QUESTION 8____ are considered to be product or service market stak.pdf
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QUESTION 8 ____ are considered to be product or service market stakeholders. Employees and customers Customers and vendors Owners and managers Government and banks Employees and customers Customers and vendors Owners and managers Government and banks Solution Suppliers and customers are considered as Product or Service market stakeholders. Employees, Owners and Managers are Internal Stakeholders Government is considered as External Stakeholders Banks are considered as Capital market stakeholders So Answer is Customers and Vendors..
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How is a Decision Support Systems (DSS) different from a Management .pdf
How is a Decision Support Systems (DSS) different from a Management Information Systems(
MIS)?
Solution
MIS provides information about the performance of an organization
Think of entire company (the firm) as
a system.
An MIS provides management with feedback.
MIS
Periodic reports
Pre-specified, generic reports.
DSS
Special reports that may only be generated once
May not know what kind of report to generate until the problem surfaces; specialized reports.
In a DSS, a manager generates the report through an interactive interface
Flexible & Adaptable reports
DSS Reporting is produced through analytical modeling, not just computing an average, or
plotting a graph.
Business Models are programmed into a DSS.
I am trying to create a program That works with two other programs i.pdf
The document describes a problem the user is having with a C++ program that uses multiple files and a makefile. Specifically, the Shape.cpp file is not working correctly and is giving errors. The user needs someone to fix the Shape.cpp file and explain what changes were made.
How does anti-malware software detect virusesWhat techniques are .pdf
How does anti-malware software detect viruses?
What techniques are available, and how do they differ?
Solution
Generally Anti-malware software runs in the background on your computer, checking every file
you open. This is generally known as on-access scanning, background scanning, resident
scanning, real-time protection, or something else, depending on your anti-malware program.
When you double-click an EXE file, it may seem like the program launches immediately – but it
doesn’t. Your anti-malware software checks the program first, comparing it to known viruses,
worms, and other types of malware. Your anti-malware software also does “heuristic” checking,
checking programs for types of bad behavior that may indicate a new, unknown virus. It’s
possible to use an anti-malware without on-access scanning, but this generally isn’t a good idea –
viruses that exploit security holes in programs wouldn’t be caught by the scanner. After a virus
has infected your system, it’s much harder to remove..
How can I upload a picture in here I tried with image properties ic.pdf
How can I upload a picture in here? I tried with image properties icon and connected the image
to the URL which is from \'picture painting\' or \'microsoft word\' program.
When I am uploading a picture, the picture is shown before posting question but it\'s not shown
after I posted question.
Solution
Take screenshots of that question. Go to \"Ask a study question\" and post question over there.
You can a attach a file upto 10MB. So once again try it..
Explain the role of ATP in the action of Hsp70 and GroELES.Solu.pdf
Explain the role of ATP in the action of Hsp70 and GroEL/ES.
Solution
The action of GroEL is in the structure of the monomer. The Hsp60 monomer has three unique
areas isolated by two pivot locations.
The apical area contains countless restricting destinations for unfurled protein substrates.Several
globular proteins won\'t tie to the apical area on the grounds as their hydrophobic parts are
gathered inside, far from the fluid medium since this is the thermodynamically ideal compliance.
Consequently, these \"substrate locales\" will just tie to proteins which are not ideally collapsed.
The apical space additionally has restricting destinations for the Hsp10 monomers of GroES.The
tropical space has an opening close to the pivot point for restricting ATP, and in addition two
connection focuses for the other portion of the GroEL particle. Whatever is left of the central
segment is tolerably hydrophilic.
The ATP addition and GroES drastically affects the compliance of the cis space. This impact is
caused by flexion and revolution at the two pivot focuses on the Hsp60 monomers.
The middle domain area overlays down and internal around 25° on the lower pivot. This impact,
duplicated through the helpful flexing of all monomers, builds the central breadth of the GroEL
confine. In any case, the apical area turns an entire 60° up and out on the upper pivot, and
furthermore turns 90° around the pivot hub.
This movement opens the pen broadly at the highest point of the cis area, yet totally expels the
substrate restricting destinations from within the confine..
