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a. sp3 b. sp2 Solution a. sp3 b. sp2.
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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.
Properties of enantiomers Their NMR and IR spec.pdf
Properties of enantiomers Their NMR and IR spec.pdf
anushasarees
O2 will be released as Na+ will not get reduce but OH^- will Solution O2 will be released as Na+ will not get reduce but OH^- will.
O2 will be released as Na+ will not get reduce bu.pdf
O2 will be released as Na+ will not get reduce bu.pdf
anushasarees
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..
Huntingtons disease and other hereditary diseas.pdf
Huntingtons disease and other hereditary diseas.pdf
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ionic character: BaF > MgO > FeO > SO2 > N2 Solution ionic character: BaF > MgO > FeO > SO2 > N2.
ionic character BaF MgO FeO SO2 N2 .pdf
ionic character BaF MgO FeO SO2 N2 .pdf
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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.
Nitrogen can hold up to 4 bonds. In sodium amide.pdf
Nitrogen can hold up to 4 bonds. In sodium amide.pdf
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C. hydrogen bonding. between N and H of different ammonia molecules Solution C. hydrogen bonding. between N and H of different ammonia molecules.
C. hydrogen bonding. between N and H of differen.pdf
C. hydrogen bonding. between N and H of differen.pdf
anushasarees
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.
import java.util.;import acm.program.;public class FlightPla.pdf
import java.util.;import acm.program.;public class FlightPla.pdf
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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..
We Know that Amines are generally basic in naturebecause of th.pdf
We Know that Amines are generally basic in naturebecause of th.pdf
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Recommended
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.
Properties of enantiomers Their NMR and IR spec.pdf
Properties of enantiomers Their NMR and IR spec.pdf
anushasarees
O2 will be released as Na+ will not get reduce but OH^- will Solution O2 will be released as Na+ will not get reduce but OH^- will.
O2 will be released as Na+ will not get reduce bu.pdf
O2 will be released as Na+ will not get reduce bu.pdf
anushasarees
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..
Huntingtons disease and other hereditary diseas.pdf
Huntingtons disease and other hereditary diseas.pdf
anushasarees
ionic character: BaF > MgO > FeO > SO2 > N2 Solution ionic character: BaF > MgO > FeO > SO2 > N2.
ionic character BaF MgO FeO SO2 N2 .pdf
ionic character BaF MgO FeO SO2 N2 .pdf
anushasarees
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.
Nitrogen can hold up to 4 bonds. In sodium amide.pdf
Nitrogen can hold up to 4 bonds. In sodium amide.pdf
anushasarees
C. hydrogen bonding. between N and H of different ammonia molecules Solution C. hydrogen bonding. between N and H of different ammonia molecules.
C. hydrogen bonding. between N and H of differen.pdf
C. hydrogen bonding. between N and H of differen.pdf
anushasarees
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.
import java.util.;import acm.program.;public class FlightPla.pdf
import java.util.;import acm.program.;public class FlightPla.pdf
anushasarees
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..
We Know that Amines are generally basic in naturebecause of th.pdf
We Know that Amines are generally basic in naturebecause of th.pdf
anushasarees
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..
There are so many java Input Output classes that are available in it.pdf
There are so many java Input Output classes that are available in it.pdf
anushasarees
Three are ways to protect unused switch ports : Option B,D and E is correct choice. Solution Three are ways to protect unused switch ports : Option B,D and E is correct choice..
Three are ways to protect unused switch ports Option B,D and E is.pdf
Three are ways to protect unused switch ports Option B,D and E is.pdf
anushasarees
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..
The water turns green because the copper(II)sulfate is breaking apar.pdf
The water turns green because the copper(II)sulfate is breaking apar.pdf
anushasarees
The mutation is known as inversion. In this a segment from one chromosome gets detached and is transferred to another chromosome where it is attached in an inverted manner. Solution The mutation is known as inversion. In this a segment from one chromosome gets detached and is transferred to another chromosome where it is attached in an inverted manner..
