P1.javaimport java.util.Scanner;keyboard inputting package pub.pdf

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P1.java import java.util.Scanner;//keyboard inputting package public class P1 {//main class public static void main(String[] args) {//main method Scanner input = new Scanner (System.in); //User will input values System.out.print(\"Enter your weght in pounds:\"); int weight = input.nextInt();//keyboard inputting //User will input values System.out.print(\"Enter your feet and inches:\"); int feet = input.nextInt();//keyboard inputting int inches=input.nextInt(); int totalheightininches=0; totalheightininches=(feet*12)+inches;//converting feet and inches into total inches float bmi; float constant=703f; bmi = (float) (weight /Math.pow(totalheightininches,2))*constant;//bmi calculation //Print out results on display System.out.println(\"BMI is\" + bmi); if(bmi <18.5) System.out.println(\"Underweight\"); else if (bmi <25) System.out.println(\"Normal\"); else if (bmi < 30) System.out.println(\"Overweight\"); else System.out.println(\"Obese\"); } } output Enter your weght in pounds:150 Enter your feet and inches:5 5 BMI is24.95858 Normal Solution P1.java import java.util.Scanner;//keyboard inputting package public class P1 {//main class public static void main(String[] args) {//main method Scanner input = new Scanner (System.in); //User will input values System.out.print(\"Enter your weght in pounds:\"); int weight = input.nextInt();//keyboard inputting //User will input values System.out.print(\"Enter your feet and inches:\"); int feet = input.nextInt();//keyboard inputting int inches=input.nextInt(); int totalheightininches=0; totalheightininches=(feet*12)+inches;//converting feet and inches into total inches float bmi; float constant=703f; bmi = (float) (weight /Math.pow(totalheightininches,2))*constant;//bmi calculation //Print out results on display System.out.println(\"BMI is\" + bmi); if(bmi <18.5) System.out.println(\"Underweight\"); else if (bmi <25) System.out.println(\"Normal\"); else if (bmi < 30) System.out.println(\"Overweight\"); else System.out.println(\"Obese\"); } } output Enter your weght in pounds:150 Enter your feet and inches:5 5 BMI is24.95858 Normal.

P1.java
import java.util.Scanner;//keyboard inputting package
public class P1 {//main class
public static void main(String[] args) {//main method
Scanner input = new Scanner (System.in);
//User will input values
System.out.print("Enter your weght in pounds:");
int weight = input.nextInt();//keyboard inputting
//User will input values
System.out.print("Enter your feet and inches:");
int feet = input.nextInt();//keyboard inputting
int inches=input.nextInt();
int totalheightininches=0;
totalheightininches=(feet*12)+inches;//converting feet and inches into total inches
float bmi;
float constant=703f;
bmi = (float) (weight /Math.pow(totalheightininches,2))*constant;//bmi calculation
//Print out results on display
System.out.println("BMI is" + bmi);
if(bmi <18.5)
System.out.println("Underweight");
else if (bmi <25)
System.out.println("Normal");
else if (bmi < 30)
System.out.println("Overweight");
else
System.out.println("Obese");
}
}
output
Enter your weght in pounds:150
Enter your feet and inches:5 5
BMI is24.95858
Normal

Solution
P1.java
import java.util.Scanner;//keyboard inputting package
public class P1 {//main class
public static void main(String[] args) {//main method
Scanner input = new Scanner (System.in);
//User will input values
System.out.print("Enter your weght in pounds:");
int weight = input.nextInt();//keyboard inputting
//User will input values
System.out.print("Enter your feet and inches:");
int feet = input.nextInt();//keyboard inputting
int inches=input.nextInt();
int totalheightininches=0;
totalheightininches=(feet*12)+inches;//converting feet and inches into total inches
float bmi;
float constant=703f;
bmi = (float) (weight /Math.pow(totalheightininches,2))*constant;//bmi calculation
//Print out results on display
System.out.println("BMI is" + bmi);
if(bmi <18.5)
System.out.println("Underweight");
else if (bmi <25)
System.out.println("Normal");
else if (bmi < 30)
System.out.println("Overweight");
else
System.out.println("Obese");
}
}
output
Enter your weght in pounds:150
Enter your feet and inches:5 5

BMI is24.95858
Normal

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1) MODERN TIMES is a very well prepared movie that describes the miserable conditions of humar resource during the period of industrialization era or machine age worlwide. It is a mix of social drama and comedy. The content is presented in a satire. How did people survived the great depression of the world is being documented here. The conditions werre miserable like acute shortage of work, factory closings, labor strikes and political unrest etc. It is a symbolic representation of what happens to a human being caught up in the mechanical bureaucracy of the intellectual state. 2) Economically it could be understood that what miser situations could depression bring up. Food and jobs comes hard in hand. To which a labor is convict to do under skill jobs as well for survival. At the time of the great depression Americans had to fight for what little they had, and as the movie showed many people went hungry. Modern Times looks toward the future, but not with enthusiasm. Often described as a satire of the machine age, It has in fact a broader theme: the dehumanizing effects of many aspects of modernity, including industrialization, bureaucracy, urbanization, and law enforcement. 3) From the movie a prospect employee should not forget about the social ills and failings of own day even if present age is seems far better than that was shown in the film. A dilbert-like early scene shows the Tramp trying to take a break during work hours in the washroom — only to have a big-screen image of the boss’s head suddenly appear on the washroom wall and order him to quit stalling and get back to work. Modern audiences watching this scene may reflect ruefully on the everyday reality of electronic surveillance in the workplace. Now, obviously there’s nothing new about widespread poverty and want, but unemployment as we know it today is a comparatively recent development — a byproduct of the modern workforce economy that requires everyone to \"get a job.\" 4) The major ethical dilemma consists whether out of povertty stealing is right or wrong. However this movie glorifies the struggle of the poor rather than vilifying them for stealing food to survive. By glorifying the struggle the movie takes on yet another social issue: when the poor steal to eat it is not criminal, just survival. Minor ones include Solution 1) MODERN TIMES is a very well prepared movie that describes the miserable conditions of humar resource during the period of industrialization era or machine age worlwide. It is a mix of social drama and comedy. The content is presented in a satire. How did people survived the great depression of the world is being documented here. The conditions werre miserable like acute shortage of work, factory closings, labor strikes and political unrest etc. It is a symbolic representation of what happens to a human being caught up in the mechanical bureaucracy of the intellectual state. 2) Economically it could be understood that what miser situations could depress.