Files Please respond to the following •Suppose you are creating.pdf
\"Files\" Please respond to the following: •Suppose you are creating a program to monitor a
vehicle’s location, a feature that a Global Positioning System (GPS) provides. Suggest one (1)
method for saving and organizing the data you received from GPS so that you may analyze that
data during the creation of the program in question. Provide a rationale to support your
suggestion. •Using the scenario from Part I of this discussion, give your opinion as to why the
Java language chooses to treat file data as a stream rather than as a single object, and specify the
main benefits of this approach. Next, analyze the main reasons why the Java language chooses to
treat file data as a stream rather than as a single object. Justify your response.
Solution
Data can be stored as a pair of coordinate Map. Pair is a simple class with coordinates as its
element. This will help in quick lookup of a point and cooresponding location and will help in
quickly performing search operation while analysing data.
why the Java language chooses to treat file data as a stream rather than as a single object,
Files can be very large more than even RAM size and loading such a big file is not always
possible but streams can always be read in small chunks without running into memroy issues.
This is the main reason for java to choose file as a stream rather than a single object.
Main benefit for this it helps your program not to crash while loading huge amount of data..
evil_server.cpp#include string #include cstdlib #include.pdf
//evil_server.cpp
#include
#include
#include
#include \"evil_server.h\"
using namespace std;
EvilServer :: EvilServer()
{
hacked[0] = hacked[1] = hacked[2] = false;
passwords[agent_index(MrMean)] = random_pw(MrMean);
passwords[agent_index(MsChief)] = random_pw(MsChief);
passwords[agent_index(DrEvil)] = random_pw(DrEvil);
}
void EvilServer :: change_pw(EvilAgent agent, string new_pw)
{
int index = agent_index(agent);
if (new_pw == passwords[index])
return;
hacked[index] = false;
passwords[index] = new_pw;
}
string EvilServer :: random_pw(EvilAgent agent)
{
string password;
int length;
switch (agent)
{
case DrEvil:
length = 4 + (rand() % 5);
for (int i = 0; i < length - 4; ++i)
password += random_char();
if (rand() % 2)
password = \"gato\" + password;
else
password += \"gato\";
break;
case MsChief:
length = 4 + (rand() % 5);
password = \"haha\";
for (int i = 5; i <= length; ++i)
password += random_char();
break;
case MrMean:
length = 1 + rand() % 8;
for (int i = 0; i < length; ++i)
if (rand() % 2)
password += \'g\';
else
password += \'r\';
break;
}
return password;
}
char EvilServer :: random_char()
{
char c = rand() % 36;
if (c < 10)
c += 48;
else
c += (97-10);
return c;
}
string EvilServer :: agent_name(EvilAgent agent)
{
switch(agent)
{
case DrEvil:
return \"Dr. Evil\";
case MsChief:
return \"Ms. Chief\";
case MrMean:
return \"Mr. Mean\";
}
}
int EvilServer :: agent_index(EvilAgent agent)
{
switch(agent)
{
case DrEvil:
return 0;
case MsChief:
return 1;
case MrMean:
return 2;
}
}
bool EvilServer :: guess(EvilAgent agent, string guess)
{
string name = agent_name(agent);
int index = agent_index(agent);
if (guess != passwords[index] || hacked[index])
return false;
hacked[index] = true;
return true;
}
bool EvilServer :: is_hacked(EvilAgent agent)
{
return hacked[agent_index(agent)];
}
//***************************************************************************
//evil_server.h
#ifndef EVILSERVER_H
#define EVILSERVER_H
#include
using namespace std;
enum EvilAgent {DrEvil, MsChief, MrMean};
class EvilServer
{
public:
EvilServer();
bool guess(EvilAgent agent, string guess);
void change_pw(EvilAgent agent, string new_pw);
bool is_hacked(EvilAgent agent);
private:
static string agent_name(EvilAgent agent);
static int agent_index(EvilAgent agent);
static string random_pw(EvilAgent agent);
static char random_char();
bool hacked[3];
string passwords[3];
};
#endif
//************************************************************
//hack.h
#include \"evil_server.h\"
void hack(EvilServer &es);
//*******************************************************
//main.cpp
#include
#include
#include
#include
#include
#include \"evil_server.h\"
#include \"hack.h\"
using namespace std;
clock_t test_start;
bool test_time_exceeds(int seconds)
{
clock_t cur = clock();
return seconds <= static_cast(cur-test_start)/CLOCKS_PER_SEC;
}
int main()
{
// Initialize random number generation, i.e. rand(), using current millisecond
srand(2017);
test_start = clock();
// Hack something
Evil.
ecorrect Question 32 0 1 pts Referencing the diagram above, use the .pdf
ecorrect Question 32 0 /1 pts Referencing the diagram above, use the following information to
answer the question. You\'re an archaeologist working in Peru, South America and you find a of
skeleton in layer B. You it undiscovered primate species because its opposable thumbs and lack
of a tall You\'re trying to decide what type of dating method is appropriate. You recall finding a
fossil in layer C known to be 200,000 years old. You discover another fossil in layer D that dates
back to 2 million years ago. Given this information, which method would you use to date the
skeleton?
