Programimport java.util.Scanner; public class SumOfDigits He.pdf

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Program: import java.util.Scanner; public class SumOfDigits // Here we are taking class name as SumOfDigits { public static void main(String[] args) { Scanner input = new Scanner(System.in); // scanning the input from the user System.out.println(\"Enter number to find it\'s sum of Digits : \"); long n = input.nextLong(); int sum = sumDigits(n); // calling sumDigits(long n) method System.out.println(\"The sum of Digits of a number is: \" + sum);// printing value of sum } public static int sumDigits(long n) { int num = (int) (n); int sum = 0; // initializing value of sum to zero while (num > 0) { sum += num % 10; // Partition the given number into individual numbers num = num / 10; // divides until all digits in a number has been covered } //End of while loop return sum; }// End of sumDigits Method } Output: Enter number to find it\'s sum of Digits : 456 The sum of Digits of a number is: 15 Solution Program: import java.util.Scanner; public class SumOfDigits // Here we are taking class name as SumOfDigits { public static void main(String[] args) { Scanner input = new Scanner(System.in); // scanning the input from the user System.out.println(\"Enter number to find it\'s sum of Digits : \"); long n = input.nextLong(); int sum = sumDigits(n); // calling sumDigits(long n) method System.out.println(\"The sum of Digits of a number is: \" + sum);// printing value of sum } public static int sumDigits(long n) { int num = (int) (n); int sum = 0; // initializing value of sum to zero while (num > 0) { sum += num % 10; // Partition the given number into individual numbers num = num / 10; // divides until all digits in a number has been covered } //End of while loop return sum; }// End of sumDigits Method } Output: Enter number to find it\'s sum of Digits : 456 The sum of Digits of a number is: 15.

Program:
import java.util.Scanner;
public class SumOfDigits // Here we are taking class name as SumOfDigits
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in); // scanning the input from the user
System.out.println("Enter number to find it's sum of Digits : ");
long n = input.nextLong();
int sum = sumDigits(n); // calling sumDigits(long n) method
System.out.println("The sum of Digits of a number is: " + sum);// printing value of sum
}
public static int sumDigits(long n)
{
int num = (int) (n);
int sum = 0; // initializing value of sum to zero
while (num > 0)
{
sum += num % 10; // Partition the given number into individual numbers
num = num / 10; // divides until all digits in a number has been covered
} //End of while loop
return sum;
}// End of sumDigits Method
}
Output:
Enter number to find it's sum of Digits :
456
The sum of Digits of a number is: 15
Solution
Program:
import java.util.Scanner;
public class SumOfDigits // Here we are taking class name as SumOfDigits
{

public static void main(String[] args)
{
Scanner input = new Scanner(System.in); // scanning the input from the user
System.out.println("Enter number to find it's sum of Digits : ");
long n = input.nextLong();
int sum = sumDigits(n); // calling sumDigits(long n) method
System.out.println("The sum of Digits of a number is: " + sum);// printing value of sum
}
public static int sumDigits(long n)
{
int num = (int) (n);
int sum = 0; // initializing value of sum to zero
while (num > 0)
{
sum += num % 10; // Partition the given number into individual numbers
num = num / 10; // divides until all digits in a number has been covered
} //End of while loop
return sum;
}// End of sumDigits Method
}
Output:
Enter number to find it's sum of Digits :
456
The sum of Digits of a number is: 15