1) MODERN TIMES is a very well prepared movie that describes the mis.pdf

1) MODERN TIMES is a very well prepared movie that describes the mis.pdf

1) MODERN TIMES is a very well prepared movie that describes the mis.pdf

1. Identify five differences between DNA replication and gene transcription. DNA replication gene transcription. occurs in preparation for cell division, Produces two two copies of DNA, double stranded Uses DNA as template to form DNA Catalyzed by DNA polymerase the end result is two daughter cells. happens in preparation for protein translation. Produces mRNA, single stranded Uses template DNA as a template and forms RNA Catalyzed by RNA polymerase End result is mRNA which is then exported out of the nucleus for translation 2. Identify four differences between DNA and RNA. DNA RNA Deoxyribonucleic acid is the hereditary material in most of the organisms Made of 2-deoxyribose sugar Contains thymine as one of the 4 nitrogen base It is double stranded Ribonucleic acid transfers genetic code needed for the creation of proteins from the nucleus to the ribosome. Made of ribose sugar Has Uracil as the nitrogen base instead of thymine Single stranded DNA replication gene transcription. occurs in preparation for cell division, Produces two two copies of DNA, double stranded Uses DNA as template to form DNA Catalyzed by DNA polymerase the end result is two daughter cells. happens in preparation for protein translation. Produces mRNA, single stranded Uses template DNA as a template and forms RNA Catalyzed by RNA polymerase End result is mRNA which is then exported out of the nucleus for translation Solution 1. Identify five differences between DNA replication and gene transcription. DNA replication gene transcription. occurs in preparation for cell division, Produces two two copies of DNA, double stranded Uses DNA as template to form DNA Catalyzed by DNA polymerase the end result is two daughter cells. happens in preparation for protein translation. Produces mRNA, single stranded Uses template DNA as a template and forms RNA Catalyzed by RNA polymerase End result is mRNA which is then exported out of the nucleus for translation 2. Identify four differences between DNA and RNA. DNA RNA Deoxyribonucleic acid is the hereditary material in most of the organisms Made of 2-deoxyribose sugar Contains thymine as one of the 4 nitrogen base It is double stranded Ribonucleic acid transfers genetic code needed for the creation of proteins from the nucleus to the ribosome. Made of ribose sugar Has Uracil as the nitrogen base instead of thymine Single stranded DNA replication gene transcription. occurs in preparation for cell division, Produces two two copies of DNA, double stranded Uses DNA as template to form DNA Catalyzed by DNA polymerase the end result is two daughter cells. happens in preparation for protein translation. Produces mRNA, single stranded Uses template DNA as a template and forms RNA Catalyzed by RNA polymerase End result is mRNA which is then exported out of the nucleus for translation.

1. Identify five differences between DNA replication and gene transc.pdf

1. Identify five differences between DNA replication and gene transc.pdf

1. Identify five differences between DNA replication and gene transc.pdf

1.1 Year cash flow present value of cash inflow = cash flow/(1+r)^n r= 5% 1 75 71.42857 2 75 68.02721 3 75 64.78782 4 1075 884.4052 value of bond sum of present value of cash flow 1088.649 1.2 Year cash flow present value of cash inflow = cash flow/(1+r)^n r= 5.2% 1 75 71.29278 2 75 67.7688 3 75 64.41901 4 1075 877.6988 value of bond sum of present value of cash flow 1081.179 1.3 % change in bond price if YTM increases by .2% (1081.18-1088.65)/1088.65 -0.69% 1.4 Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s duration, the greater its sensitivity to the change (also know as volatility) and vice versa. 1.4 Year cash flow Year*cash flow present value of Year*cash inflow = cash flow/(1+r)^n r= 5% 1 75 75 71.42857 2 75 150 136.0544 3 75 225 194.3635 4 1075 4300 3537.621 total 3939.467 Macculay duration in years total/face value of bond 3939.467/1000 3.939467 1.4 modified duration macculay duration/(1+YTM) 3.94/1.05 3.752381 1.5 change in price if YTM increases by .2% modified duration*change in YTM 3.75*.2% 0.75% 1.1 Year cash flow present value of cash inflow = cash flow/(1+r)^n r= 5% 1 75 71.42857 2 75 68.02721 3 75 64.78782 4 1075 884.4052 value of bond sum of present value of cash flow 1088.649 1.2 Year cash flow present value of cash inflow = cash flow/(1+r)^n r= 5.2% 1 75 71.29278 2 75 67.7688 3 75 64.41901 4 1075 877.6988 value of bond sum of present value of cash flow 1081.179 1.3 % change in bond price if YTM increases by .2% (1081.18-1088.65)/1088.65 -0.69% 1.4 Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s duration, the greater its sensitivity to the change (also know as volatility) and vice versa. 1.4 Year cash flow Year*cash flow present value of Year*cash inflow = cash flow/(1+r)^n r= 5% 1 75 75 71.42857 2 75 150 136.0544 3 75 225 194.3635 4 1075 4300 3537.621 total 3939.467 Macculay duration in years total/face value of bond 3939.467/1000 3.939467 1.4 modified duration macculay duration/(1+YTM) 3.94/1.05 3.752381 1.5 change in price if YTM increases by .2% modified duration*change in YTM 3.75*.2% 0.75% Solution 1.1 Year cash flow present value of cash inflow = cash flow/(1+r)^n r= 5% 1 75 71.42857 2 75 68.02721 3 75 64.78782 4 1075 884.4052 value of bond sum of present value of cash flow 1088.649 1.2 Year cash flow present value of cash inflow = cash flow/(1+r)^n r= 5.2% 1 75 71.29278 2 75 67.7688 3 75 64.41901 4 1075 877.6988 value of bond sum of present value of cash flow 1081.179 1.3 % change in bond price if YTM increases by .2% (1081.18-1088.65)/1088.65 -0.69% 1.4 Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s duration, the greater its sensitivity to the change (also know as volatility) and vice versa. 1.4 Year cash flow Year*cash flow present value of Year*cash inflow = cash flow/(1+r)^n r= 5% 1 75 75 71.42857 2 75 150 136.0544 3 75 225 194.3635 4 1075 4300 3537.621 total 3939.467 M.