The mutation is known as inversion. In this a segment from one chrom.pdf
The mutation is known as inversion. In this a segment from one chrom.pdf
anushasarees
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 .
The main organelles in protein sorting and targeting are Rough endop.pdf
The main organelles in protein sorting and targeting are Rough endop.pdf
anushasarees
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 .
Successfully supporting managerial decision-making is critically dep.pdf
Successfully supporting managerial decision-making is critically dep.pdf
anushasarees
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..
SolutionTo know that the team has identified all of the significa.pdf
SolutionTo know that the team has identified all of the significa.pdf
anushasarees
Solution : a) Maximum bus speed = bus driver delay + propagation delay + address decoder delay + time to fetch the requested data + set up time =2 +10 + 6 + 25 +1.5 = 44.5 ns per cycle Ans: maximum speed of the bus: 44.5 ns b) Number of clock cycles needed to complete the input operation is 4. Because a new transfer is started in clock cycle 4.
Solutiona) Maximum bus speed = bus driver delay + propagation del.pdf
Solutiona) Maximum bus speed = bus driver delay + propagation del.pdf
anushasarees
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..
Solution Polymerase chain reaction is process in which several co.pdf
Solution Polymerase chain reaction is process in which several co.pdf
anushasarees
Doubling [NO] would quadruple the rate Solution Doubling [NO] would quadruple the rate.
Doubling [NO] would quadruple the rate .pdf
Doubling [NO] would quadruple the rate .pdf
anushasarees
Correct answer: F)4.0 Solution Correct answer: F)4.0.
Correct answer F)4.0 .pdf
Correct answer F)4.0 .pdf
anushasarees
D.) The system is neither at steady state or equilibrium because the mass of the limes tone rock is slowly changing meaning there is an on going chemical reaction. Solution D.) The system is neither at steady state or equilibrium because the mass of the limes tone rock is slowly changing meaning there is an on going chemical reaction..
D.) The system is neither at steady state or equi.pdf
D.) The system is neither at steady state or equi.pdf
anushasarees
public class Team { //Attributes private String teamId; private String name; private String firstName; private String lastName; private String phone; private String email; /** * Constructor * param teamId * param name * param firstName * param lastName * param phone * param email */ public Team(String teamId, String name, String firstName, String lastName, String phone, String email) { this.teamId = teamId; this.name = name; this.firstName = firstName; this.lastName = lastName; this.phone = phone; this.email = email; } /** * return the teamId */ public String getTeamId() { return teamId; } /** * return the name */ public String getName() { return name; } /** * return the firstName */ public String getFirstName() { return firstName; } /** * return the lastName */ public String getLastName() { return lastName; } /** * return the phone */ public String getPhone() { return phone; } /** * return the email */ public String getEmail() { return email; } Override public String toString() { return \"Team {teamId=\" + teamId + \", name=\" + name + \", firstName=\" + firstName + \", lastName=\" + lastName + \", phone=\" + phone + \", email=\" + email + \"}\"; } Override public boolean equals(Object obj) { if(obj == null) return false; else { if(!(obj instanceof Team)) return false; else { Team other = (Team)obj; if(this.teamId.equalsIgnoreCase(other.teamId) && this.name.equalsIgnoreCase(other.name) && this.firstName.equalsIgnoreCase(other.firstName) && this.lastName.equalsIgnoreCase(other.lastName) && this.phone.equalsIgnoreCase(other.phone) && this.email.equalsIgnoreCase(other.email)) return true; else return false; } } } /** * Checks if the email id contains \"(at sign)\". * Is yes returns true else returns false */ public boolean isValidEmail() { return this.email.contains(\"(at sign)\"); } } public class Game { //Attributes private String gameID; private String homeTeamId; private String guestTeamId; private String gameDate; private int homeTeamScore; private int guestTeamScore; /** * Constructor * param gameID * param homeTeamId * param guestTeamId * param gameDate * param homeTeamScore * param guestTeamScore */ public Game(String gameID, String homeTeamId, String guestTeamId, String gameDate, int homeTeamScore, int guestTeamScore) { this.gameID = gameID; this.homeTeamId = homeTeamId; this.guestTeamId = guestTeamId; this.gameDate = gameDate; this.homeTeamScore = homeTeamScore; this.guestTeamScore = guestTeamScore; } /** * return the gameID */ public String getGameID() { return gameID; } /** * return the homeTeamId */ public String getHomeTeamId() { return homeTeamId; } /** * return the guestTeamId */ public String getGuestTeamId() { return guestTeamId; } /** * return the gameDate */ public String getGameDate() { return gameDate; } /** * return the homeTeamScore */ public int getHomeTeamScore() { return homeTeamScore; } /** * return the guestTeamScore */ public int getGuestTeamScore() { return guestTeamScore; } Override public String toString() { ret.
