Elements or steps of acceptance testingSolutionElements or ste.pdf

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Elements or steps of acceptance testing Solution Elements or steps of acceptance testing.

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Long term capital = Common stock + shareholders equity + Retained ea.pdf

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hmmSolutionhmm.pdf

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Form1.cs using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; using System.Drawing; using System.Linq; using System.Text; using System.Threading.Tasks; using System.Windows.Forms; namespace LotteryGUI { public partial class Form1 : Form { private int[] iGuesses = new int[3]; // declared at form level private int[] iSortedGuesses = new int[3]; public Form1() { InitializeComponent(); } private void textBox1_Leave(object sender, EventArgs e) { if (textBox1.Text == \"\") { MessageBox.Show(\"You haven\'t chosen your first number yet\"); textBox1.Focus(); return; } int iGuess; int.TryParse(textBox1.Text, out iGuess); if (iGuess < 1 || iGuess > 4) { MessageBox.Show(\"You have entered an incorrect number please try again\"); textBox1.Focus(); return; } iGuesses[0] = iGuess; iSortedGuesses[0] = iGuess; } private void textBox2_Leave(object sender, EventArgs e) { if (textBox2.Text == \"\") { MessageBox.Show(\"You haven\'t chosen your second number yet\"); textBox2.Focus(); return; } int iGuess; int.TryParse(textBox2.Text, out iGuess); if (iGuess < 1 || iGuess > 4) { MessageBox.Show(\"You have entered an incorrect number please try again\"); textBox2.Focus(); return; } iGuesses[1] = iGuess; iSortedGuesses[1] = iGuess; } private void textBox3_Leave(object sender, EventArgs e) { if (textBox3.Text == \"\") { MessageBox.Show(\"You haven\'t chosen your third number yet\"); textBox3.Focus(); return; } int iGuess; int.TryParse(textBox3.Text, out iGuess); if (iGuess < 1 || iGuess > 4) { MessageBox.Show(\"You have entered an incorrect number please try again\"); textBox3.Focus(); return; } iGuesses[2] = iGuess; iSortedGuesses[2] = iGuess; } private void button1_Click(object sender, EventArgs e) { if (textBox1.Text == \"\" || textBox2.Text == \"\" || textBox3.Text == \"\") { MessageBox.Show(\"You haven\'t yet chosen three numbers\"); return; } // declaring variables integer type int iNum1; int iNum2; int iNum3; // declaring the constants const int iMATCHONE = 10; const int iMATCHTWO = 100; const int iMATCHTHREE = 1000; const int iMATCHFOUR = 10000; // declaring array int[] iSortedNums = new int[3]; // generating random numbers Random randomnumber = new Random(); iNum1 = randomnumber.Next(1, 5); // number between 1 and 4 iNum2 = randomnumber.Next(1, 5); iNum3 = randomnumber.Next(1, 5); iSortedNums[0] = iNum1; iSortedNums[1] = iNum2; iSortedNums[2] = iNum3; Array.Sort(iSortedNums); // sort random numbers Array.Sort(iSortedGuesses); // sort the guesses label4.Text = String.Format(\"The random numbers are : \" + iNum1 + \", \" + iNum2 + \", \" + iNum3); label5.Text = String.Format(\"The numbers you chose are : \" + iGuesses[0] + \", \" + iGuesses[1] + \", \" + iGuesses[2]); if (iGuesses[0] == iNum1 && iGuesses[1] == iNum2 && iGuesses[2] == iNum3) { MessageBox.Show(\"you won $\" + iMATCHFOUR); } else if (iSortedGuesses[0] == iSortedNums[0] && iSortedGuesses[1] == iSortedNums[1] && iSortedGuesses[2] == iSortedNums[2]) { Message.