1.1Yearcash flowpresent value of cash inflow = cash flow(1+r).pdf

1.1Yearcash flowpresent value of cash inflow = cash flow(1+r).pdf

1.1Yearcash flowpresent value of cash inflow = cash flow(1+r).pdf

this is my code to count the frequency of words in a text file #define _GNU_SOURCE #include #include #include #include /****************************** here is the Structure for storing words received from each thread ******************************/ struct return_vall{ char wordlist[100][100]; int count[100]; } *arr; pthread_t *threads; int lines; /***************************** here are 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 thta are executed by each thread on seperate line. **************************/ void *word_count(void* num) { int *ln = num; unsigned int line_number = *ln; line_number++; //printf(\"Thread %d\ \",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); //printf(\"%s\ \",command); //fflush(stdout); 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); //printf(\"%d %s\",arr[line_number-1].count[counter],arr[line_number- 1].wordlist[counter]); 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_vall*) calloc(lines, sizeof(struct return_vall) * 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; } Solution this is my code to count the frequency of words in a text file #define _GNU_SOURCE #include #include #include #include /****************************** here is the Structure for storing words received from each thread ******************************/ struct return_vall{ char wordlist[100][100]; .

this is my code to count the frequency of words in a text file#.pdf

this is my code to count the frequency of words in a text file#.pdf

this is my code to count the frequency of words in a text file#.pdf

The ice allows the refluxed materials to crystall.pdf

The ice allows the refluxed materials to crystall.pdf

The ice allows the refluxed materials to crystall.pdf

1. b 2. c (bed bug) 3. d (caterpillars) 4. c (lyme disease) 5. e (chagas disease) 6. a (hives) 7. e (short lived) 8. d (plants) 9. a (1892) 10. e (parvoviridae) 11. mosquito 12. a (flies) 13. c (lyme disease) 14. 15. --- holometabolous is the type of metamorphosis in beetles 16. oligocene 17. e (6 or more) 18. e Solution 1. b 2. c (bed bug) 3. d (caterpillars) 4. c (lyme disease) 5. e (chagas disease) 6. a (hives) 7. e (short lived) 8. d (plants) 9. a (1892) 10. e (parvoviridae) 11. mosquito 12. a (flies) 13. c (lyme disease) 14. 15. --- holometabolous is the type of metamorphosis in beetles 16. oligocene 17. e (6 or more) 18. e.

1. b2. c (bed bug)3. d (caterpillars)4. c (lyme disease)5. e.pdf

1. b2. c (bed bug)3. d (caterpillars)4. c (lyme disease)5. e.pdf

1. b2. c (bed bug)3. d (caterpillars)4. c (lyme disease)5. e.pdf

#In this project you will write a program play TicTacToe #using two players (labels 0,1) or one play (label 0) playing with the machine (label 1). #The TicTacToe board has 9 integers board = [1,2,3,4,5,6,7,8,9]. The following # are the modules for the program # #def reset() resets the board to the original values # board = [1,2,3,4,5,6,7,8,9] # #def printBoard() print the current state of the board using the format # #The current TicTacToe Board # | 1 | 2 | 3 | # | 4 | 5 | 6 | # | 7 | O | 9 | # #The current TicTacToe Board # | X | 2 | 3 | # | 4 | 5 | 6 | # | 7 | O | 9 | #Note from the above that player 0 and 1 have played numbers 8 and 1 #respectively and the board display O for player 0 and X from player 1 # #def changeBoard(num1, player) using the chosen box number to change #the value of the box to 0 or -1 depending on whether the player is 0 or 1, #respectively. # #def play(player) prints the player number (0 or 1) and prompts the player # to enter a box value that have not changed to \'O\' or \'X\' # #def checkRows(value) checks to see which of the rows of the board # has the same value and returns True, otherwise, returns False # #def checkCols(value) checks to see which of the cols of the board # has the same value and returns True, otherwise, returns False # #def checkDiagonal(value) checks to see which of the diagonals of the board # has the same value and returns True, otherwise, returns False # #def win(player) checks if a player wins the game, returns True of the player wins # and False otherwise # #def machinePlay(player) plays the role of the player using random number. # this function generates numbers in the interval [1,9] and uses the first # generated random number that has not been used to play the game. The # #def ticTacToe(numPlayers) accepts the number of players and simulates the #ticTapToe, asking players to enter unused box numbers. # #def main() is the driver module that accepts the number of players from the user # and calls the ticTacToe module #Assignment: Complete the follwoing modules: #checkCols #checkRows #checkDiagonal #win #Sample of the output is # #Project #This game can be played by one or two players #Enter the number of players, 1/2 for one/two players: 1 # #The current TicTapToe Board #| 1 | 2 | 3 | #| 4 | 5 | 6 | #| 7 | 8 | 9 | #Player 0 Enter a box value: 9 # #The current TicTapToe Board #| 1 | 2 | 3 | #| 4 | 5 | 6 | #| 7 | 8 | O | #Player 1 ***Computer*** playing #The current TicTapToe Board #| 1 | 2 | 3 | #| 4 | 5 | 6 | #| 7 | X | O | #Player 0 Enter a box value: 1 # #The current TicTapToe Board #| O | 2 | 3 | #| 4 | 5 | 6 | #| 7 | X | O | #Player 1 ***Computer*** playing # #The current TicTapToe Board #| O | X | 3 | #| 4 | 5 | 6 | #| 7 | X | O | #Player 0 Enter a box value: 5 # #The current TicTapToe Board #| O | X | 3 | #| 4 | O | 6 | #| 7 | X | O | #Player 0 Wins #Do you want to quit? #Begin Program #import random number generator #from random library #from random import randint, seed #Global va.