Solution
I ave explained te answeer in a little detail since te points needed explaination. The conventional
belief is that carbon datin is the metod but here we see that there are two types of age
determinations. Geologists in the late 18th and early 19th century studied rock layers and the
fossils in them to determine relative age. For this question we will not employ the carbon dating
method ,used for determining the age of an object containing organic material by using the
properties of radiocarbon (14
C), a radioactive isotope of carbon. because the half-life of 14
C (the period of time after which half of a given sample will have decayed) is about 5,730 years,
the oldest dates that can be reliably measured by this process date to around 50,000 years ago,
although special preparation methods occasionally permit accurate analysis of older samples.
I will list it as two parts which contain informmation to determine the age of rocks and fossils.
PART 1: DETERMINING RELATIVE AGE OF ROCKS
there are two basic principles used by geologists to determine the sequence of ages of rocks.
They are:
Principle of superposition: Younger sedimentary rocks are deposited on top of older sedimentary
rocks.
Principle of cross-cutting relations: Any geologic feature is younger than anything else that it
cuts across.
PART 2: RADIOMETRIC AGE-DATING
Some elements have forms (called isotopes) with unstable atomic nuclei that have a tendency to
change, or decay. For example, U-235 is an unstable isotope of uranium that has 92 protons and
143 neutrons in the nucleus of each atom. Through a series of changes within the nucleus, it
emits several particles, ending up with 82 protons and 125 neutrons. This is a stable condition,
and there are no more changes in the atomic nucleus. A nucleus with that number of protons is
called lead (chemical symbol Pb). The protons (82) and neutrons (125) total 207. This particular
form (isotope) of lead is called Pb-207. U-235 is the parent isotope of Pb-207, which is the
daughter isotope.
Many rocks contain small amounts of unstable isotopes and the daughter isotopes into which
they decay. Where the amounts of parent and daughter isotopes can be accurately measured, the
ratio can be used to determine how old the rock is.
At any moment there is a small chance that each of the nuclei of U-235 will suddenly decay.
That chance of decay is very small, but it is alw.
Recommended
How is a Decision Support Systems (DSS) different from a Management .pdf
How is a Decision Support Systems (DSS) different from a Management Information Systems(
MIS)?
Solution
MIS provides information about the performance of an organization
Think of entire company (the firm) as
a system.
An MIS provides management with feedback.
MIS
Periodic reports
Pre-specified, generic reports.
DSS
Special reports that may only be generated once
May not know what kind of report to generate until the problem surfaces; specialized reports.
In a DSS, a manager generates the report through an interactive interface
Flexible & Adaptable reports
DSS Reporting is produced through analytical modeling, not just computing an average, or
plotting a graph.
Business Models are programmed into a DSS.
I am trying to create a program That works with two other programs i.pdf
The document describes a problem the user is having with a C++ program that uses multiple files and a makefile. Specifically, the Shape.cpp file is not working correctly and is giving errors. The user needs someone to fix the Shape.cpp file and explain what changes were made.
How does anti-malware software detect virusesWhat techniques are .pdf
How does anti-malware software detect viruses?
What techniques are available, and how do they differ?
Solution
Generally Anti-malware software runs in the background on your computer, checking every file
you open. This is generally known as on-access scanning, background scanning, resident
scanning, real-time protection, or something else, depending on your anti-malware program.
When you double-click an EXE file, it may seem like the program launches immediately – but it
doesn’t. Your anti-malware software checks the program first, comparing it to known viruses,
worms, and other types of malware. Your anti-malware software also does “heuristic” checking,
checking programs for types of bad behavior that may indicate a new, unknown virus. It’s
possible to use an anti-malware without on-access scanning, but this generally isn’t a good idea –
viruses that exploit security holes in programs wouldn’t be caught by the scanner. After a virus
has infected your system, it’s much harder to remove..