Form1.csusing System; using System.Collections.Generic; using .pdf

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Influenza A: The recently raised awareness of the threat of a new influenza pandemic has stimulated interest in the detection of influenza A viruses in human as well as animal secretions. Virus isolation alone is unsatisfactory for this purpose because of its inherent limited sensitivity and the lack of host cells that are universally permissive to all influenza A viruses. Previously described PCR methods are more sensitive but are targeted predominantly at virus strains currently circulating in humans, since the sequences of the primer sets display considerable numbers of mismatches to the sequences of animal influenza A viruses. Therefore, a new set of primers, based on highly conserved regions of the matrix gene, was designed for single-tube reverse transcription-PCR for the detection of influenza A viruses from multiple species. This PCR proved to be fully reactive with a panel of 25 genetically diverse virus isolates that were obtained from birds, humans, pigs, horses, and seals and that included all known subtypes of influenza A virus. It was not reactive with the 11 other RNA viruses tested. Comparative tests with throat swab samples from humans and fecal and cloacal swab samples from birds confirmed that the new PCR is faster and up to 100-fold more sensitive than classical virus isolation procedures. Various Methods of Detection: Specimens. Cloacal swab specimens were collected from ducks (widgeon [Mareca penelope], gadwall [Mareca strepera], and mallard [Anas plathyrhynchos]) at a marshaling lake in Lekkerkerk, The Netherlands, and droppings as well as cloacal swab specimens were collected from geese (greylag goose [Anser anser], white- fronted goose [Anser albifrons albifrons], barnacle goose [Branta leucopsis], and brent goose [Branta bernicla]) in Groningen and Eemdijk, The Netherlands, between 1997 and 1999. Cloacal swab specimens and droppings were collected from shorebirds at Öland, Sweden, in the spring of 1999. Cotton swabs were used for sampling and were subsequently stored in transport medium. Throat swab specimens collected from humans were also stored in transport medium. The samples were stored at 4°C for a few days, at 20°C for less than a week, or at 70°C for extended periods of time. Transport medium consisted of Hanks balanced salt solution supplemented with 10% glycerol, 200 U of penicillin per ml, 200 µg of streptomycin per ml, 100 U of polymyxin B sulfate per ml, 250 µg of gentamicin per ml, and 50 U of nystatin per ml (all from ICN, Zoetermeer, The Netherlands). RNA isolation. RNA was isolated with a high pure RNA isolation kit (Roche Molecular Biochemicals) according to the instructions from the manufacturer, with minor modifications. A 0.2-ml sample was homogenized by vortexing and was subsequently lysed with 0.4 ml of lysis-binding buffer to which poly(A) (Roche Molecular Biochemicals) was added as a carrier to 1 µg/ml. After binding to the column, DNase I digestion, and washing, the RNA was eluted in 50 µl o.

Influenza A The recently raised awareness of th.pdf

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I believe it\'s named Methyl iodide. You wouldn\'t number the C that the Iodine is on because there is only one carbon. Also i think that because there is only one carbon the base name is Iodide instead of the longest carbon chain. Hope this helps! Enjoy. Solution I believe it\'s named Methyl iodide. You wouldn\'t number the C that the Iodine is on because there is only one carbon. Also i think that because there is only one carbon the base name is Iodide instead of the longest carbon chain. Hope this helps! Enjoy..

I believe its named Methyl iodide. You wouldn.pdf

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Data warehousing has quickly evolved into a unique and popular business application class. Early builders of data warehouses already consider their systems to be key components of their IT strategy and architecture. Numerous examples can be cited of highly successful data warehouses developed and deployed for businesses of all sizes and all types. Hardware and software vendors have quickly developed products and services that specifically target the data warehousing market. This paper will introduce key concepts surrounding the data warehousing systems. What is a data warehouse? A simple answer could be that a data warehouse is managed data situated after and outside the operational systems. A complete definition requires discussion of many key attributes of a data warehouse system. Later in Section 2, we will identify these key attributes and discuss the definition they provide for a data warehouse. Section 3 briefly reviews the activity against a data warehouse system. Initially in Section 1, however, we will take a brief tour of the traditions of managing data after it passes through the operational systems and the types of analysis generated from this historical data. Evolution of an application class This section reviews the historical management of the analysis data and the factors that have led to the evolution of the data warehousing application class. Traditional approaches to historical data In reviewing the development of data warehousing, we need to begin with a review of what had been done with the data before of evolution of data warehouses. Let us first look at how the kind of data that ends up in today\'s data warehouses had been managed historically. Throughout the history of systems development, the primary emphasis had been given to the operational systems and the data they process. It is not practical to keep data in the operational systems indefinitely; and only as an afterthought was a structure designed for archiving the data that the operational system has processed. The fundamental requirements of the operational and analysis systems are different: the operational systems need performance, whereas the analysis systems need flexibility and broad scope. It has rarely been acceptable to have business analysis interfere with and degrade performance of the operational systems. Data from legacy systems In the 1970s virtually all business system development was done on the IBM mainframe computers using tools such as Cobol, CICS, IMS, DB2, etc. The 1980s brought in the new mini- computer platforms such as AS/400 and VAX/VMS. The late eighties and early nineties made UNIX a popular server platform with the introduction of client/server architecture. Despite all the changes in the platforms, architectures, tools, and technologies, a remarkably large number of business applications continue to run in the mainframe environment of the 1970s. By some estimates, more than 70 percent of business data for large corporations still resi.