#In this project you will write a program play TicTacToe #using tw.pdf

#In this project you will write a program play TicTacToe #using tw.pdf

#In this project you will write a program play TicTacToe #using tw.pdf

Web sever environment: A Web server is a program that uses HTTP (Hypertext Transfer Protocol) to serve the files that form Web pages to users, in response to their requests, which are forwarded by their computers\' HTTP clients. Dedicated computers and appliances may be referred to as Web servers as well.The process is an example of the client/server model. All computers that host Web sites must have Web server programs. Leading Web servers include Apache (the most widely- installed Web server), Microsoft\'s Internet Information Server (IIS) and nginx (pronouncedengine X) from NGNIX. Other Web servers include Novell\'s NetWare server, Google Web Server (GWS) and IBM\'s family of Domino servers. Web servers often come as part of a larger package of Internet- and intranet-related programs for serving email, downloading requests for File Transfer Protocol (FTP) files, and building and publishing Web pages. Considerations in choosing a Web server include how well it works with the operating system and other servers, its ability to handle server-side programming, security characteristics, and the particular publishing, search engine and site building tools that come with it. Advantages of using a web server within your development environment: Problems posed by web server environment and methods to solve: Various high-profile hacking attacks have proven that web security remains the most critical issue to any business that conducts its operations online. Web servers are one of the most targeted public faces of an organization, because of the sensitive data they usually host. Securing a web server is as important as securing the website or web application itself and the network around it. If you have a secure web application and an insecure web server, or vice versa, it still puts your business at a huge risk. Your company’s security is as strong as its weakest point. Although securing a web server can be a daunting operation and requires specialist expertise, it is not an impossible task. Long hours of research and an overdose of coffee and take away food, can save you from long nights at the office, headaches and data breaches in the future. Irrelevant of what web server software and operating system you are running, an out of the box configuration is usually insecure. Therefore one must take some necessary steps in order to increase web server security. Below is a list of tasks one should follow when securing a web server. 1. Remove Unnecessary Services Default operating system installations and configurations, are not secure. In a typical default installation, many network services which won’t be used in a web server configuration are installed, such as remote registry services, print server service, RAS etc. The more services running on an operating system, the more ports will be left open, thus leaving more open doors for malicious users to abuse. Switch off all unnecessary services and disable them, so next time the server is rebooted, they .

Web sever environmentA Web server is a program that uses HTTP (Hy.pdf

Web sever environmentA Web server is a program that uses HTTP (Hy.pdf

Web sever environmentA Web server is a program that uses HTTP (Hy.pdf

We ca use HashSet data structure to remove duplicats from an array in O(n) time. HashSet does not allow duplicate elements. So add all elements from an array to HashSet. While adding first check whether current element is already in Set or not. If it is already there then remove from an array. PSUDO CODE: void removeDuplicateFromArray(int []array, int size): 1. create a HashSet: HashSet set = new HashSet(); 2. Itereate give array and add element in Set one by one: int k = 0; for i=0 to size-1: if(!set.contains(array[i]) // if current element is not in Set, then add it in set and include in array also array[k++] = array[i]; set.add(array[i]) Runnning TIme: O(n) Solution We ca use HashSet data structure to remove duplicats from an array in O(n) time. HashSet does not allow duplicate elements. So add all elements from an array to HashSet. While adding first check whether current element is already in Set or not. If it is already there then remove from an array. PSUDO CODE: void removeDuplicateFromArray(int []array, int size): 1. create a HashSet: HashSet set = new HashSet(); 2. Itereate give array and add element in Set one by one: int k = 0; for i=0 to size-1: if(!set.contains(array[i]) // if current element is not in Set, then add it in set and include in array also array[k++] = array[i]; set.add(array[i]) Runnning TIme: O(n).

We ca use HashSet data structure to remove duplicats from an array i.pdf

We ca use HashSet data structure to remove duplicats from an array i.pdf

We ca use HashSet data structure to remove duplicats from an array i.pdf

this eample is of t-test for comparing two means here mean1 = mean of FIFO and mean 2 : Mean of LIFO so here we use student t-test for comparision of 2 means to evaluate both methods in this procedure ans: Student t-test for comparision of means Solution this eample is of t-test for comparing two means here mean1 = mean of FIFO and mean 2 : Mean of LIFO so here we use student t-test for comparision of 2 means to evaluate both methods in this procedure ans: Student t-test for comparision of means.