public class Team {Attributes private String teamId; private.pdf
public class Team {Attributes private String teamId; private.pdf
anushasarees
Pr(E) shows the probability of only event E happening. Pr(F) shows the probability of only event F happening. Pr(EUF) shows the probability of happening of either one of them, which would be, only E happening + only F happening, - both E and F happenig at the same time(which is Pr(EnF). Reason for subtractinig Pr(EnF) is, because these are common elements, and hence are already counted twice in only E happening, and only F happening, so we had to subtract it once. Thus Pr(EUF) = Pr(E) + Pr(F) - Pr(EnF) putting values, 0.7 = 0.3 + 0.6 - Pr(EnF) which gives, Pr(EnF) = 0.2 Solution Pr(E) shows the probability of only event E happening. Pr(F) shows the probability of only event F happening. Pr(EUF) shows the probability of happening of either one of them, which would be, only E happening + only F happening, - both E and F happenig at the same time(which is Pr(EnF). Reason for subtractinig Pr(EnF) is, because these are common elements, and hence are already counted twice in only E happening, and only F happening, so we had to subtract it once. Thus Pr(EUF) = Pr(E) + Pr(F) - Pr(EnF) putting values, 0.7 = 0.3 + 0.6 - Pr(EnF) which gives, Pr(EnF) = 0.2.
Pr(E) shows the probability of only event E happening.Pr(F) shows .pdf
Pr(E) shows the probability of only event E happening.Pr(F) shows .pdf
anushasarees
Pneumothorax : The presence of abnorma air in the chest cavity outside the lungs is known as Pneumothorax. In this condition, air leaks out of the lungs in to the pleural cavity leading to difficulty in breathing, chest pain and fainting. Atelectasis: Atelectasis is also known as a lung collapse. In the case of atelectasis, alveoli ( balloon-like structures where the gas excahnge takes place) deflates due to airway blockage or loss of elasticity and lose air leading to airlessness. This can result in difficulty in breathing and chest pain. SInce Tyler suffered a pneumothorax and atelectasis, his respiratory distress include difficulty in breathing, shortness of breath, tightness in the chest and chest pain. Treatment of pneumothorax and atelectasis: The treatment could include puncturing of the chest to release air from the pleural cavity, surgery to repair the damage in this chest wall or lungs, oxygen therapy and bronchoscopy to remove the airway blockage. Solution Pneumothorax : The presence of abnorma air in the chest cavity outside the lungs is known as Pneumothorax. In this condition, air leaks out of the lungs in to the pleural cavity leading to difficulty in breathing, chest pain and fainting. Atelectasis: Atelectasis is also known as a lung collapse. In the case of atelectasis, alveoli ( balloon-like structures where the gas excahnge takes place) deflates due to airway blockage or loss of elasticity and lose air leading to airlessness. This can result in difficulty in breathing and chest pain. SInce Tyler suffered a pneumothorax and atelectasis, his respiratory distress include difficulty in breathing, shortness of breath, tightness in the chest and chest pain. Treatment of pneumothorax and atelectasis: The treatment could include puncturing of the chest to release air from the pleural cavity, surgery to repair the damage in this chest wall or lungs, oxygen therapy and bronchoscopy to remove the airway blockage..