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a) What is the role of Trizol in RNA extraction? Trizol reagent is used for RNA isolation from cells and tissues. This reagent is used as it helps in maintaining RNA integrity during tissue homogenization, while at the same time disrupting and breaking down cells and cell components. The RNA remains only in the aqueous phase(Amrita). b) What was the purpose of performing a reverse transcription reaction. What were the products of thesereaction? Reverse transcription polymerase chain reaction (RT-PCR), a variant of polymerase chain reaction (PCR), is a technique commonly used in molecular biology to detect RNA expression incontrast the traditional PCR is used to exponentially amplify target DNA sequences, (Freeman, 1999). The product of this reaction will be RNA Solution a) What is the role of Trizol in RNA extraction? Trizol reagent is used for RNA isolation from cells and tissues. This reagent is used as it helps in maintaining RNA integrity during tissue homogenization, while at the same time disrupting and breaking down cells and cell components. The RNA remains only in the aqueous phase(Amrita). b) What was the purpose of performing a reverse transcription reaction. What were the products of thesereaction? Reverse transcription polymerase chain reaction (RT-PCR), a variant of polymerase chain reaction (PCR), is a technique commonly used in molecular biology to detect RNA expression incontrast the traditional PCR is used to exponentially amplify target DNA sequences, (Freeman, 1999). The product of this reaction will be RNA.

a) What is the role of Trizol in RNA extractionTrizol reagent is .pdf

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A) odd numbers case: Sequential or linear search is a searching mechanism where every element one by one is searched until the result is found. where l is list and x is our search item. Minimum number of sequential search comparisons = 1 Maximum number of sequential search comparisons = N Average number of sequential search comparisons for successful searches = ½ N To find the middle element we can use (biggest - smallest + 1)/ 2. 2)EVEN number case:. if the array has even number of elements then divide the list into two subparts of equal size name array1 array 2, the element which is immediately after array 1 is considered as mid element. so the complexity even in this case if n/2. 3) If we find the element in 1st position then comparisons = 1 2nd positions then 2. 3rd postion 3..... nth position = n so the average no of comparisons in linear search is (1+2+3+4+5+6.....+n) = n(n+1)/2 Solution A) odd numbers case: Sequential or linear search is a searching mechanism where every element one by one is searched until the result is found. where l is list and x is our search item. Minimum number of sequential search comparisons = 1 Maximum number of sequential search comparisons = N Average number of sequential search comparisons for successful searches = ½ N To find the middle element we can use (biggest - smallest + 1)/ 2. 2)EVEN number case:. if the array has even number of elements then divide the list into two subparts of equal size name array1 array 2, the element which is immediately after array 1 is considered as mid element. so the complexity even in this case if n/2. 3) If we find the element in 1st position then comparisons = 1 2nd positions then 2. 3rd postion 3..... nth position = n so the average no of comparisons in linear search is (1+2+3+4+5+6.....+n) = n(n+1)/2.

