energy synthesis. The major domian of life differences with morphological relations in the evolution of vertebrate animal life (at the phylum level & domain level) from invertebrates (from four -limb species): The major features of embryonic vertebrate zygote formation & followed by eight-cell stage of embryo finally blastula stage and gastrulation. This stage is absent in the archae & prokaryotes species. There are total 2 types of groups of species that are divided based on presence of morphological innovation that is \"presence of vertebral column\" in animal domain & prokaryote. Those living species which do not have vertebra\" are referred as \"invertebrates\" and those which possess vertebrae\" are referred as “chordates\" or \"vertebrates\"..

classify domian of life differencesSolutionClassify the domain.pdf

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Compare and contrast the development of a WBS in traditional project management versus development in an agile environment. Solution An Agile based WBS is organized around units of end user functionality - commonly referred to as user stories. Each user story represents the work to deliver that increment of functionality. In Agile based WBS, Features are decomposed into Epics (large user stories) and Epics into User Stories. User stories are decomposed to represent functionality that can be completed within one iteration - typically one to four weeks depending on the team/project. Each leaf level WBS element represent the working software to be produced through analysis, design, coding and testing of the requirement in the user story. Because each user story is atomic, stories can be re-prioritized, added and or removed without changing the scope, provided that the total size/effort of stories within the scope of the project does not change. Additionally, iterative development facilitates multiple releases (deployments) within a project. So, user stories are typically scheduled within an iteration and a release. Note, that until an iteration has started, the stories within that iteration can be changed, as part of iteration planning. In case of traditional project management WBS is organized around the primary products of the project (or planned outcomes) instead of the work needed to produce the products (planned actions). Since the planned outcomes are the desired ends of the project, they form a relatively stable set of categories in which the costs of the planned actions needed to achieve them can be collected. A well-designed WBS makes it easy to assign each project activity to one and only one terminal element of the WBS. In addition to its function in cost accounting, the WBS also helps map requirements from one level of system specification to another, for example a requirements cross reference matrix mapping functional requirements to high level or low level design documents..

Compare and contrast the development of a WBS in traditional project.pdf

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B. You are an evolutionary biologist in a heated argument with a creation scientist who makes the statement—\"Mathematically, it is inconceivable that anything as complex as a protein, let alone a living cell or a human, could spring up by chance\". 1. Make a supporting statement for chance. 2. Now play the other side and respond to the supporting statement as you think a creation scientist would respond.? Solution You are an evolutionary biologist in a heated argument with a creation scientist who makes the statement— \"Mathematically, it is inconceivable that anything as complex as a protein, let alone a living cell or a human, could spring up by chance\". 1. Make a supporting statement for chance. Chance: It plays a part in evolution (for example, random mutations in a gene pool can give rise to few new traits), but evolution does not depend ONLY on chance in creating new organisms, proteins or any other entities. This is quite opposite to natural selection. Natural selection is the principal which is known for the mechanism of evolution. By chance some mutation caused in the gene pool and it gave to new acceptable traits. If they found adjustable to the environment, they continues as a new species. The \"undesirable\" (nonadaptive) ones will be thus elimated from that gene pool. As long as the forces of selection stay constant, natural selection pushes the evolution in only one direction and produce new desired structures. Thus chance plays a part in evolution. Nature plays a major role in this. Fossil studies support this theory/ process. Where we can see many intermediate or failed species in between major species. Archaeopteryx, a link between reptiles and birds, extinctinted soon is a good example of intermediate species. Haldane and Oparin proposed a theory of the origin of life on Earth through the transformation, during the gradual chemical evolution of particles that contain carbon in the primordial soup shows an example of Evolution. 2. Now play the other side and respond to the supporting statement as you think a creation scientist would respond? As a creation scientist i do not agree with science and evolution part, all this is created by God and thus it follows biblical roots. All that it is in the nature is creation of God, all of a sudden it appeared on the earth. According to us GOD created all species like birds,animals etc., later he wanted to create a genius brained animal, so he created humans. And God creation according to us all this is due to CHOICE not because of CHANCE.Thus it involved logical and systematic process..