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ConvertingSeconds.java import java.util.Scanner; public class ConvertingSeconds { public static void main(String[] args) { //Declaring variables int fromseconds,hours,minutes,seconds,toSeconds; //Scanner class Object is used to read the inputs entered by the user Scanner sc=new Scanner(System.in); //getting the number of seconds entered by the user System.out.print(\"Enter Total Number of Seconds :\"); fromseconds=sc.nextInt(); //calling the method secondTime() by passing the seconds as arguments String str=secondTime(fromseconds); //Converting the string to String array where \',\' is the delimeter String arr[]=str.split(\",\"); //Converting the String into integer hours=Integer.parseInt(arr[0]); minutes=Integer.parseInt(arr[1]); seconds=Integer.parseInt(arr[2]); //calling the method inSeconds by passing the hours,minutes,seconds as arguments toSeconds=inSeconds(hours,minutes,seconds); //Displaying the hours,minutes,seconds to seconds System.out.println(hours+\" hours \"+minutes+\" minutes \"+seconds+\" seconds corresponds to \"+toSeconds+\" seconds\"); } /*This method will converts the hours,minutes,seconds to seconds * Params:hours,minutes,seconds * Return:totalSeconds */ private static int inSeconds(int hours, int minutes, int seconds) { //Converting the hours,minutes and seconds to totalseconds int totSeconds=hours*3600+minutes*60+seconds; return totSeconds; } /*This method will converts the seconds to hours,minutes,seconds * Params:total Seconds * Return:String */ private static String secondTime(int totseconds) { //Declaring variable int secs=totseconds; //calculating the number of hours int hours=totseconds/3600; totseconds=totseconds-(hours*3600); //calculating the number of minutes int minutes=totseconds/60; totseconds=totseconds-(minutes*60); //Calculating the number of seconds int seconds=totseconds; //displaying the total seconds to hours,minutes,seconds System.out.println(secs+\" Seconds corresponds to \"+hours+\" hours \"+minutes+\" minutes \"+seconds+\" seconds\"); return hours+\",\"+minutes+\",\"+seconds; } } _____________________________________________ Output: Enter Total Number of Seconds :10000 10000 Seconds corresponds to 2 hours 46 minutes 40 seconds 2 hours 46 minutes 40 seconds corresponds to 10000 seconds _____________________________________________ Output1: Enter Total Number of Seconds :15000 15000 Seconds corresponds to 4 hours 10 minutes 0 seconds 4 hours 10 minutes 0 seconds corresponds to 15000 seconds _________________________________________Thank You Solution ConvertingSeconds.java import java.util.Scanner; public class ConvertingSeconds { public static void main(String[] args) { //Declaring variables int fromseconds,hours,minutes,seconds,toSeconds; //Scanner class Object is used to read the inputs entered by the user Scanner sc=new Scanner(System.in); //getting the number of seconds entered by the user System.out.print(\"Enter Total Number of Seconds :\"); fromseconds=sc.nextInt(); //calling the method seco.

ConvertingSeconds.javaimport java.util.Scanner;public class Conv.pdf

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Properties of enantiomers?: Their NMR and IR spectra are identical. However, enantiomers behave differently in the presence of other chiral molecules or objects. For example, enantiomers do not migrate identically on chiral chromatographic media, such as quartz or standard media that have been chirally modified. The NMR spectra of enantiomers are affected differently by single-enantiomer chiral additives such as Eufod. Chiral compounds rotate plane polarized light. Each enantiomer will rotate the light in a different sense, clockwise or counterclockwise. Molecules that do this are said to be optically active. Characteristically, different enantiomers of chiral compounds often taste and smell differently and have different effects as drugs – see below. These effects reflect the chirality inherent in biological systems. One chiral \'object\' that interacts differently with the two enantiomers of a chiral compound is circularly polarised light: An enantiomer will absorb left- and right-circularly polarised light to differing degrees. This is the basis of circular dichroism (CD) spectroscopy. Usually the difference in absorptivity is relatively small (parts per thousand). CD spectroscopy [12] is a powerful analytical technique for investigating the secondary structure of proteins and for determining the absolute configurations of chiral compounds, in particular, transition metal complexes. CD spectroscopy is replacing polarimetry as a method for characterising chiral compounds, although the latter is still popular with sugar chemists. Solution Properties of enantiomers?: Their NMR and IR spectra are identical. However, enantiomers behave differently in the presence of other chiral molecules or objects. For example, enantiomers do not migrate identically on chiral chromatographic media, such as quartz or standard media that have been chirally modified. The NMR spectra of enantiomers are affected differently by single-enantiomer chiral additives such as Eufod. Chiral compounds rotate plane polarized light. Each enantiomer will rotate the light in a different sense, clockwise or counterclockwise. Molecules that do this are said to be optically active. Characteristically, different enantiomers of chiral compounds often taste and smell differently and have different effects as drugs – see below. These effects reflect the chirality inherent in biological systems. One chiral \'object\' that interacts differently with the two enantiomers of a chiral compound is circularly polarised light: An enantiomer will absorb left- and right-circularly polarised light to differing degrees. This is the basis of circular dichroism (CD) spectroscopy. Usually the difference in absorptivity is relatively small (parts per thousand). CD spectroscopy [12] is a powerful analytical technique for investigating the secondary structure of proteins and for determining the absolute configurations of chiral compounds, in particular, transition metal complexes. CD spectroscopy is re.