Pneumothorax The presence of abnorma air in the chest cavity outsi.pdf
Pneumothorax The presence of abnorma air in the chest cavity outsi.pdf
anushasarees
Perhaps too much has been written about the debate surrounding Nicholas Carr\'s [2003] article, IT Doesn\'t Matter, in the Harvard Business Review. In lieu of a debate, what management needs now are some answers on how to measure IT value and how to develop an appropriate investment strategy for the IT function. Many believe that for too long the IT function went 444 Communications of the Association for Information Systems (Volume 14, 2004)443-455 From the CIO Point of View: The “IT Doesn’t Matter” Debate by L. DeJarnette, R. Laskey, and H. Edgar Trainor without a rational, effective investment strategy. Boiled down that is the net, net of the Carr debate. Turning back to the debate itself, in the main, each position centers on an individual\'s perspective and actual experience with the IT function. Simply put, an individual opinion is in the eye of the beholder. Many people with bad IT experiences are aggressively in the pro-Carr camp. Others have an opposite experience and reaction. For many, the debate can trigger an emotional rather than a rational response. Personally, I have seen almost 40 years of IT history elapse. Certainly this experience does not constitute totality but it is enough to weigh in. This individual perspective, involving both positive and negative experiences, was shaped by experiences in three roles. Initially as a CIO for a Fortune 100 and then, for a Fortune 25 companies. Second, as a Big-4 Partner and consultant to IT and general management with experience with over 100 clients. Last, as a non-IT executive who was critically dependent on IT performance on multiple occasions. At the Southern California chapter of the Society of Information Management (SCSIM), the panel was organized from a perspective of black hat, white hat and grey hat. My draw for the event was the black hat, speaking from the IT doesn’t matter or pro-Carr camp. After some soul searching and reflection, I found the role easier to accept than I initially believed possible. This article is based in part on the SCSIM panel but is augmented by the reaction by others, largely CIOs at the event, shared in private at the conclusion of the chapter meeting. What is the appropriate way to deal with the issues raised in the Carr debate? Hopefully, this article takes some positive steps to deal with the debate premises in a constructive manner. HISTORY REPEATS ITSELF, PERHAPS IT SHOULD AGAIN In IT Doesn’t Matter, the Carr position, as pointed out in Larry DeJarnett\'s article (which follows this one) is that scarcity governs whether a resource is truly strategic. Carr\'s perception is that IT is a commodity, thus making it non-strategic by my definition. A historical view puts Carr\'s premise into perspective. From the beginning and into the mid-to-late 1990\'s IT uniqueness was pervasive because most systems were either custom or significantly modified application packages. The era of competitive advantage systems arose with firms such as Merrill Lynch, Am.
Perhaps too much has been written about the debate surrounding Nicho.pdf
Perhaps too much has been written about the debate surrounding Nicho.pdf
anushasarees
NADH produced in the cytoplasm produces two to three ATP by the electron transport system FADH2 adds its electrons to the electron transport system at a lower level than NADH, so it produces approximately two ATP. Solution NADH produced in the cytoplasm produces two to three ATP by the electron transport system FADH2 adds its electrons to the electron transport system at a lower level than NADH, so it produces approximately two ATP..
NADH produced in the cytoplasm produces two to th.pdf
NADH produced in the cytoplasm produces two to th.pdf
anushasarees
If top of the stack will be at position 0, then time complexity for operations: 1. push() : O(n) => you need to pop all elements then push at postion 0 then again insert all poped elements 2. pop() : O(n) => first go to postion 0 then push all other elemnts 3. top() : O(n) => pop all other elements to get 0th element Solution If top of the stack will be at position 0, then time complexity for operations: 1. push() : O(n) => you need to pop all elements then push at postion 0 then again insert all poped elements 2. pop() : O(n) => first go to postion 0 then push all other elemnts 3. top() : O(n) => pop all other elements to get 0th element.
