Nitrogen can hold up to 4 bonds. In sodium amide.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.
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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](https://image.slidesharecdn.com/nitrogencanholdupto4bonds-230412110845-6463a5c8/85/Nitrogen-can-hold-up-to-4-bonds-In-sodium-amide-pdf-1-320.jpg)
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Properties of enantiomers Their NMR and IR spec.pdf
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.
O2 will be released as Na+ will not get reduce bu.pdf
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.
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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
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ionic character BaF MgO FeO SO2 N2 .pdf
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C. hydrogen bonding. between N and H of differen.pdf
C. hydrogen bonding. between N and H of different ammonia molecules
Solution
C. hydrogen bonding. between N and H of different ammonia molecules.
import java.util.;import acm.program.;public class FlightPla.pdf
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.
We Know that Amines are generally basic in naturebecause of th.pdf
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..
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. 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
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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
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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..
Recommended
Properties of enantiomers Their NMR and IR spec.pdf
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
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complexes. CD spectroscopy is re.
O2 will be released as Na+ will not get reduce bu.pdf
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.
Huntingtons disease and other hereditary diseas.pdf
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..
ionic character BaF MgO FeO SO2 N2 .pdf
This document lists compounds in order of their ionic character from most ionic to least ionic: barium fluoride (BaF) is the most ionic, followed by magnesium oxide (MgO), then iron(II) oxide (FeO), then sulfur dioxide (SO2), and finally nitrogen (N2) is the least ionic.
C. hydrogen bonding. between N and H of differen.pdf
C. hydrogen bonding. between N and H of different ammonia molecules
Solution
C. hydrogen bonding. between N and H of different ammonia molecules.
import java.util.;import acm.program.;public class FlightPla.pdf
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.
We Know that Amines are generally basic in naturebecause of th.pdf
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..
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. 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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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..
The water turns green because the copper(II)sulfate is breaking apar.pdf
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 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 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 main organelles in protein sorting and targeting are Rough endop.pdf
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 .
Successfully supporting managerial decision-making is critically dep.pdf
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 .
SolutionTo know that the team has identified all of the significa.pdf
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..
Solutiona) Maximum bus speed = bus driver delay + propagation del.pdf
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.
Solution Polymerase chain reaction is process in which several co.pdf
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..
Doubling [NO] would quadruple the rate .pdf
Doubling a substance would quadruple its rate. The document suggests doubling the level of a substance would have the effect of quadrupling its rate. In 3 sentences, the document states that doubling one factor would cause the rate to increase fourfold.
Correct answer F)4.0 .pdf
The document provides the correct answer to a question, which is choice F) 4.0. It also labels this as the solution but does not provide any explanation or work shown to support how this answer was obtained. The summary is intended to capture the key information and focus of the document in 3 sentences or less.
D.) The system is neither at steady state or equi.pdf
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..
public class Team {Attributes private String teamId; private.pdf
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.
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 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.
Pneumothorax The presence of abnorma air in the chest cavity outsi.pdf
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..
Perhaps too much has been written about the debate surrounding Nicho.pdf
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.
NADH produced in the cytoplasm produces two to th.pdf
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..
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 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.
I keep getting C... .pdf
This short document discusses a student who keeps getting C grades on their assignments and exams. The student is advised to study more each day and practice working through problems without looking up the answers in order to improve their understanding and raise their grades. Regular practice and self-testing is recommended as the best way for the student to learn the material more thoroughly and start achieving higher scores.
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 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%.
Generally polar compounds are soluble in polar solvents and non-pola.pdf
Generally polar compounds are soluble in polar solvents and non-polar compounds are soluble in
non-polar solvents.
Water is polar solvent and most of the organic solvents are non-polar.
So mostly ionic compounds are soluble in water.
Solution
Generally polar compounds are soluble in polar solvents and non-polar compounds are soluble in
non-polar solvents.
Water is polar solvent and most of the organic solvents are non-polar.
So mostly ionic compounds are soluble in water..
Hindi varnamala | hindi alphabet PPT.pdf
हिंदी वर्णमाला पीपीटी, hindi alphabet PPT presentation, hindi varnamala PPT, Hindi Varnamala pdf, हिंदी स्वर, हिंदी व्यंजन, sikhiye hindi varnmala, dr. mulla adam ali, hindi language and literature, hindi alphabet with drawing, hindi alphabet pdf, hindi varnamala for childrens, hindi language, hindi varnamala practice for kids, https://www.drmullaadamali.com
Chapter 4 - Islamic Financial Institutions in Malaysia.pptx
Chapter 4 - Islamic Financial Institutions in Malaysia.pptxMohd Adib Abd Muin, Senior Lecturer at Universiti Utara Malaysia
This slide is special for master students (MIBS & MIFB) in UUM. Also useful for readers who are interested in the topic of contemporary Islamic banking.
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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 correct choice.
Solution
Three are ways to protect unused switch ports :
Option B,D and E is correct choice..
The water turns green because the copper(II)sulfate is breaking apar.pdf
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 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 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 main organelles in protein sorting and targeting are Rough endop.pdf
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 .
Successfully supporting managerial decision-making is critically dep.pdf
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 .
SolutionTo know that the team has identified all of the significa.pdf
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..
Solutiona) Maximum bus speed = bus driver delay + propagation del.pdf
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.
Solution Polymerase chain reaction is process in which several co.pdf
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..
Doubling [NO] would quadruple the rate .pdf
Doubling a substance would quadruple its rate. The document suggests doubling the level of a substance would have the effect of quadrupling its rate. In 3 sentences, the document states that doubling one factor would cause the rate to increase fourfold.
Correct answer F)4.0 .pdf
The document provides the correct answer to a question, which is choice F) 4.0. It also labels this as the solution but does not provide any explanation or work shown to support how this answer was obtained. The summary is intended to capture the key information and focus of the document in 3 sentences or less.
D.) The system is neither at steady state or equi.pdf
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..
public class Team {Attributes private String teamId; private.pdf
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.
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 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.
Pneumothorax The presence of abnorma air in the chest cavity outsi.pdf
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..
Perhaps too much has been written about the debate surrounding Nicho.pdf
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.
NADH produced in the cytoplasm produces two to th.pdf
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..
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 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.
I keep getting C... .pdf
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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 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%.
Generally polar compounds are soluble in polar solvents and non-pola.pdf
Generally polar compounds are soluble in polar solvents and non-polar compounds are soluble in
non-polar solvents.
Water is polar solvent and most of the organic solvents are non-polar.
So mostly ionic compounds are soluble in water.
Solution
Generally polar compounds are soluble in polar solvents and non-polar compounds are soluble in
non-polar solvents.
Water is polar solvent and most of the organic solvents are non-polar.
So mostly ionic compounds are soluble in water..
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- 1. 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