B. You are an evolutionary biologist in a heated argument with a cre.pdf

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Assume Hashtable is a simple array of size 8, with indices 0..7. Numeric keys are mapped by a Hashfunction that gives the mod(8,n) value for any key “n” yielding the Hashtable index for that key (0..7). A Hashtable entry is null unless a key exists with that index as its hashed index; if so, the Hashtable entry points to the first node of a linked list of keys with that hash index. The last node on this linked list has a null reference for the next referenced node. Assume the occurrence of a linked list node is represented by the object “Node” and its “Data” and “NextRef” attributes. Create pseudocode for the \"Remove\" operation. Solution Removal of a Node from a Singly Linked List To complete deletion of firstNode in the list we have to change Head pointing to NextRef of firstNode. Pseudocode: firstNode = Head Head = firstNode->NextRef free firstNode Complexity: Time Complexity: O(1) Space Complexity: O(1) Sourcecode: int delNodeData(int num) { struct Node *prev_ptr, *node; node=Head; while(cur_ptr != NULL) { if(cur_ptr->Data == num) { if(cur_ptr==Head) { Head=cur_ptr->NextRef; free(node); return 0; } else { prev_ptr->NextRef=node->NextRef; free(node); return 0; } } else { prev_ptr=node; node=node->NextRef; } } printf(\"\ Element %d is not found in the List\", num); return 1; } Deleting Last Node in the Singly Linked List Traverse to Last Node in the List using two pointers namely prevNode and node. Once node reaches the last Node in the list point NextRef in prevNode to NULL and free the node. Pseudocode: node = head forever: if node->NextRef == NULL break prevNode = node node = node->NextRef prevNode->NextRef = NULL free Node Complexity: Time Complexity: O(n) Space Complexity: O(1) Deleting Node from position \'p\' in the List To delete a Node at the position \'p\' we have to first traverse the list until we reach the position \'p\'. For this case have to maintain two pointers namely prevNode and Node. Since Singly Linked Lists are uni-directional we have to maintain the information about previous Node in prevNode. Once we reach the position \'p\' we have to modify prevNode NextRef pointing to Node NextRef and free Node. Pseudocode: curNode = head curPos = 1 forever: if curPos == P || Node == NULL break prevNode = node node = node->NextRef curPos++ if noode != NULL: prevNode->NextRef = node->NextRef free curNode Complexity: Time Complexity: O(n) worst case Space Complexity: O(3) Sourcecode: int delNodeLoc(int loc) { struct Node *prev_ptr, *node; int i; node=Head; if(loc > (length()) || loc <= 0) { printf(\"\ Deletion of Node at given location is not possible\ \"); } else { // If the location is starting of the list if (loc == 1) { Head=node->NextRef; free(node); return 0; } else { for(i=1;iNextRef; } prev_ptr->NextRef=node->NextRef; free(node); } } return 1; }.

Assume Hashtable is a simple array of size 8, with indices 0..7. Num.pdf

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Amon the following, which has the lowest levels of dissolved iron? a. Atlantic Ocean b. Indian Ocean c. Pacific Ocean d. The temperate regions of all oceans Solution The correct option is Pacific ocean because of the low dust sedimentation and iron is readily soluble in low oxygen regions and the continental margins. It was believed that iron remained in the localized areas but didn\'t contribute to the overall iron content of the ocean. As it is explained by Nielsen, as soon as the iron-rich fluids hit the seawater with high oxygen content, the iron gets dumped out and is not contributed to the overall amount of the dissolved iron..

Amon the following, which has the lowest levels of dissolved iron.pdf

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A box contains 10 red balls and 40 black balls. Two balls are drawn .pdf