this eample is of t-test for comparing two meanshere mean1 = m.pdf

this eample is of t-test for comparing two meanshere mean1 = m.pdf

this eample is of t-test for comparing two meanshere mean1 = m.pdf

The middle ear bones in mammals are derived from bones in the dentary bone (mandible) of Therapsids. In the Therapsid dentary, there were three bones used in the jaw hinge: the angular, articular and quadrate. The tympanic, malleus, and incus are homologous to the angular, articular, and quadrate Tympanic = Angular Malleus = Articular Incus = Quadrate In early Therapsids, the quadrate, angular and articular were directly connected to the dentary bone in the jaw hinge. These bones were essential for jaw movement. These bones were also utilized for crude hearing capabilities. The next stage of evolution came with the separation of these bones and the dentary. After the bones separated from the jaw, they formed the postdentary rod. It is believed that these the mammalian middle ear bones evolved at least twice. It is believed that monotreme and therian ancestors evolved very similar middle ear bones, but in two separate paths. In monotreme ancestors, there is a mandibular trough, that is believed to have accommodated the postdentary rod. This rod is believed to have been connected to the trough by ligaments. This rod was used for hearing and the quadrate, angular, and articular were no longer needed in the jaw hinge. In therians, the mandibular trough no longer held the quadrate, angular, and articular. These bones were held to the jaw by ossified Meckel’s cartilage. This bridge of cartilage and bone extended away from the dentary and approached the area that would make up the modern ear cavity. The evolution of the mammalian middle ear and jaw joint were pivotal steps in the evolution of mammals. It is also a great example of how classical comparative anatomy, paleontology and developmental biology have come together to piece together how this remarkable transformation of jaw joint to ear ossicles was able to come about. The homologies of the malleus, incus and stapes to the articular, quadrate and columella, and tympanic ring and gonial to the angular and prearticular suggested by comparative anatomy 175 years ago have been recently confirmed by molecular and developmental biology. The recent discovery of new mammali form fossils has allowed careful documentation of the shift from primary to secondary jaw articulation, creating an opportunity to follow the transformation of the post-dentary skeletal elements. This fossil data has been complemented by the study of marsupial development, providing insight into the changing role of the malleus and incus, and the relationship of the primary and secondary jaw joints. We now have a number of unanswered questions. What are the signalling molecules involved in the interactions between the condylar process and the glenoid fossa that create the novel mammalian jaw joint? What are the mechanisms that lead to transformation of Meckel’s cartilage into a ligament allowing isolation of the ear from the jaw? What controls the distribution of secondary cartilages? What controls the timing of differentiation an.

The middle ear bones in mammals are derived from bones in the dentar.pdf

The middle ear bones in mammals are derived from bones in the dentar.pdf

The middle ear bones in mammals are derived from bones in the dentar.pdf

There is an old saying “Do not put all eggs in one basket” Why? Because risk is high.By keeping eggs in different baskets, we reduce risk We try to do same thing through diversification. By investing our capital in different assets , we try to reduce risk. Diversified folios reduce the risk and also the ratio of Risk to reward. If w1, w2 , w3 …wn are weight in the portfolio for assets 1, 2,3 ….n Then,w1+w2+w3+……………………+wn=1 R1, R2,R3,…….Rn are the return of the assets 1, 2 , 3 ….n S1, S2, S3……Sn are the standard deviation of the assets 1, 2, 3 …n Portfolio Return=w1R1+w2R2+w3R3+…….+wnRn Portfolio Variance=(w1^2)*(S1^2)+(w2^2)(S2^2)+………….(wn^2)*(Sn^2)+2w1w2*Cov(1,2)+2w1w3* Cov(1,3)+………+w(n-1)wn*Cov(n,(n-1) Cov(1,2)=Covariance of returns of asset1 and asset2 Portfolio Standard Deviation =Square root of Portfolio variance Risk of a stock is measured by standard deviation. Hence reduction of standard deviation through diversification means reduction of risk. We can take a simple example of two assets 1 and 2 Return of asset1=R1=15% Return of asset2=R2=12% Standard deviation of asset 1=S1=10% Standard deviation of asset 2=S2=8% Correlation of asset 1 and 2=Corr(1,2)=0.1 Covariance(1,2)=Corr(1,2)*S1*S2=0.1*10*8=8 Assume for simplicity, equal amount is invested in asset 1 and asset 2 Hence, w1=w2=0.5 Portfolio Return; 0.5*15+0.5*12=13.5% Portfolio Variance=(0.5^2)*(10^2)+(0.5^2)*(8^2)+2*0.5*0.5*8=45 Portfolio Standard Deviation=Square root of Variance=(45^0.5)= 6.708204 We can see, the risk of portfolio as measured by Standard Deviation has reduced significantly to 6,7 whereas the assets in the portfolio had standard deviation of 10 and 8 Risk / Return ratio of the portfolio=6.7/13.5=0.496 Risk/Return ratio of asset1=10/15= 0.666667 Risk/Return ratio of asset2=8/12= 0.666667 Risk return ratio of the portfolio is lower WHY INVEST IN DIFFERENT INDUSTRIES TO DIVERSIFY? Return of same types of assets are highly correlated. Suppose, if you invest in 10 different Technology Companies, you may not be able to reduce risk, because all the companies will have similar return and highly correlated. We have seen in the equation of Portfolio Variance, one factor=2 w1*w2*Cov(1,2) If there is high correlation between return of asset1 and 2, the Covariance(1,2) will be high . Hence this factor 2w1w2Cov(1,2) will be high. As a result portfolio standard deviation will be high and diversification will not serve any purpose. Hence, in order to get benefit of diversification, we need to invest in different type of industries so that the correlation between different assets are low. Consequently, the portfolio risk (measured through standard deviation) is low. Solution There is an old saying “Do not put all eggs in one basket” Why? Because risk is high.By keeping eggs in different baskets, we reduce risk We try to do same thing through diversification. By investing our capital in different assets , we try to reduce risk. Diversified folios reduce the risk and also the rati.