If top of the stack will be at position 0, then time complexity for .pdf
If top of the stack will be at position 0, then time complexity for .pdf
anushasarees
I keep getting C... Solution I keep getting C....
I keep getting C... .pdf
I keep getting C... .pdf
anushasarees
HA <=> H+ + A- pH = -log[H+] = 4.0 [H+] = 10-pH = 10-4 M Percent ionization = [H+]/[HA] x 100% = 10-4/0.6 x 100 = 0.017% Solution HA <=> H+ + A- pH = -log[H+] = 4.0 [H+] = 10-pH = 10-4 M Percent ionization = [H+]/[HA] x 100% = 10-4/0.6 x 100 = 0.017%.
HA = H+ + A-pH = -log[H+] = 4.0[H+] = 10-pH = 10-4 MPercent .pdf
HA = H+ + A-pH = -log[H+] = 4.0[H+] = 10-pH = 10-4 MPercent .pdf
anushasarees
This presentation covers the essential parameters of Unit 2 Operations Processes of the subject Operations & Supply Chain Management. Topics Covered: Volume Variety and Flow. Types of Processes and Operations Systems - Continuous Flow system and intermittent flow systems.Job Production, Batch Production, Assembly line and Continuous Flow, Process and Product Layout. Design of Service Systems, Service Blueprinting.
OSCM Unit 2_Operations Processes & Systems
OSCM Unit 2_Operations Processes & Systems
Sandeep D Chaudhary
Stay updated with the latest academic trends: Access the 2024 list of Scopus-indexed journals for cutting-edge research across disciplines. Visit us at: https://iscopepublication.com/scopus-indexed-journals.php
Scopus Indexed Journals 2024 - ISCOPUS Publications
Scopus Indexed Journals 2024 - ISCOPUS Publications
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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..
There are so many java Input Output classes that are available in it.pdf
There are so many java Input Output classes that are available in it.pdf
anushasarees
Three are ways to protect unused switch ports : Option B,D and E is correct choice. Solution Three are ways to protect unused switch ports : Option B,D and E is correct choice..
Three are ways to protect unused switch ports Option B,D and E is.pdf
Three are ways to protect unused switch ports Option B,D and E is.pdf
anushasarees
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..
The water turns green because the copper(II)sulfate is breaking apar.pdf
The water turns green because the copper(II)sulfate is breaking apar.pdf
anushasarees
The mutation is known as inversion. In this a segment from one chromosome gets detached and is transferred to another chromosome where it is attached in an inverted manner. Solution The mutation is known as inversion. In this a segment from one chromosome gets detached and is transferred to another chromosome where it is attached in an inverted manner..
The mutation is known as inversion. In this a segment from one chrom.pdf
The mutation is known as inversion. In this a segment from one chrom.pdf
anushasarees
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 .
The main organelles in protein sorting and targeting are Rough endop.pdf
The main organelles in protein sorting and targeting are Rough endop.pdf
anushasarees
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 .
Successfully supporting managerial decision-making is critically dep.pdf
Successfully supporting managerial decision-making is critically dep.pdf
anushasarees
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..
SolutionTo know that the team has identified all of the significa.pdf
SolutionTo know that the team has identified all of the significa.pdf
anushasarees
Solution : a) Maximum bus speed = bus driver delay + propagation delay + address decoder delay + time to fetch the requested data + set up time =2 +10 + 6 + 25 +1.5 = 44.5 ns per cycle Ans: maximum speed of the bus: 44.5 ns b) Number of clock cycles needed to complete the input operation is 4. Because a new transfer is started in clock cycle 4.
Solutiona) Maximum bus speed = bus driver delay + propagation del.pdf
Solutiona) Maximum bus speed = bus driver delay + propagation del.pdf
anushasarees
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..