How can I upload a picture in here I tried with image properties ic.pdf
How can I upload a picture in here? I tried with image properties icon and connected the image
to the URL which is from \'picture painting\' or \'microsoft word\' program.
When I am uploading a picture, the picture is shown before posting question but it\'s not shown
after I posted question.
Solution
Take screenshots of that question. Go to \"Ask a study question\" and post question over there.
You can a attach a file upto 10MB. So once again try it..
Explain the role of ATP in the action of Hsp70 and GroELES.Solu.pdf
Explain the role of ATP in the action of Hsp70 and GroEL/ES.
Solution
The action of GroEL is in the structure of the monomer. The Hsp60 monomer has three unique
areas isolated by two pivot locations.
The apical area contains countless restricting destinations for unfurled protein substrates.Several
globular proteins won\'t tie to the apical area on the grounds as their hydrophobic parts are
gathered inside, far from the fluid medium since this is the thermodynamically ideal compliance.
Consequently, these \"substrate locales\" will just tie to proteins which are not ideally collapsed.
The apical space additionally has restricting destinations for the Hsp10 monomers of GroES.The
tropical space has an opening close to the pivot point for restricting ATP, and in addition two
connection focuses for the other portion of the GroEL particle. Whatever is left of the central
segment is tolerably hydrophilic.
The ATP addition and GroES drastically affects the compliance of the cis space. This impact is
caused by flexion and revolution at the two pivot focuses on the Hsp60 monomers.
The middle domain area overlays down and internal around 25° on the lower pivot. This impact,
duplicated through the helpful flexing of all monomers, builds the central breadth of the GroEL
confine. In any case, the apical area turns an entire 60° up and out on the upper pivot, and
furthermore turns 90° around the pivot hub.
This movement opens the pen broadly at the highest point of the cis area, yet totally expels the
substrate restricting destinations from within the confine..
Files Please respond to the following •Suppose you are creating.pdf
\"Files\" Please respond to the following: •Suppose you are creating a program to monitor a
vehicle’s location, a feature that a Global Positioning System (GPS) provides. Suggest one (1)
method for saving and organizing the data you received from GPS so that you may analyze that
data during the creation of the program in question. Provide a rationale to support your
suggestion. •Using the scenario from Part I of this discussion, give your opinion as to why the
Java language chooses to treat file data as a stream rather than as a single object, and specify the
main benefits of this approach. Next, analyze the main reasons why the Java language chooses to
treat file data as a stream rather than as a single object. Justify your response.
Solution
Data can be stored as a pair of coordinate Map. Pair is a simple class with coordinates as its
element. This will help in quick lookup of a point and cooresponding location and will help in
quickly performing search operation while analysing data.
why the Java language chooses to treat file data as a stream rather than as a single object,
Files can be very large more than even RAM size and loading such a big file is not always
possible but streams can always be read in small chunks without running into memroy issues.
This is the main reason for java to choose file as a stream rather than a single object.
Main benefit for this it helps your program not to crash while loading huge amount of data..
evil_server.cpp#include string #include cstdlib #include.pdf
//evil_server.cpp
#include
#include
#include
#include \"evil_server.h\"
using namespace std;
EvilServer :: EvilServer()
{
hacked[0] = hacked[1] = hacked[2] = false;
passwords[agent_index(MrMean)] = random_pw(MrMean);
passwords[agent_index(MsChief)] = random_pw(MsChief);
passwords[agent_index(DrEvil)] = random_pw(DrEvil);
}
void EvilServer :: change_pw(EvilAgent agent, string new_pw)
{
int index = agent_index(agent);
if (new_pw == passwords[index])
return;
hacked[index] = false;
passwords[index] = new_pw;
}
string EvilServer :: random_pw(EvilAgent agent)
{
string password;
int length;
switch (agent)
{
case DrEvil:
length = 4 + (rand() % 5);
for (int i = 0; i < length - 4; ++i)
password += random_char();
if (rand() % 2)
password = \"gato\" + password;
else
password += \"gato\";
break;
case MsChief:
length = 4 + (rand() % 5);
password = \"haha\";
for (int i = 5; i <= length; ++i)
password += random_char();
break;
case MrMean:
length = 1 + rand() % 8;