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As late as 1992, the United States was running budget deficits of nearly $300 billion. During the remainder of the 1990\'s, deficits declined and became surpluses. As the new century began, these surpluses again turned into deficits. •Explain the decline in deficits and subsequent surpluses in the late 1990\'s. •Explain the return to deficit spending since the turn of the century. •Consider the causes of the deficits and surpluses and provide your own insight as to whether these surpluses or deficits have a \"positive\" or \"negative\" effect on our economy Solution As the 1990s came to an end, virtually all financial-market observers and politicians rejoiced that budget deficits were gone and that the U.S. had entered a period of fiscal stability. Politicians on both sides of the aisle viewed the new world of budget surpluses as an opportunity to save Social Security, to cut Federal debt, and to lower taxes. Budget deficits or budget surpluses are neither good nor bad, they are fiscal policy options that can stabilize or de-stabilize an economy. The U.S. economy, under Reagan, Bush, and early Clinton, experienced higher not lower rates of economic growth at a time when budget deficits were at record levels and tax rates were on the decline. Over the past 20 years a period of record budget deficits the stock market increased from 950 to over 11,000, providing ample evidence that budget deficits are not necessarily bad for an economy.On the other hand, budget surpluses are not necessarily good. As the budget surpluses began to expand dramatically in late 1999 and into 2000, the stock market appeared to be reversing the record up-trend of the 1980s and \'90s. One possible reason for the reversal: the growing budget surplus was sapping the savings of the private sector. Consumers were able to maintain their spending only through expanding debt to record level. In the late 90s, the confluence of strong economic growth, decelerating federal government spending, and substantial increases in tax rates (revenues) had the unexpected effect of producing enormous budget surpluses. As unexpected budget surpluses burst on the scene in the late 1990s, most mainstream economists weren’t sensitive to the contractionary impact that surpluses can have on the economy. The reason was simple that none of their economic models ever included the forecast of a substantial surplus. Ronald Reagan’s was the vice president of the USA for eight years and knew firsthand how difficult it was to bring the federal budget deficit down from 6 percent of GDP in 1983 to 2.8 percent in Reagan’s last budget. Bush knew that the many budget deals of the 1980s had been hard fought and always involved higher taxes as part of the deficit reduction, Reagan signed into law 11 major tax increases that raised taxes by $133 billion in 1988 or 2.7 percent of GDP.Bush administration economists believed that an easier monetary policy was the key to stimulating growth. But Federal Reserve Board.

As late as 1992, the United States was running budget deficits of ne.pdf

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A bacteriophage population is introduced to a bacterial colony that is in stationary phase with rapidly depleting nutrients. Appraise the following scenarios and select which one would favor long-term survival of the bacteriophage. A. The bacteriophages infect the remaining viable cells and integrate their genetic material into the host cell genome thereby preserving their genome. B. The bacteriophages do not infect the bacterial cells since they are dying; they remain inert outside of the cells. C. The bacteriophages infect the bacterial cells and go through all stages of development apart from release, thus keeping the progeny phage from entering an inhospitable environment. Solution Answer : C. The bacteriophages infect the bacterial cells and go through all stages of development apart from release, thus keeping the progeny phage from entering an inhospitable environment..

A bacteriophage population is introduced to a bacterial colony that .pdf

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Xavier and Yolanda both have classes that end at noon and they agree to meet every day after class. They arrive at the coffee shop independently. Xavier\'s arrival time is X and Yolanda\'s arrival time is Y, where X and Y are measured in minutes after noon. The individual density functions are given. (Xavier arrives sometime after noon and is more likely to arrive promptly than late. Yolanda always arrives by 12:10 PM and is more likely to arrive late than promptly.) After Yolanda arrives, she\'ll wait for up to 20 minutes for Xavier, but he won\'t wait for her. Find the probability that they meet. (Round your answer to three decimal places.) f1(x) = {e^-x if x gt o and o if x Solution.