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Huntington\'s disease and other hereditary diseases impact people\'s lives, because they are transferred by defects or incorrect expressions of genetic material. The transfer of dominant and recessive genes has impacted people\'s mating decisions for millenia, which can have very strong impacts on people\'s lives. Solution Huntington\'s disease and other hereditary diseases impact people\'s lives, because they are transferred by defects or incorrect expressions of genetic material. The transfer of dominant and recessive genes has impacted people\'s mating decisions for millenia, which can have very strong impacts on people\'s lives..

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Nitrogen can hold up to 4 bonds. In sodium amide, [NH2]– ion has two N-H covalent bonds. In ammonia, [NH3] molecule has three N-H covalent bonds. In ammonium, [NH4]+ ion has four N-H covalent bonds. Elements that do not follow the octet rule are H, He, B, and Be Solution Nitrogen can hold up to 4 bonds. In sodium amide, [NH2]– ion has two N-H covalent bonds. In ammonia, [NH3] molecule has three N-H covalent bonds. In ammonium, [NH4]+ ion has four N-H covalent bonds. Elements that do not follow the octet rule are H, He, B, and Be.

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import java.util.*; import acm.program.*; public class FlightPlanner extends ConsoleProgram { /* Private instance variables */ private FlightDB flights; //creates a new database private ArrayList enteredCities = new ArrayList(); //keeps track of entered cities private String firstCity; //keeps track of the first city entered by the user public void init() { //passes the text file to the database to read and parse flights = new FlightDB(\"flights.txt\"); } public void run() { welcome(); askForFistCity(); askForMoreCities(); printFinalRoute(); } /* Welcomes the user */ private void welcome() { println(\"Welcome to Flight Planner\"); println(\"Here is a list of all the cities in our database\"); Iterator it = flights.getCities(); while(it.hasNext()) { println(\" \" + it.next()); } println(\"Let\'s plan a round-trip route!\"); } /* asks the user for the starting city and prints out * all the possible destination cities for that city */ private void askForFistCity() { while(true) { firstCity = readLine(\"Enter the starting city: \"); if(flights.ContainsKey(firstCity)) { enteredCities.add(firstCity); break; } else{ println(\"You can\'t get to that city by a direct flight.\"); println(\"Here is a list of all the cities in our database\"); Iterator it = flights.getCities(); while(it.hasNext()) { println(\" \" + it.next()); } } } println(\"From \" + firstCity + \" you can fly directly to:\"); Iterator it = flights.findRoute(firstCity); while(it.hasNext()) { println(\" \" + it.next()); } } /* asks the user for the cities he/she wants to fly to, * and prints out possible destination cities for each city * until the user enters the starting city */ private void askForMoreCities() { String city = firstCity; String lastCity = city; while(true) { city = readLine(\"Where do you want to go from \" + city + \"? \"); if(city.equals(firstCity)) { break; } if(flights.ContainsKey(city) == true) { lastCity = city; enteredCities.add(city); } else{ city = lastCity; println(\"You can\'t get to that city by a direct flight.\"); } println(\"From \" + city + \" you can fly directly to:\"); Iterator it = flights.findRoute(city); while(it.hasNext()) { println(\" \" + it.next()); } } } /* prints out the chosen route */ private void printFinalRoute() { println(\"The route you\'ve chosen is\"); String route = enteredCities.get(0); for(int i = 1; i \" + enteredCities.get(i); } route += \" -> \" + enteredCities.get(0); println(route); } } import acm.program.*; public class FlightPlanner extends ConsoleProgram { /* Private instance variables */ private FlightDB flights; //creates a new database private ArrayList enteredCities = new ArrayList(); //keeps track of entered cities private String firstCity; //keeps track of the first city entered by the user public void init() { //passes the text file to the database to read and parse flights = new FlightDB(\"flights.txt\"); } public void run() { welcome(); askForFistCity(); askForMoreCities(); printFinalRoute(); } /* Welcomes the use.