for (int i = 0; i < length; ++i)
if (rand() % 2)
password += \'g\';
else
password += \'r\';
break;
}
return password;
}
char EvilServer :: random_char()
{
char c = rand() % 36;
if (c < 10)
c += 48;
else
c += (97-10);
return c;
}
string EvilServer :: agent_name(EvilAgent agent)
{
switch(agent)
{
case DrEvil:
return \"Dr. Evil\";
case MsChief:
return \"Ms. Chief\";
case MrMean:
return \"Mr. Mean\";
}
}
int EvilServer :: agent_index(EvilAgent agent)
{
switch(agent)
{
case DrEvil:
return 0;
case MsChief:
return 1;
case MrMean:
return 2;
}
}
bool EvilServer :: guess(EvilAgent agent, string guess)
{
string name = agent_name(agent);
int index = agent_index(agent);
if (guess != passwords[index] || hacked[index])
return false;
hacked[index] = true;
return true;
}
bool EvilServer :: is_hacked(EvilAgent agent)
{
return hacked[agent_index(agent)];
}
//***************************************************************************
//evil_server.h
#ifndef EVILSERVER_H
#define EVILSERVER_H
#include
using namespace std;
enum EvilAgent {DrEvil, MsChief, MrMean};
class EvilServer
{
public:
EvilServer();
bool guess(EvilAgent agent, string guess);
void change_pw(EvilAgent agent, string new_pw);
bool is_hacked(EvilAgent agent);
private:
static string agent_name(EvilAgent agent);
static int agent_index(EvilAgent agent);
static string random_pw(EvilAgent agent);
static char random_char();
bool hacked[3];
string passwords[3];
};
#endif
//************************************************************
//hack.h
#include \"evil_server.h\"
void hack(EvilServer &es);
//*******************************************************
//main.cpp
#include
#include
#include
#include
#include
#include \"evil_server.h\"
#include \"hack.h\"
using namespace std;
clock_t test_start;
bool test_time_exceeds(int seconds)
{
clock_t cur = clock();
return seconds <= static_cast(cur-test_start)/CLOCKS_PER_SEC;
}
int main()
{
// Initialize random number generation, i.e. rand(), using current millisecond
srand(2017);
test_start = clock();
// Hack something
Evil.
ecorrect Question 32 0 1 pts Referencing the diagram above, use the .pdf
ecorrect Question 32 0 /1 pts Referencing the diagram above, use the following information to
answer the question. You\'re an archaeologist working in Peru, South America and you find a of
skeleton in layer B. You it undiscovered primate species because its opposable thumbs and lack
of a tall You\'re trying to decide what type of dating method is appropriate. You recall finding a
fossil in layer C known to be 200,000 years old. You discover another fossil in layer D that dates
back to 2 million years ago. Given this information, which method would you use to date the
skeleton?
Solution
I ave explained te answeer in a little detail since te points needed explaination. The conventional
belief is that carbon datin is the metod but here we see that there are two types of age
determinations. Geologists in the late 18th and early 19th century studied rock layers and the
fossils in them to determine relative age. For this question we will not employ the carbon dating
method ,used for determining the age of an object containing organic material by using the
properties of radiocarbon (14
C), a radioactive isotope of carbon. because the half-life of 14
C (the period of time after which half of a given sample will have decayed) is about 5,730 years,
the oldest dates that can be reliably measured by this process date to around 50,000 years ago,
although special preparation methods occasionally permit accurate analysis of older samples.
I will list it as two parts which contain informmation to determine the age of rocks and fossils.
PART 1: DETERMINING RELATIVE AGE OF ROCKS
there are two basic principles used by geologists to determine the sequence of ages of rocks.
They are:
Principle of superposition: Younger sedimentary rocks are deposited on top of older sedimentary
rocks.
Principle of cross-cutting relations: Any geologic feature is younger than anything else that it
cuts across.
PART 2: RADIOMETRIC AGE-DATING
Some elements have forms (called isotopes) with unstable atomic nuclei that have a tendency to
change, or decay. For example, U-235 is an unstable isotope of uranium that has 92 protons and
143 neutrons in the nucleus of each atom. Through a series of changes within the nucleus, it
emits several particles, ending up with 82 protons and 125 neutrons. This is a stable condition,
and there are no more changes in the atomic nucleus. A nucleus with that number of protons is
called lead (chemical symbol Pb). The protons (82) and neutrons (125) total 207. This particular
form (isotope) of lead is called Pb-207. U-235 is the parent isotope of Pb-207, which is the
daughter isotope.