Xavier and Yolanda both have classes that end at noon and they agree .pdf

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1. Low platelet count is a recessively inherited trait. Reevaluation of the genetic screen indicates that Nikoleta is homozygous dominant normal for a trait that causes low platelet count and her husband is homozygous recessive. If the F1 offspring has normal platelets and beta-thalassemia (15pts). a. What are the parental gametes? (2pts) b. Based on the F1 progeny phenotype given and the parental gametes, write the two possible genotypes. (2 pts) c. What is the gamete frequency if the two genotypes in (b) were crossed with each other? (4pts)? (4pts) d. What is the phenotype frequency for the cross in (c)? SHOW ALL WORK. (7 pts) Solution Both the conditions are recessive. Nikoleta: Homozygous dominant for trait that causes low platelet, and homozygous recessive for beta thalessemia. Genotype is LLbb Husband: Homozygous recessive for low platelet count, and heterozygous dominant for beta- thalessemia. Genotype is llBb The possible gametes produced by these parents: Nikoleta: Lb Husband: lB, lb Possible genotype of the F1 offspring: LlBb or Llbb Since the F1 offspring are given to be affected with beta-talassemai, the F1 offspring genotype should be Llbb. Possible gametes are LB, lb, lB, and lb (if genotype is LlBb) / Lb, lb (if genotype is Llbb). Therefore, the frequency of the gametes is LB = 1, lb = 2, lB =1, and Lb = 2 The following is the phenotype frequency: LLBb, LlBb, LlBb (Normal platelet count and no beta thallasemia) = 3 LLbb, Llbb, Llbb (Normal platelet count and beta thallasemia) = 3 llBb (Low platelet count and no beta thallasemia) = 1 llbb (Low platelet count and beta thallasemia) = 1 Thus phenotypic ratio stands as: Normal platelet count and no beta thallasemia : Normal platelet count and beta thallasemia : Low platelet count and no beta thallasemia : Low platelet count and beta thallasemia = 3 : 3 : 1 : 1.

1. Low platelet count is a recessively inherited trait. Reevaluation.pdf

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1. An important contribution of Fiedler\'s research on the contingency model of leadership is that it: a. uses weights which are determined in an arbitrary manner in order to determine situation favorability. b. goes beyond the notion of leadership styles to try to show how styles fit a situation. c. recommends medium least preferred coworker leaders as they are more effective in many situations. d. is clear about how the model will work over time. 2. Which of the following characteristics was explored by the studies evaluating the characteristics of the leader-member exchange relationship? a. Income b. Educational qualification c. Value agreement d. Recruitment process 3. Which of the following is a characteristic of entrepreneurial leaders? a. They are concerned with innovation and creativity. b. They are interested in maintaining the status quo. c. They avoid taking risks and are unwilling to stretch themselves. d. They tend to be dependent on their followers. 4. An important aspect of leadership is: a. dictating changes required in an organization. b. maintaining a passive relationship with followers. c. maintaining the status quo and not creating change. d. influencing others to come together around a common vision. 5. It becomes easier for people to see their own opportunities for leadership when they: a. start seeing leaders as individuals who are above others. b. understand that the concept of leadership cannot continue to develop. c. stop equating leadership with greatness and public visibility. d. comprehend that leadership involves maintaining the status quo. 6. Erica, a team manager, was known for heading projects with diligence and conscientiousness. However, she was insensitive and excessively ambitious. She did not take feedback in a constructive manner. As a result, this prevented her from moving forward in terms of her career since she had problems communicating with her team members and motivating them. In the context of the given scenario, which of the following phenomenon does Erica experience? a. Communication apprehension b. Fear-based motivation c. Paradigm change d. Derailment Name: Class: Date: Quiz 1: Chp 1-3 (Fall 2015) 2 7. Which of the following statements is true of the contingency approach to leadership? a. A leadership style that works in one situation will work in all situations. b. These approaches seek to investigate traits that can improve performance in all situations. c. The needs, maturity, and cohesiveness of followers have been identified as a key contingency. d. These approaches hold that behavior that is effective in some circumstances is effective under all conditions. 8. Stage two in the development of the individualized leadership theory explored the _____ in more detail. a. vertical dyad linkage model b. partnership building c. leader–member exchange d. entrepreneurship model 9. Which of the following statements is true of leaders who work collaboratively? a. They learn to keep the lines of commu.