A) odd numbers caseSequential or linear search is a searching mec.pdf

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#include #include #include #include /*Structure for storing words received from each thread*/ struct return_val{ char wordlist[100][100]; int count[100]; } *arr; pthread_t *threads; int lines; /*Aggregates Frequency Sum of different threads*/ int aggregate(char *a) { int sum = 0; int i=0; while(1) { if(i==lines) break; int count = 0; while(1) { if(arr[i].count[count] == -1) { break; } if(strcmp(a,arr[i].wordlist[count]) == 0) { sum += arr[i].count[count]; arr[i].count[count] = 0; } count++; } i++; } return sum; } /*Function executed by each thread on separate line*/ void *word_count(void* num) { int *ln = num; unsigned int line_number = *ln; line_number++; char cmd_p1[9] = \"sed -n \'\\0\"; char cmd_p2[2]; sprintf(cmd_p2,\"%d\",line_number); char cmd_p3[14] = \"p\' \'file.txt\'\\0\"; char command[100]; command[0] = \'\\0\'; strcat(command,cmd_p1); strcat(command,cmd_p2); strcat(command,cmd_p3); //usleep(line_number); char cmd[100] = \" | tr [:space:] \'\\\ \' | grep -v \'^\\\\s*$\' | sort | uniq -c | sort\\0\"; strcat(command,cmd); FILE *in; in= popen(command, \"r\"); rewind(in); char buff[50]; int counter = 0; while(fgets(buff,sizeof(buff),in)) { char c=\' \'; int i = 0; int cnt = atoi(buff); arr[line_number-1].count[counter] = cnt; while(c!=\'\\0\') { c=buff[i]; buff[i]=buff[i+6]; i++; } int cnnt = 0; while(c!=\' \') { c = buff[cnnt]; cnnt++; } i=0; while(c!=\'\\0\') { c=buff[i]; buff[i]=buff[i+cnnt]; i++; } sprintf(arr[line_number-1].wordlist[counter],\"%s\",buff); counter++; } arr[line_number-1].count[counter] = -1; fclose(in); return NULL; } int main(void) { FILE *fp; fp = fopen(\"file.txt\",\"r\"); char a[200]; lines = 0; void *status = NULL; while(fgets(a,sizeof(a),fp) !=NULL) { lines++; } fclose(fp); arr = (struct return_val*) calloc(lines, sizeof(struct return_val) * lines); threads = malloc(sizeof(pthread_t)*lines); int i; for(i=0;i= 0) { if(arr[i].count[ctr] > 0) { strcpy(final[final_ctr],arr[i].wordlist[ctr]); final_cnt[final_ctr] = arr[i].count[ctr]; arr[i].count[ctr] = 0; int sum = aggregate(arr[i].wordlist[ctr]); final_cnt[final_ctr] += sum; printf(\"%d %s\",final_cnt[final_ctr],final[final_ctr]); } ctr++; final_ctr++; } i++; if(i==lines) { break; } } free(arr); return 0; } #include #include #include #include /*Structure for storing words received from each thread*/ struct return_val{ char wordlist[100][100]; int count[100]; } *arr; pthread_t *threads; int lines; /*Aggregates Frequency Sum of different threads*/ int aggregate(char *a) { int sum = 0; int i=0; while(1) { if(i==lines) break; int count = 0; while(1) { if(arr[i].count[count] == -1) { break; } if(strcmp(a,arr[i].wordlist[count]) == 0) { sum += arr[i].count[count]; arr[i].count[count] = 0; } count++; } i++; } return sum; } /*Function executed by each thread on separate line*/ void *word_count(void* num) { int *ln = num; unsigned int line_number = *ln; line_number++; char cmd_p1[9] = \"sed -n \'\\0\"; char cmd_p2[2]; sprintf(cmd_p2,\"%d\",line_number); char cmd_p3[14] .