There is an old saying “Do not put all eggs in one basket”WhyBe.pdf

There is an old saying “Do not put all eggs in one basket”WhyBe.pdf

There is an old saying “Do not put all eggs in one basket”WhyBe.pdf

The ER Model is focussed to be a description of real-world entities. Although its development/construction is done in such a way as to allow fair and easy translation to the relational schema model, this is not fully trivial process. The ER diagram represents the conceptual level of database design while the relational schema represents the logical level for the database design. Following simple rules are followed: ADDRESS( cityAddress, city, state ) candidate key:cityAddress, city , state PERSON( pid,name ) candidate key: pid OBSERVATION (oid, date, temperature, humidity, precip. ,city), candidate key: oid foreign key: city OBSERVER ( pid ), candidate key: pid METEREOLOGIST( pid ), candidate key: pid CONSUMER( pid ), candidate key: pid WARNINGS( wwid,description,date,city) candidate key: wwid FORECAST( fid,city,date,low,high) candidate key: fid WEATHERPROGRAM( wid,name ) candidatekey; wid LIVESAT(city, streerAddress, state,pid) RECORDS(oid,pid,method) ISSUES(wwid.pid) PRESENTS(fid,pid) VIEWS(pid,fid) MAKES(wid,fid,oid) 15 tables will be created in all along with relationship tables and only 9 tables if relationships are shown as foreign keys Solution The ER Model is focussed to be a description of real-world entities. Although its development/construction is done in such a way as to allow fair and easy translation to the relational schema model, this is not fully trivial process. The ER diagram represents the conceptual level of database design while the relational schema represents the logical level for the database design. Following simple rules are followed: ADDRESS( cityAddress, city, state ) candidate key:cityAddress, city , state PERSON( pid,name ) candidate key: pid OBSERVATION (oid, date, temperature, humidity, precip. ,city), candidate key: oid foreign key: city OBSERVER ( pid ), candidate key: pid METEREOLOGIST( pid ), candidate key: pid CONSUMER( pid ), candidate key: pid WARNINGS( wwid,description,date,city) candidate key: wwid FORECAST( fid,city,date,low,high) candidate key: fid WEATHERPROGRAM( wid,name ) candidatekey; wid LIVESAT(city, streerAddress, state,pid) RECORDS(oid,pid,method) ISSUES(wwid.pid) PRESENTS(fid,pid) VIEWS(pid,fid) MAKES(wid,fid,oid) 15 tables will be created in all along with relationship tables and only 9 tables if relationships are shown as foreign keys.

The ER Model is focussed to be a description of real-world entities..pdf

The ER Model is focussed to be a description of real-world entities..pdf

The ER Model is focussed to be a description of real-world entities..pdf

Please find the answers below: Answer 16: Option A (IEP). (IEP or individual educational plan is an official document which carries the information depicting the personal/specific educatinal goals of children with special needs. It is not only a document, but refers to a protocol/programme which clearly demarcates the specific needs of a student. It has some features such as specilization, personalized, validity, usability as a shared document, information about requirement of a specific teacher/mentor or subject expert etc.) Answer 17: The statement is true. Technically, based upon the physical, physiological and psychological development of children, the age group of 6-11 has been globally accepted as the time of adolescence for growing children. This age group is associated with discrete phsycial growth (puberty), psycholgoical growth (feelings, emotions etc) and physiological growth (activation of reproductive organs). Solution Please find the answers below: Answer 16: Option A (IEP). (IEP or individual educational plan is an official document which carries the information depicting the personal/specific educatinal goals of children with special needs. It is not only a document, but refers to a protocol/programme which clearly demarcates the specific needs of a student. It has some features such as specilization, personalized, validity, usability as a shared document, information about requirement of a specific teacher/mentor or subject expert etc.) Answer 17: The statement is true. Technically, based upon the physical, physiological and psychological development of children, the age group of 6-11 has been globally accepted as the time of adolescence for growing children. This age group is associated with discrete phsycial growth (puberty), psycholgoical growth (feelings, emotions etc) and physiological growth (activation of reproductive organs)..

Please find the answers belowAnswer 16 Option A (IEP). (IEP or i.pdf

Please find the answers belowAnswer 16 Option A (IEP). (IEP or i.pdf

Please find the answers belowAnswer 16 Option A (IEP). (IEP or i.pdf

Question 1 answer is A Global trade allows wealthy countries to use their resources - whether labor, technology or capital - more efficiently. Because countries are endowed with different assets and natural resources (land, labor, capital and technology), some countries may produce the same good more efficiently and therefore sell it more cheaply than other countries. If a country cannot efficiently produce an item, it can obtain the item by trading with another country that can. This is known as specialization in international trade. Question 2 answer is C All options are correct Question 3 answer is D Revenue tariffs are levied to raise money. Question 4 answer is D.(cannot promote third world development, unless governments offer some incentive for importing from third world economies). Question 5 answer is A Question 6 answer is C Solution Question 1 answer is A Global trade allows wealthy countries to use their resources - whether labor, technology or capital - more efficiently. Because countries are endowed with different assets and natural resources (land, labor, capital and technology), some countries may produce the same good more efficiently and therefore sell it more cheaply than other countries. If a country cannot efficiently produce an item, it can obtain the item by trading with another country that can. This is known as specialization in international trade. Question 2 answer is C All options are correct Question 3 answer is D Revenue tariffs are levied to raise money. Question 4 answer is D.(cannot promote third world development, unless governments offer some incentive for importing from third world economies). Question 5 answer is A Question 6 answer is C.