1. An important contribution of Fiedlers research on the contingen.pdf

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^^^ Discuss about Header Node? And also write a program for unordered single linked list and linked implementation of QUEUE? Solution Header node: Sometimes it is desire to keep an extra node at the front of a List. Such a node doesn’t represent an item in the List and is called a Header Node or a List Header. The Information field of such Header node might be unused but the Next field maintains the first node address. More often the information of such a node could be used to keep Global Information about the Entire List. One of the applications of Header node is the information portion of the Header node contains the number of nodes (not including Header) in the List. In this Structure the Header must be adjusted the total number of nodes when we apply addition or deletion operation on the List. We can directly obtained the total number of nodes without traversing the entire List. By using this application, we can develop Data Structure Stack easily. Since Header node information field maintains the number of elements that the Stack contains. Another application is we can simply develop the Data Structure Queue. Until now, two external pointers, Front & Rear, were necessary for a List to represent a Queue. However, now only a single external pointer is sufficient to maintain a Queue that is Head node. Because the Header node next field behave like a Front and information field of Header node maintains last node address so that it is behave like a Rear . Another possibility for the use of the information portion of a List Header is as a pointer to a Curr node in the List during a traversal process. This would be eliminate the node for an external pointer during traversal. #include #include #include #include struct node { char info; struct node *next; }; typedef struct node sl; void display(sl *); sl * create(sl *root) { sl *curr,*new; char val; printf(\"\ Enter $ to STOP --- Otherwise Continue : \"); fflush(stdin); scanf(\"%c\",&val); while(val != \'$\') { new=(sl *) malloc(sizeof(sl)); new->info = val; new->next = NULL; if(root == NULL) root = new; else { curr = root; while( curr->next != NULL ) curr = curr->next; curr->next = new; } printf(\" Enter $ to STOP --- Otherwise Continue : \"); fflush(stdin); scanf(\"%c\",&val); } printf(\"\ The Created Linked List Is \ \"); display(root); return root; } sl * delete(sl *root,char val) { sl *temp,*curr,*rear; curr = root; rear = NULL; while( curr != NULL && val != curr->info ) { rear = curr; curr = curr->next; } if(curr == NULL) { if( rear == NULL) { printf(\"\ Linked List is Empty\"); printf(\"\ Deletion is not Possible \"); } else { printf(\"\ deleted Node does not exist in List\"); printf(\"\ Deletion is not Possible \"); } } else { if( rear == NULL) { temp = root; root = root->next; } else { temp = curr; rear->next = curr->next; } printf(\"\ Node is deleted \"); printf(\"\ Deleted Node information is %c\",temp->info); free(temp); } display(root); return root; } sl * insert(sl *root,char val) {.

^^^ Discuss about Header Node And also write a program for unorder.pdf

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Write the code above and the ones below in netbeans IDE 8.1 3. (Eliminate duplicates) Write a method that returns a new array by eliminating the duplicate values in the array using the following method header: public static int[] eliminateDuplicates(int[] list) Write a test program that reads in ten integers, invokes the method, and displays the result. Here is the sample run of the program: Enter ten numbers: 1 2 3 2 1 6 3 4 5 2 [Enter] The distinct numbers are: 1 2 3 6 4 5 4. (Sort students) Write a program that prompts the user to enter the number of students, the students’ names, and their scores, and prints student names in decreasing order of their scores. Solution Please follow the code and comments for description : a) CODE : import java.util.Scanner; public class MeanSD { public static double deviation (double[] x) { // method to return the deviation double sum = 0, mean = 0, deviation = 0; // required initialisations double[] temp = new double[10]; // temporary array for (int i = 0; i < 10; i++) //calculate the standard deviation { temp[i] = Math.pow((x[i] - mean), 2); // assigning the value to the array sum += temp[i]; // adding up the values } mean = sum / 10; // getting the mean deviation = Math.sqrt(mean); // calculating the deviation return deviation; // returning the deviation } public static double mean (double[] x) { // method to return themean of the values entered double sum = 0, mean; // required initialisations for (int i = 0; i < 10; i++) //Take input in the array { sum += x[i]; //sum of all elements } mean = sum / 10; // calculating the mean return mean; // returning the mean value } public static void main(String[] args) { // driver method System.out.println(\"Enter the 10 numbers.\"); // prompt to enter the data Scanner in = new Scanner(System.in); // scanner class to get the data from the user double[] arr = new double[10]; // array that saves the data double sum = 0, mean = 0, deviation = 0; // required initialisations for (int i = 0; i < 10; i++) //Take input in the array { System.out.print(\"Enter a number : \"); // prompt to enter the number arr[i] = in.nextDouble(); // getting the data from the console } mean = mean(arr); // calling the method to return the mean value System.out.println(\"Mean : \" + mean); //Display mean of all elements deviation = deviation(arr); // calling the method to return teh deviation value System.out.println(\"Deviation : \" + deviation); // display the result } } OUPTPUT : Enter the 10 numbers. Enter a number : 1 Enter a number : 2 Enter a number : 3 Enter a number : 4 Enter a number : 5 Enter a number : 6 Enter a number : 7 Enter a number : 8 Enter a number : 9 Enter a number : 10 Mean : 5.5 Deviation : 6.2048368229954285 b) CODE : import java.util.Scanner; public class MyDuplicateElements { public static int[] eliminateDuplicates(int[] input) { // method to remove the duplicates int j = 0; // required initialisations int i = 1; //return if the array length is less than 2 if (input..