Solution Polymerase chain reaction is process in which several co.pdf
Solution Polymerase chain reaction is process in which several co.pdf
anushasarees
Doubling [NO] would quadruple the rate Solution Doubling [NO] would quadruple the rate.
Doubling [NO] would quadruple the rate .pdf
Doubling [NO] would quadruple the rate .pdf
anushasarees
Correct answer: F)4.0 Solution Correct answer: F)4.0.
Correct answer F)4.0 .pdf
Correct answer F)4.0 .pdf
anushasarees
D.) The system is neither at steady state or equilibrium because the mass of the limes tone rock is slowly changing meaning there is an on going chemical reaction. Solution D.) The system is neither at steady state or equilibrium because the mass of the limes tone rock is slowly changing meaning there is an on going chemical reaction..
D.) The system is neither at steady state or equi.pdf
D.) The system is neither at steady state or equi.pdf
anushasarees
public class Team { //Attributes private String teamId; private String name; private String firstName; private String lastName; private String phone; private String email; /** * Constructor * param teamId * param name * param firstName * param lastName * param phone * param email */ public Team(String teamId, String name, String firstName, String lastName, String phone, String email) { this.teamId = teamId; this.name = name; this.firstName = firstName; this.lastName = lastName; this.phone = phone; this.email = email; } /** * return the teamId */ public String getTeamId() { return teamId; } /** * return the name */ public String getName() { return name; } /** * return the firstName */ public String getFirstName() { return firstName; } /** * return the lastName */ public String getLastName() { return lastName; } /** * return the phone */ public String getPhone() { return phone; } /** * return the email */ public String getEmail() { return email; } Override public String toString() { return \"Team {teamId=\" + teamId + \", name=\" + name + \", firstName=\" + firstName + \", lastName=\" + lastName + \", phone=\" + phone + \", email=\" + email + \"}\"; } Override public boolean equals(Object obj) { if(obj == null) return false; else { if(!(obj instanceof Team)) return false; else { Team other = (Team)obj; if(this.teamId.equalsIgnoreCase(other.teamId) && this.name.equalsIgnoreCase(other.name) && this.firstName.equalsIgnoreCase(other.firstName) && this.lastName.equalsIgnoreCase(other.lastName) && this.phone.equalsIgnoreCase(other.phone) && this.email.equalsIgnoreCase(other.email)) return true; else return false; } } } /** * Checks if the email id contains \"(at sign)\". * Is yes returns true else returns false */ public boolean isValidEmail() { return this.email.contains(\"(at sign)\"); } } public class Game { //Attributes private String gameID; private String homeTeamId; private String guestTeamId; private String gameDate; private int homeTeamScore; private int guestTeamScore; /** * Constructor * param gameID * param homeTeamId * param guestTeamId * param gameDate * param homeTeamScore * param guestTeamScore */ public Game(String gameID, String homeTeamId, String guestTeamId, String gameDate, int homeTeamScore, int guestTeamScore) { this.gameID = gameID; this.homeTeamId = homeTeamId; this.guestTeamId = guestTeamId; this.gameDate = gameDate; this.homeTeamScore = homeTeamScore; this.guestTeamScore = guestTeamScore; } /** * return the gameID */ public String getGameID() { return gameID; } /** * return the homeTeamId */ public String getHomeTeamId() { return homeTeamId; } /** * return the guestTeamId */ public String getGuestTeamId() { return guestTeamId; } /** * return the gameDate */ public String getGameDate() { return gameDate; } /** * return the homeTeamScore */ public int getHomeTeamScore() { return homeTeamScore; } /** * return the guestTeamScore */ public int getGuestTeamScore() { return guestTeamScore; } Override public String toString() { ret.