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We Know that: Amines are generally basic in naturebecause of their ability to donate the lone pair of electrons onthe Nitrogen atom. the presence of electronwithdrawinggroup decreases the basicity of the amines. For the given aminesp-Nitro aniline is least basic as the pKaindicates that p-Nitroaniline is Acidic in nature. Solution We Know that: Amines are generally basic in naturebecause of their ability to donate the lone pair of electrons onthe Nitrogen atom. the presence of electronwithdrawinggroup decreases the basicity of the amines. For the given aminesp-Nitro aniline is least basic as the pKaindicates that p-Nitroaniline is Acidic in nature..

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There are so many java Input Output classes that are available in it. In this some of them are: a) Writer: It describes Abstract clss for writing to character streams. b) BufferedInputStream: It describes a BufferedInputStream adds functionality to another another input stream namely it shows the ability to support the input and make the changes if any formats happened. c) BufferOutputStream: it describes, This class implements a buffered output stream. d) File: An abstract representation of files and directory pathnames. Let us take a File class in deep discussion: public class File extends objects - This is the basic syntax for this input output file User interfaces and operating systems use system-dependent pathname strings to name files and directories. This class presents an abstract, system-independent view of hierarchical pathnames. An abstract pathname has two components A file system may implement restrictions to certain operations on the actual file-system object, such as reading, writing, and executing. These restrictions are collectively known as access permissions. The file system may have multiple sets of access permissions on a single object. For example, one set may apply to the object\'s owner, and another may apply to all other users. The access permissions on an object may cause some methods in this class to fail. Instances of the File class are immutable; that is, once created, the abstract pathname represented by a File object will never change. Solution There are so many java Input Output classes that are available in it. In this some of them are: a) Writer: It describes Abstract clss for writing to character streams. b) BufferedInputStream: It describes a BufferedInputStream adds functionality to another another input stream namely it shows the ability to support the input and make the changes if any formats happened. c) BufferOutputStream: it describes, This class implements a buffered output stream. d) File: An abstract representation of files and directory pathnames. Let us take a File class in deep discussion: public class File extends objects - This is the basic syntax for this input output file User interfaces and operating systems use system-dependent pathname strings to name files and directories. This class presents an abstract, system-independent view of hierarchical pathnames. An abstract pathname has two components A file system may implement restrictions to certain operations on the actual file-system object, such as reading, writing, and executing. These restrictions are collectively known as access permissions. The file system may have multiple sets of access permissions on a single object. For example, one set may apply to the object\'s owner, and another may apply to all other users. The access permissions on an object may cause some methods in this class to fail. Instances of the File class are immutable; that is, once created, the abstract pathname represented by a File object will never change..

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The water turns green because the copper(II)sulfate is breaking apart; the green that you see is the copper. It starts attaching to the magnesium because it starts forming a complex ion with the magnesium and the sulfate; the magnesium and sulfate will bond because they form an insoluble compound, and the copper can bond onto it, forming complex ion. Solution The water turns green because the copper(II)sulfate is breaking apart; the green that you see is the copper. It starts attaching to the magnesium because it starts forming a complex ion with the magnesium and the sulfate; the magnesium and sulfate will bond because they form an insoluble compound, and the copper can bond onto it, forming complex ion..