Write the code above and the ones below in netbeans IDE 8.13. (Eli.pdf

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Which statement about gills is false? They contain the environmental medium. The respiratory membrane has a large surface area. They are surrounded by the environmental medium. Solution Which statement about gills is false? They contain the environmental medium.---------answer The respiratory membrane has a large surface area. They are surrounded by the environmental medium. ====================================================== EXPLANATION GILLS Gills significantly increment the surface area for exchange of gases. They happen in an assortment of creature gatherings including arthropods , annelids, fish, and creatures of land and water. Gills normally are convoluted outgrowths containing veins secured by a thin epithelial layer. Commonly gills are sorted out into a progression of plates and might be interior (as in crabs and fish) or outer to the body (as in a few creatures of land and water). Gills are extremely proficient at expelling oxygen from water: there is just 1/20 the measure of oxygen present in water as in a similar volume of air. Water streams over gills in one course while blood streams the other way through gill vessels. This countercurrent stream augments oxygen exchange..

Which statement about gills is falseThey contain the environmenta.pdf

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Why do Boolean functions play a central role in hardware architectures? Describe Composite Gates. Describe Multiplexor’s and Demultiplexor’s? What is the importance of the use of the selection bit and the data bits? Solution In electronics, a multiplexer (or mux) is a device that selects one of several analog or digital input signals and forwards the selected input into a single line. A multiplexer of 2n inputs has n select lines, which are used to select which input line to send to the output.Multiplexers are mainly used to increase the amount of data that can be sent over the network within a certain amount of time and bandwidth. A multiplexer is also called a data selector. Multiplexers can also be used to implement Boolean functions of multiple variables. An electronic multiplexer makes it possible for several signals to share one device or resource, for example one A/D converter or one communication line, instead of having one device per input signal. Conversely, a demultiplexer (or demux) is a device taking a single input signal and selecting one of many data-output-lines, which is connected to the single input. A multiplexer is often used with a complementary demultiplexer on the receiving end. An electronic multiplexer can be considered as a multiple-input, single-output switch, and a demultiplexer as a single-input, multiple-output switch.The schematic symbol for a multiplexer is an isosceles trapezoid with the longer parallel side containing the input pins and the short parallel side containing the output pin.The schematic on the right shows a 2-to-1 multiplexer on the left and an equivalent switch on the right. The wire connects the desired input to the output..

Why do Boolean functions play a central role in hardware architectur.pdf

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Which of the following is not a primary tissue category? a. Epithelial b. Simple Squamous c. Muscle d. Nerve What type of tissue in the liver functions to secrete bile? a. Epithelial b. Connective c. Muscle d. Nerve What type of tissue in the pancreas holds epithelial cells in place? a. Epithelial b. Connective c. Muscle d. Nerve What would be a good function of stratified squamous epithelium? a. Absorption b. Secretion c. Contraction d. Protection Where would you find transitional epithelium? a. The trachea b. The lungs c. The uterus d. The skin If I saw multiple layers of tall cells, how could I describe them? a. Simple Squamous b. Stratified cuboidal c. Stratified columnar d. Simple columnar Solution 1. b. Simple squamous The primary tissues are - Epithelial, connective, muscular and nervous. 2. a. Epithelial Bile is secreted by Cholangiocytes which are epithelial cells. 3. b. Connective Connective tissue in pancreas holds the epithelial cells. 4. d. Protection The major function of stratified squamous epithelium is protection.