public class Team {Attributes private String teamId; private.pdf
public class Team {Attributes private String teamId; private.pdf
anushasarees
Pr(E) shows the probability of only event E happening. Pr(F) shows the probability of only event F happening. Pr(EUF) shows the probability of happening of either one of them, which would be, only E happening + only F happening, - both E and F happenig at the same time(which is Pr(EnF). Reason for subtractinig Pr(EnF) is, because these are common elements, and hence are already counted twice in only E happening, and only F happening, so we had to subtract it once. Thus Pr(EUF) = Pr(E) + Pr(F) - Pr(EnF) putting values, 0.7 = 0.3 + 0.6 - Pr(EnF) which gives, Pr(EnF) = 0.2 Solution Pr(E) shows the probability of only event E happening. Pr(F) shows the probability of only event F happening. Pr(EUF) shows the probability of happening of either one of them, which would be, only E happening + only F happening, - both E and F happenig at the same time(which is Pr(EnF). Reason for subtractinig Pr(EnF) is, because these are common elements, and hence are already counted twice in only E happening, and only F happening, so we had to subtract it once. Thus Pr(EUF) = Pr(E) + Pr(F) - Pr(EnF) putting values, 0.7 = 0.3 + 0.6 - Pr(EnF) which gives, Pr(EnF) = 0.2.
Pr(E) shows the probability of only event E happening.Pr(F) shows .pdf
Pr(E) shows the probability of only event E happening.Pr(F) shows .pdf
anushasarees
Pneumothorax : The presence of abnorma air in the chest cavity outside the lungs is known as Pneumothorax. In this condition, air leaks out of the lungs in to the pleural cavity leading to difficulty in breathing, chest pain and fainting. Atelectasis: Atelectasis is also known as a lung collapse. In the case of atelectasis, alveoli ( balloon-like structures where the gas excahnge takes place) deflates due to airway blockage or loss of elasticity and lose air leading to airlessness. This can result in difficulty in breathing and chest pain. SInce Tyler suffered a pneumothorax and atelectasis, his respiratory distress include difficulty in breathing, shortness of breath, tightness in the chest and chest pain. Treatment of pneumothorax and atelectasis: The treatment could include puncturing of the chest to release air from the pleural cavity, surgery to repair the damage in this chest wall or lungs, oxygen therapy and bronchoscopy to remove the airway blockage. Solution Pneumothorax : The presence of abnorma air in the chest cavity outside the lungs is known as Pneumothorax. In this condition, air leaks out of the lungs in to the pleural cavity leading to difficulty in breathing, chest pain and fainting. Atelectasis: Atelectasis is also known as a lung collapse. In the case of atelectasis, alveoli ( balloon-like structures where the gas excahnge takes place) deflates due to airway blockage or loss of elasticity and lose air leading to airlessness. This can result in difficulty in breathing and chest pain. SInce Tyler suffered a pneumothorax and atelectasis, his respiratory distress include difficulty in breathing, shortness of breath, tightness in the chest and chest pain. Treatment of pneumothorax and atelectasis: The treatment could include puncturing of the chest to release air from the pleural cavity, surgery to repair the damage in this chest wall or lungs, oxygen therapy and bronchoscopy to remove the airway blockage..