#include stdio.h#include stdlib.h#include string.h#inclu.pdf

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#include #include #include static int read_data(char *buffer, size_t buflen); static int direction_searching(char arr[18][18], char *str, int len, int m, int n, int dx, int dy); static int searchData_from(char arr[18][18], char *str, int len, int m, int n); int main(void) { int m, n, len; char arr[18][18], str[50]; printf(\"Welcome to word search puzzle CTRL-C TO QUIT\ \ \"); FILE *fileData = fopen(\"dataSearch.txt\", \"r\"); if (!fileData) { printf(\"Error occured in opening file\ \"); return 1; } for (n = 0; n < 18; n++) { for (m = 0; m < 18; m++) { fscanf(fileData, \" %c\", &arr[n][m]); arr[n][m] = toupper(arr[n][m]); } } fclose(fileData); printf(\" \"); for (m = 0; m < 18; m++) printf(\"%-2d\", m); printf(\"\ ______________________________________\ \"); for (n = 0; n < 18; n++) { printf(\"%-2d|\", n); for (m = 0; m < 18; m++) printf(\"%c \", arr[n][m]); printf(\"\ \"); } while ((len = read_data(str, sizeof(str))) != EOF) { printf(\"Searching for: [%s]\ \", str); int totalCount = 0; for (n = 0; n < 18; n++) { for (m = 0; m < 18; m++) { if (arr[n][m] == (str[0]) && searchData_from(arr, str, len, m, n)) totalCount++; } } printf(\"Found %s %d times\ \", str, totalCount); } printf(\"\ \"); return 0; } // reading data from file static int read_data(char *buffer, size_t buflen) { printf(\"\ Please enter the word to be searched: \"); if (fgets(buffer, buflen, stdin) == 0) return EOF; size_t len = strlen(buffer); if (buffer[len-1] == \'\ \') buffer[--len] = \'\\0\'; if (len == 0) return EOF; for (size_t i = 0; i < len; i++) buffer[i] = toupper(buffer[i]); return len; } // searching data from array static int searchData_from(char arr[18][18], char *str, int len, int m, int n) { struct yx { int dy; int dx; } puzzleDirection[] = { { +1, 0 }, { -1, 0 }, { +1, +1 }, { -1, +1 }, { 0, +1 }, { 0, -1 }, { -1, -1 }, { +1, -1 }, }; enum { num_directions = sizeof(puzzleDirection) / sizeof(puzzleDirection[0]) }; int totalCount = 0; for (int i = 0; i < num_directions; i++) { if (direction_searching(arr, str, len, m, n, puzzleDirection[i].dx, puzzleDirection[i].dy)) totalCount++; } return totalCount; } // checking in given direction static int direction_searching(char arr[18][18], char *str, int len, int m, int n, int dx, int dy) { for (int i = 1; i < len; i++) { int x = m + i * dx; int y = n + i * dy; if (x < 0 || x >= 18 || y < 0 || y >= 18) return 0; if (arr[y][x] != str[i]) return 0; } printf(\"word Found %s starting at (%d,%d) to (%d,%d)\ \", str, n, m, n + (len - 1) * dy, m + (len - 1) * dx); /* checking words*/ char *pad = \"\"; for (int i = 0; i < len; i++) { int x = m + i * dx; int y = n + i * dy; printf(\"%s%c (%d,%d)\", pad, arr[y][x], y, x); pad = \", \"; } putchar(\'\ \'); return 1; } Solution #include #include #include static int read_data(char *buffer, size_t buflen); static int direction_searching(char arr[18][18], char *str, int len, int m, int n, int dx, int dy); static int searchData_from(char arr[18][18], char *str, int len, int m, int n); in.

#include ctype.h #include stdio.h #include string.hstati.pdf

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10. False. Asexually produced organisms are similar as well as identical to their parents. Because the lack of linkage and crossing over. 11. False. Prokaryotes have single free floating nucleoide. 12. False. Human cells are Diploid. 13. False. During S phase. 14. True. 15. True. 16. False. Prophase 1 and Metaphase 1 crossing over occurs. not in meiosis 2. 17. False. 18. True. 19. False. 20. True. Solution 10. False. Asexually produced organisms are similar as well as identical to their parents. Because the lack of linkage and crossing over. 11. False. Prokaryotes have single free floating nucleoide. 12. False. Human cells are Diploid. 13. False. During S phase. 14. True. 15. True. 16. False. Prophase 1 and Metaphase 1 crossing over occurs. not in meiosis 2. 17. False. 18. True. 19. False. 20. True..

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