Question 1 answer is A Global trade allows wealthy countries to use .pdf

Question 1 answer is A Global trade allows wealthy countries to use .pdf

Question 1 answer is A Global trade allows wealthy countries to use .pdf

Precipitation hardening, or age hardening, provides one of the most widely used mechanisms for the strengthening of metal alloys. The fundamental understanding and basis for this technique was established in early work at the U. S. Bureau of Standards on Duralumin. The importance of theoretical suggestion for the development of new alloys is clear from the historical record. At the end of the 19th century, cast iron was the only important commercial alloy not already known to western technology at the time of the Romans. When age hardening of aluminum was discovered accidentally by Wilm, during the years 1903 -1911, it quickly became an important commercial alloy under the trade name Duralumin. The strength and hardness of some metal alloys may be enhanced by the formation of extremely small uniformly dispersed second-phase particles within the original phase matrix in a process known as precipitation or age hardening. The precipitate particles act as obstacles to dislocation movement and thereby strengthen the heat-treated alloys. Many aluminum based alloys, copper- tin, certain steels, nickel based super-alloys and titanium alloys can be strengthened by age hardening processes. In order for an alloy system to be able to be precipitation-strengthened, there must be a terminal solid solution that has a decreasing solid solubility as the temperature decreases. The Al-Cu (Duralumin is an aluminum alloy of 2XXX group) phase diagram shown in Figure 1 shows this type of decrease along the solvus between the and + regions. Consider a 96wt%Al – 4wt%Cu alloy which is chosen since there is a large degrease in the solid solubility of solid solution in decreasing the temperature from 550°C to 75°C. Solution Precipitation hardening, or age hardening, provides one of the most widely used mechanisms for the strengthening of metal alloys. The fundamental understanding and basis for this technique was established in early work at the U. S. Bureau of Standards on Duralumin. The importance of theoretical suggestion for the development of new alloys is clear from the historical record. At the end of the 19th century, cast iron was the only important commercial alloy not already known to western technology at the time of the Romans. When age hardening of aluminum was discovered accidentally by Wilm, during the years 1903 -1911, it quickly became an important commercial alloy under the trade name Duralumin. The strength and hardness of some metal alloys may be enhanced by the formation of extremely small uniformly dispersed second-phase particles within the original phase matrix in a process known as precipitation or age hardening. The precipitate particles act as obstacles to dislocation movement and thereby strengthen the heat-treated alloys. Many aluminum based alloys, copper- tin, certain steels, nickel based super-alloys and titanium alloys can be strengthened by age hardening processes. In order for an alloy system to be able to be precipitation-strengthene.

Precipitation hardening, or age hardening, provides one of the most .pdf

Precipitation hardening, or age hardening, provides one of the most .pdf

Precipitation hardening, or age hardening, provides one of the most .pdf

Protists are beneficial to the ecosystem in generating oxygen as they are photosynthetic & single celled in aquatic system result in higher dissolved oxygen finally other trophic levels are in equilibrium Some protists are natural scavangers & profoundly promote \"decomposition of natural organic matter & dead bodies\" therefore nitrogen balance in the ecosystem Some protist are harmful to the ecosystem and they cause disease to various animal and humans result in ecological imbalance in trophic levels of food web Economic importance of protists: Bio-medical importance of protists: Ecological importance of protists: Ques-4: Green algae: These are the algae, which were originated predominantly from the endosymbiotic cyanobacteria, and these algae possess both plastids and chloroplasts for photosynthesis to perform their nutrients synthesis. Volvox is multicellular green algae and possesses cell differentiation as an evolutionary reproductive adaptation. The cells are associated with flagella outside for motility and vacuoles & chloroplasts inside. They mediate cell –to- cell communication through sensory perception obtained from “eyespot” finally neighboring cells join through “cytoplasmic strands” that promote cell –to-cell communication. Recent research has illustrated that cellular adhesion molecules (Algal-CAMs) are found in multicellular Volvox. However, the bulk transport has not observed in the Volvox, a primitive feature still yet to develop in them. Brown algae are the algae, which possess secondary chloroplasts as their functional units to perform their photosynthesis and these algae are commonly originated from the endosymbiotic red algae. Examples are diatoms. Phylogenetic relationship with land plants: Red algae do not possess property of photosynthesis & as they do not contain green pigments Ques-3: Outer coverings: Sphingolipids Pellicle Phospholipids & glycolipids Solution Protists are beneficial to the ecosystem in generating oxygen as they are photosynthetic & single celled in aquatic system result in higher dissolved oxygen finally other trophic levels are in equilibrium Some protists are natural scavangers & profoundly promote \"decomposition of natural organic matter & dead bodies\" therefore nitrogen balance in the ecosystem Some protist are harmful to the ecosystem and they cause disease to various animal and humans result in ecological imbalance in trophic levels of food web Economic importance of protists: Bio-medical importance of protists: Ecological importance of protists: Ques-4: Green algae: These are the algae, which were originated predominantly from the endosymbiotic cyanobacteria, and these algae possess both plastids and chloroplasts for photosynthesis to perform their nutrients synthesis. Volvox is multicellular green algae and possesses cell differentiation as an evolutionary reproductive adaptation. The cells are associated with flagella outside for motility and vacuoles & chloroplasts inside. They mediate.

Protists are beneficial to the ecosystem in generating oxygen as the.pdf

Protists are beneficial to the ecosystem in generating oxygen as the.pdf

Protists are beneficial to the ecosystem in generating oxygen as the.pdf

Now a normal O2 molecule has 12 electrons in the valence shell so, O2+ has 11 electrons in its valence shell. The first two go in the s2s and the next two go in the s*2s. Now two go in the s2p. Now thats 6 of the electrons. 4 electrons go in the p2p level and the remaining one goes in the p*2p level. Now count that up and the bond order is 1/2 the number of bonding electrons minus antibonding electrons: Order = 1/2 (8 - 3) = 5/2 Solution Now a normal O2 molecule has 12 electrons in the valence shell so, O2+ has 11 electrons in its valence shell. The first two go in the s2s and the next two go in the s*2s. Now two go in the s2p. Now thats 6 of the electrons. 4 electrons go in the p2p level and the remaining one goes in the p*2p level. Now count that up and the bond order is 1/2 the number of bonding electrons minus antibonding electrons: Order = 1/2 (8 - 3) = 5/2.