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The main organelles in protein sorting and targeting are Rough endoplasmic reticulum. As the question is all about sorting and targeting to membrane and secretion, I will stick to only these two targeting pathways. Please note the mechansm of targeting proteins to other organelles such as mitichondria, ER, chloroplast is different. Journey of a secretory protein: The protein sorting and targeting occur in endoplasmic reticulum. Most of the secretory proteins as well as membrane proteins are translocated into ER co-translationally (Co-translational translocation). i.e, they are moved into ER while their translation is going on. this translocation process is mediated by Signal recognition particle (SRP). SRP recognizes N-terminal signal peptide on the protein that is being translated. The translation process pauses for a while when the ribosome-protein-SRP complex is being transported to SRP receptor on the ER membrane. The protein is inserted into translocon channel on ER membrane and enters ER. The signaal sequence from nascent peptide is immediately cleaved in case of secretoey proteins and type I transmembrane proteins by signal peptidase. the translation resumes directing the protein being formed into ER. Inside the ER, chaperone proteins bind the protein to guide through correct folding. Then the protein is transported by ER membrane vesicles to golgi apparatus for further processing like protein modification and glycosylation. ER--- cis-golgi---trans---golgi--- lysosomes. the proteins from lysosomes fuse with membrane and according to the signals present in nascent proteins, they either remain as transmembrane proteins or secreted out of the membrane. transmembrane G-receptors span the membrane seven times (also called serpentine receptors) donot contain signal sequence at N-terminal. The first transmembrane domain acts as the signal sequence that is recognized by SRP. Some secretory proteins as well as transmembrane proteins may undero post-translational translocation. i.e, transportation to ER after the completion of protein synthesis. the entire process is believed to be same as co-translational translocation. Solution The main organelles in protein sorting and targeting are Rough endoplasmic reticulum. As the question is all about sorting and targeting to membrane and secretion, I will stick to only these two targeting pathways. Please note the mechansm of targeting proteins to other organelles such as mitichondria, ER, chloroplast is different. Journey of a secretory protein: The protein sorting and targeting occur in endoplasmic reticulum. Most of the secretory proteins as well as membrane proteins are translocated into ER co-translationally (Co-translational translocation). i.e, they are moved into ER while their translation is going on. this translocation process is mediated by Signal recognition particle (SRP). SRP recognizes N-terminal signal peptide on the protein that is being translated. The translation process pauses for a .

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Successfully supporting managerial decision-making is critically dependent upon the availability of integrated, high quality information organized and presented in a timely and easily understood manner. Data warehouses have emerged to meet this need. They serve as an integrated repository for internal and external data—intelligence critical to understanding and evaluating the business within its environmental context. With the addition of models, analytic tools, and user interfaces, they have the potential to provide actionable information resources—business intelligence that supports effective problem and opportunity identification, critical decision-making, and strategy formulation, implementation, and evaluation. Four themes frame our analysis: integration, implementation, intelligence, and innovation. 1:four major categories of business environment factors is INTEGRATION,IMPLEMENTATION,INTELLIGENCE AND INNOVATION. Organizations use data warehousing to support strategic and mission-critical applications. Data deposited into the data warehouse must be transformed into information and knowledge and appropriately disseminated to decision-makers within the organization and to critical partners in various capacities within the organizational value chain. Crucial problems that must be addressed in this area are: the modes of dissemination of information to the end user; the development, selection, and implementation of appropriate models, analytic tools, and data mining tools; the privacy and security of data; system performance; and adequate levels of training and support. The human–computer interface is of paramount importance in the data warehouse environment and the primary determinant of success from the end-user perspective. In order to support analysis and reporting tasks, the data warehouse must have high quality data and make these data accessible through intuitive interface technologies. Data warehouse browsing tools provide star- schema query-like access through a flexible menu-based interface, with pull-down menus representing important dimensions. These types of tools are easy to use and support some ad-hoc exploration, but are usually controlled through an administrative layer that determines the data available to endusers. In developing a flexible interface, there is a tradeoff between the ability to express ad-hoc queries and the ease-of-use that results from pre-defined constructs implemented by data warehouse designers and administrators. Of course, SQL can provide an ad-hoc query facility, but its use requires some care in the data warehouse environment where the combination of very large tables and ill-formed user queries can produce some truly awful performance and potentially erroneous results. Casual users may not have sufficient understanding of SQL or of the database schema to effectively use such an interface. Typically, only trained power users (e.g., DBAs, application developers) are permitted to write SQL queries on .