Pneumothorax The presence of abnorma air in the chest cavity outsi.pdf
Pneumothorax The presence of abnorma air in the chest cavity outsi.pdf
anushasarees
Perhaps too much has been written about the debate surrounding Nicholas Carr\'s [2003] article, IT Doesn\'t Matter, in the Harvard Business Review. In lieu of a debate, what management needs now are some answers on how to measure IT value and how to develop an appropriate investment strategy for the IT function. Many believe that for too long the IT function went 444 Communications of the Association for Information Systems (Volume 14, 2004)443-455 From the CIO Point of View: The “IT Doesn’t Matter” Debate by L. DeJarnette, R. Laskey, and H. Edgar Trainor without a rational, effective investment strategy. Boiled down that is the net, net of the Carr debate. Turning back to the debate itself, in the main, each position centers on an individual\'s perspective and actual experience with the IT function. Simply put, an individual opinion is in the eye of the beholder. Many people with bad IT experiences are aggressively in the pro-Carr camp. Others have an opposite experience and reaction. For many, the debate can trigger an emotional rather than a rational response. Personally, I have seen almost 40 years of IT history elapse. Certainly this experience does not constitute totality but it is enough to weigh in. This individual perspective, involving both positive and negative experiences, was shaped by experiences in three roles. Initially as a CIO for a Fortune 100 and then, for a Fortune 25 companies. Second, as a Big-4 Partner and consultant to IT and general management with experience with over 100 clients. Last, as a non-IT executive who was critically dependent on IT performance on multiple occasions. At the Southern California chapter of the Society of Information Management (SCSIM), the panel was organized from a perspective of black hat, white hat and grey hat. My draw for the event was the black hat, speaking from the IT doesn’t matter or pro-Carr camp. After some soul searching and reflection, I found the role easier to accept than I initially believed possible. This article is based in part on the SCSIM panel but is augmented by the reaction by others, largely CIOs at the event, shared in private at the conclusion of the chapter meeting. What is the appropriate way to deal with the issues raised in the Carr debate? Hopefully, this article takes some positive steps to deal with the debate premises in a constructive manner. HISTORY REPEATS ITSELF, PERHAPS IT SHOULD AGAIN In IT Doesn’t Matter, the Carr position, as pointed out in Larry DeJarnett\'s article (which follows this one) is that scarcity governs whether a resource is truly strategic. Carr\'s perception is that IT is a commodity, thus making it non-strategic by my definition. A historical view puts Carr\'s premise into perspective. From the beginning and into the mid-to-late 1990\'s IT uniqueness was pervasive because most systems were either custom or significantly modified application packages. The era of competitive advantage systems arose with firms such as Merrill Lynch, Am.
Perhaps too much has been written about the debate surrounding Nicho.pdf
Perhaps too much has been written about the debate surrounding Nicho.pdf
anushasarees
NADH produced in the cytoplasm produces two to three ATP by the electron transport system FADH2 adds its electrons to the electron transport system at a lower level than NADH, so it produces approximately two ATP. Solution NADH produced in the cytoplasm produces two to three ATP by the electron transport system FADH2 adds its electrons to the electron transport system at a lower level than NADH, so it produces approximately two ATP..
NADH produced in the cytoplasm produces two to th.pdf
NADH produced in the cytoplasm produces two to th.pdf
anushasarees
If top of the stack will be at position 0, then time complexity for operations: 1. push() : O(n) => you need to pop all elements then push at postion 0 then again insert all poped elements 2. pop() : O(n) => first go to postion 0 then push all other elemnts 3. top() : O(n) => pop all other elements to get 0th element Solution If top of the stack will be at position 0, then time complexity for operations: 1. push() : O(n) => you need to pop all elements then push at postion 0 then again insert all poped elements 2. pop() : O(n) => first go to postion 0 then push all other elemnts 3. top() : O(n) => pop all other elements to get 0th element.
If top of the stack will be at position 0, then time complexity for .pdf
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anushasarees
I keep getting C... Solution I keep getting C....
I keep getting C... .pdf
I keep getting C... .pdf
anushasarees
HA <=> H+ + A- pH = -log[H+] = 4.0 [H+] = 10-pH = 10-4 M Percent ionization = [H+]/[HA] x 100% = 10-4/0.6 x 100 = 0.017% Solution HA <=> H+ + A- pH = -log[H+] = 4.0 [H+] = 10-pH = 10-4 M Percent ionization = [H+]/[HA] x 100% = 10-4/0.6 x 100 = 0.017%.
HA = H+ + A-pH = -log[H+] = 4.0[H+] = 10-pH = 10-4 MPercent .pdf
HA = H+ + A-pH = -log[H+] = 4.0[H+] = 10-pH = 10-4 MPercent .pdf
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