Which of the following is not a primary tissue category a. Epitheli.pdf

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What does this figure depict? hydrogen bonding ionization, followed by ionic bonding nonpolar covalent bonding formation of a polar covalent bond Solution ANS: B. Ionization, followed by ionic bonding. Explanation: Ionic bonds: In this electrons are transferred from the metal atom to non-metal atom. Oppositely charged ions are formed which are held together by electro static attraction. The electrons on outside of the sodium are transferred to the chlorine molecule, so both atoms have a full outer shell and became ions..

What does this figure depict hydrogen bonding ionization, followed.pdf

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This problem illustrates one of the possible pitfalls of blindly applying numerical methods without paying attention to the theoretical aspects of the differential equation itself. Consider the equation ty\' + 3y - 9t^2 = 0. Use the MATLAB program myeuler.m from Chapter 8 to compute the Euler Method approximation to the solution with initial condition y(-0.5) = 3.15, using step size h = 0.2 and n = 10 steps The program will generate a lot of ordered pairs (x_i, .y_i). Use plot to graph the piecewise linear function connecting the points (x_i, y_i). Now modify the program to implement the Improved Euler Method. Can you make sense of your answers? Next, use ode45 to find an approximate solution on the interval (-0.5, 0.5), and plot it with plot. Print out the values of the solution at the points -0.06: 0.02: 0.06. What is the interval on which the approximate solution is defined? Solve the equation explicitly and graph the solutions for the initial conditions y(0) = 0, y(-0.5) = 3.15, y(0, 5) = 3.15, y(-0.5) = -3.45, and y(0.5) = -3.45. Now explain your results in (a)-(c). Could we have known, without solving the equation, whether to expect meaningful results in parts (a) and (b)? Why? Can you explain how ode45 avoids making the same mistake? Solution clc clear all close all x=input(\'Enter the first sequence: \'); l1=input(\'Enter the lower limit: \'); u1=input(\'Enter the upper limit: \'); x1=l1:1:u1; h=input(\'Enter the second sequence: \'); l2=input(\'Enter the lower limit: \'); u2=input(\'Enter the upper limit: \'); h1=l2:1:u2; l=l1+l2; u=u1+u2; n=l:1:u; s=numel(n); i=1; for i=1:s y(i)=0; for k=1:numel(x) if (i+1-k)<=0 y(i)=y(i)+(x(k)*0); else if (i+1-k)>numel(h) y(i)=y(i)+(x(k)*0); else y(i)=y(i)+(x(k)*h(i+1-k)); k=k+1; end end end i=i+1; end disp(y); subplot(2,2,1);stem(x1,x); title(\'First sequence\');xlabel(\'n\');ylabel(\'x(n)\'); subplot(2,2,2);stem(h1,h); title(\'Second Sequence\');xlabel(\'n\');ylabel(\'h(n)\'); subplot(2,2,[3 4]);stem(n,y); title(\'Convoluted sequence\');xlabel(\'n\');ylabel(\'y(n)\'); Comment only X = input(\'Enter x: \'); %input vector X and H H = input(\'Enter h: \') ; LenX = length(X); %defining their lenghts LenH = length(H); y = zeros(1,LenX+LenH); %defing vector y of zeroes and of size % lenth of X + length of H t = zeros(1,LenH); % definign a vector t of same length as H for i = 1:LenH+LenX-1 % Running a for loop from 1 to length of Y -1 if i<=LenX % till I IS Lesser then length of X i.e overlap about to begin t(1)= X(i); % put x(i) on t(1) later it is shifted forwards in the vector t i.e. later t(2)=t(1) for j = 1:LenH % in the if condition a for loop from 1 to length of H y(i) = y(i) + H(j)*t(j); % summing for all H(j)*t(j) and putting it at y(1) or y(2) or Y(i) in first iteration % i.e. for i=1 only firt multiplication would % be non zero rest all zeroes. end for k = LenH:-1:2 % shifting old value of t(i) to t(i+1) now there would me 1+ non zeroes values in t % this cycle would continue until i is lesse.

This problem illustrates one of the possible pitfalls of blindly appl.pdf

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