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Solution : To know that the team has identified all of the significant risks associated with an IT acquisition alternative we have to go through the steps that the team has followed to come up with the significant risks. For this we will match the work done by them with the steps involved in IT acquisition process. There are several that we have to follow for the IT acquisition process and they are given below: 1) First step includes to carefully identify the business objective along with proper planning and knowing the system requirements. 2) Second step involves the redesigning of the Information system architecture so that it would be easy to identify the efforts to be put to develop the specific application for the organization. 3) Third step involves the decision making done regarding the several options available for the procurement of software solutions. Options might include leasing through ASP, leasing through contract development, outsourcing, etc. Before choosing any of the option its pros and cons, both are evaluated. 4) Fourth step involves the feasibility analysis before making the final decision for any software solution. The feasibility analysis is done with respect to technical requirements, economical requirements, operational and political requirements, legal and contractual requirements, etc. 5) Fifth step includes the important decision of deciding for the best available option among several options. 6) Sixth step involves the careful evaluation of the proposals and choosing the best suited one that fulfills the requirements of the organization. 7) Seventh step is to implement the chosen solution within the organization. This helps in testing the performance of the solution as per the requirement given by the organization. 8) Eighth step is to monitor the selected solution that whether it is effective and efficient according to the user. If all these steps are followed by the team than we will come to know that they have identified all of the significant risks associated with an IT acquisition alternative..

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Solution : Polymerase chain reaction is process in which several copy of specific segment of DNA can be produced within short interval of time.PCR include various cycles of heating and cooling.DNA is double stranded structure.Initially,DNA molecule is heated to divide it into two single strands.Then under lower temperature condition short strand of DNA which start its synthesis are attached to DNA template.After that temperature is increased and new strand of DNA is formed..

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1. Program: import java.util.Scanner; public class SumOfDigits // Here we are taking class name as SumOfDigits { public static void main(String[] args) { Scanner input = new Scanner(System.in); // scanning the input from the user System.out.println("Enter number to find it's sum of Digits : "); long n = input.nextLong(); int sum = sumDigits(n); // calling sumDigits(long n) method System.out.println("The sum of Digits of a number is: " + sum);// printing value of sum } public static int sumDigits(long n) { int num = (int) (n); int sum = 0; // initializing value of sum to zero while (num > 0) { sum += num % 10; // Partition the given number into individual numbers num = num / 10; // divides until all digits in a number has been covered } //End of while loop return sum; }// End of sumDigits Method } Output: Enter number to find it's sum of Digits : 456 The sum of Digits of a number is: 15 Solution Program: import java.util.Scanner; public class SumOfDigits // Here we are taking class name as SumOfDigits {

2. public static void main(String[] args) { Scanner input = new Scanner(System.in); // scanning the input from the user System.out.println("Enter number to find it's sum of Digits : "); long n = input.nextLong(); int sum = sumDigits(n); // calling sumDigits(long n) method System.out.println("The sum of Digits of a number is: " + sum);// printing value of sum } public static int sumDigits(long n) { int num = (int) (n); int sum = 0; // initializing value of sum to zero while (num > 0) { sum += num % 10; // Partition the given number into individual numbers num = num / 10; // divides until all digits in a number has been covered } //End of while loop return sum; }// End of sumDigits Method } Output: Enter number to find it's sum of Digits : 456 The sum of Digits of a number is: 15