Now a normal O2 molecule has 12 electrons in the valence shell so, O.pdf

Now a normal O2 molecule has 12 electrons in the valence shell so, O.pdf

Now a normal O2 molecule has 12 electrons in the valence shell so, O.pdf

//operating system linux,ubuntu,Mac #include #include #include #include // for Time using namespace std; struct Node { int number; struct Node* nextNode; }; //append Method Implementation void append(struct Node** head, int num) { if (*head == NULL) { // checking if first node is NULL or not (*head) = (struct Node*)malloc(sizeof(struct Node)); (*head)->number = num; (*head)->nextNode = NULL; } else { struct Node* temp = *head; while (temp->nextNode != NULL) { temp = temp->nextNode; } struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); //dynamically allocation of memory newNode->number = num; newNode->nextNode = NULL; temp->nextNode = newNode; } } /* Getting Node with value from List*/ Node* search(struct Node** head, int value) { clock_t begin = clock(); std::cout<number == value) { clock_t end = clock(); std::cout<nextNode; } return NULL; } /*Display List Method Implementation*/ void printList(struct Node* head) { while (head != NULL) { std::cout << head->number << \" \"; head = head->nextNode; } std::cout << endl; } int main(int argc, char *argv[]) { struct Node* head = NULL,*temp; srand(time(0)); if ( argc != 3 ) // argc should be 2 for correct execution // We print error message std::cout<<\"Please Enter 3 argument\"< Solution //operating system linux,ubuntu,Mac #include #include #include #include // for Time using namespace std; struct Node { int number; struct Node* nextNode; }; //append Method Implementation void append(struct Node** head, int num) { if (*head == NULL) { // checking if first node is NULL or not (*head) = (struct Node*)malloc(sizeof(struct Node)); (*head)->number = num; (*head)->nextNode = NULL; } else { struct Node* temp = *head; while (temp->nextNode != NULL) { temp = temp->nextNode; } struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); //dynamically allocation of memory newNode->number = num; newNode->nextNode = NULL; temp->nextNode = newNode; } } /* Getting Node with value from List*/ Node* search(struct Node** head, int value) { clock_t begin = clock(); std::cout<number == value) { clock_t end = clock(); std::cout<nextNode; } return NULL; } /*Display List Method Implementation*/ void printList(struct Node* head) { while (head != NULL) { std::cout << head->number << \" \"; head = head->nextNode; } std::cout << endl; } int main(int argc, char *argv[]) { struct Node* head = NULL,*temp; srand(time(0)); if ( argc != 3 ) // argc should be 2 for correct execution // We print error message std::cout<<\"Please Enter 3 argument\"<.

operating system linux,ubuntu,Mac#include iostream #include .pdf

operating system linux,ubuntu,Mac#include iostream #include .pdf

operating system linux,ubuntu,Mac#include iostream #include .pdf

Network Protocol: 1) A Protocol is used for the communication between two or more machines. 2) We can even define the network protocol as, it is for the communication between two network devices. 3) Network protocols include some mechanisms such as to identify the devices and make connections with each other. 4) After the connection is obtained, data is packaged into messages are sent and received. 5) Some protocols even support message acknowledgement and data compression is used for high-performance. What is a network service: 1) A network service is an application. 2) This runs at application layer and above. 3) This is used in computer networking. 4) Network service provides data Storage, manipulation ,presentation , communication 5) This is implemented using client-server architecture or peer-to-peer architecture. Service Interface: 1) This interface is used to implement a Service-Oriented Architecture(SOA). 2) This is implemented to achieve interoperability among the applications among various technologies. Solution Network Protocol: 1) A Protocol is used for the communication between two or more machines. 2) We can even define the network protocol as, it is for the communication between two network devices. 3) Network protocols include some mechanisms such as to identify the devices and make connections with each other. 4) After the connection is obtained, data is packaged into messages are sent and received. 5) Some protocols even support message acknowledgement and data compression is used for high-performance. What is a network service: 1) A network service is an application. 2) This runs at application layer and above. 3) This is used in computer networking. 4) Network service provides data Storage, manipulation ,presentation , communication 5) This is implemented using client-server architecture or peer-to-peer architecture. Service Interface: 1) This interface is used to implement a Service-Oriented Architecture(SOA). 2) This is implemented to achieve interoperability among the applications among various technologies..

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P1.javaimport java.util.Scanner;keyboard inputting package pub.pdf

1. P1.java import java.util.Scanner;//keyboard inputting package public class P1 {//main class public static void main(String[] args) {//main method Scanner input = new Scanner (System.in); //User will input values System.out.print("Enter your weght in pounds:"); int weight = input.nextInt();//keyboard inputting //User will input values System.out.print("Enter your feet and inches:"); int feet = input.nextInt();//keyboard inputting int inches=input.nextInt(); int totalheightininches=0; totalheightininches=(feet*12)+inches;//converting feet and inches into total inches float bmi; float constant=703f; bmi = (float) (weight /Math.pow(totalheightininches,2))*constant;//bmi calculation //Print out results on display System.out.println("BMI is" + bmi); if(bmi <18.5) System.out.println("Underweight"); else if (bmi <25) System.out.println("Normal"); else if (bmi < 30) System.out.println("Overweight"); else System.out.println("Obese"); } } output Enter your weght in pounds:150 Enter your feet and inches:5 5 BMI is24.95858 Normal

2. Solution P1.java import java.util.Scanner;//keyboard inputting package public class P1 {//main class public static void main(String[] args) {//main method Scanner input = new Scanner (System.in); //User will input values System.out.print("Enter your weght in pounds:"); int weight = input.nextInt();//keyboard inputting //User will input values System.out.print("Enter your feet and inches:"); int feet = input.nextInt();//keyboard inputting int inches=input.nextInt(); int totalheightininches=0; totalheightininches=(feet*12)+inches;//converting feet and inches into total inches float bmi; float constant=703f; bmi = (float) (weight /Math.pow(totalheightininches,2))*constant;//bmi calculation //Print out results on display System.out.println("BMI is" + bmi); if(bmi <18.5) System.out.println("Underweight"); else if (bmi <25) System.out.println("Normal"); else if (bmi < 30) System.out.println("Overweight"); else System.out.println("Obese"); } } output Enter your weght in pounds:150 Enter your feet and inches:5 5

3. BMI is24.95858 Normal