Many rocks contain small amounts of unstable isotopes and the daughter isotopes into which
they decay. Where the amounts of parent and daughter isotopes can be accurately measured, the
ratio can be used to determine how old the rock is.
At any moment there is a small chance that each of the nuclei of U-235 will suddenly decay.
That chance of decay is very small, but it is alw.
Disneys Expedition EverestOne of the newest thrill rides to open.pdf
Disney\'s Expedition Everest
One of the newest thrill rides to open in the Walt Disney World Resort may just be the most
impressive. As Disney approached its 50th anniversary, the company wanted to celebrate in a
truly special way. What was its idea? Create a park attraction that would, in many ways, serve as
the link between Disney\'s amazing past and its prom ising future. Disney showed that it was
ready to pull out all stops in order to get everything just right.
In 2006, The Walt Disney Company introduced Expedition Everest in Disney\'s Animal
Kingdom Park at Lake Buena Vista, Florida. Expedition Everest is more than just a roller
coaster. It is the embodiment of the Disney spirit: a ride that combines Disney\'s trademark
thrills, unexpected twists and turns, incredible attention to detail, and impressive project
management skills.
First, let\'s consider some of the technical details of
Expedition Everest:
With a peak of just under 200 feet, the ride is con tained within the tallest of 18 mountains
created by Disney\'s Imagineers at Disney parks worldwide.
The ride contains nearly a mile of track, with twists, tight turns, and sudden drops.
The Disney team created a Yeti: an enormous, fur-covered, Audio-Animatronics monster pow
ered by a set of hydraulic cylinders whose com bined thrust equals that of a Boeing 747 airliner.
Through a series of sketches, computer-animated drawings, sculptures, and tests that took more
than two years to perfect, Disney created and pro grammed its Abominable Snowman to stand
over 10 feet tall and serve as the focal point of the ride.
More than 900 bamboo plants, 10 species of trees, and 110 species of shrubs were planted to re-
create the feeling of the Himalayan lowlands surrounding Mount Everest.
More than 1,800 tons of steel were used to construct the mountain. The covering of the
framework was done using more than 3,000 prefabricated \"chips\" created from 25,000
individual computer-molded pieces of steel.
To create the proper color schemes, 2,000 gallons of stain and paint were used on rockwork and
throughout the village Disney designed to serve as a backdrop for the ride.
More than 2,000 handcrafted items from Asia are used as props, cabinetry, and architectural
ornamentation.
Building an attraction does not come easily or quickly for Disney\'s Imagineers. Expedition
Everest was several years in development as Disney sent teams, including Walt Disney
Imagineering\'s Creative Executive Joe Rohde, on repeated trips to the Himalayas in Nepal to
study the lands, architecture, colors, ecol ogy, and culture in order to create the most authentic
setting for the new attraction. Disney\'s efforts reflect a desire to do much more than provide a
world-class ride experience; they demonstrate the Imagineers\' eagerness to tell a story-a story
that combines the mythology of the Yeti figure with the unique history of the Nepalese living in
the shadow of the world\'s tallest mountain. Ultimately, the attraction, with .
Discuss ONE risk that a company faces when trying to diversify inte.pdf
Discuss ONE risk that a company faces when trying to diversify internationally. Provide
examples, and also discuss how this risk can be mitigated
Solution
One very significant risk that a Company faces when trying to diversify internationally is the
cultural risk because there are certain Dominant cultures that express the core values that are
shared by a majority of the organization’s members while there are also certain subcultures
which could be regarded as the minicultures within an organization, typically defined by
department designations and geographical separation. Hence in terms of implications, both these
cultures impact the tolerance level of the people belonging to different cultures in a major way.
Here, it is a rare ability to bridge such a perception gap which has already been rooted down into
the beliefs of the people hailing from different culture. Further, the National culture has all more
greater impact than the Organizational culture on the minds of the people. In certain culture,
there is a facilitation of commitment to something larger than self-interest that enhances the
stability of the social system. However, it may not get along well with that of the other cultures
where social commitment does not form an intrinsic part of the culture. This is where the
ideologies clashes between the cultures and there arises the challenge of tolerance. Here, if the
leaders are not intellectually brilliant, the diversity management issue will fail. Further, when the
Company goes international, the diversified team ethical issues are more critical to barriers to
change and diversity which also comes in the way of welcoming or accepting another culture in
harmony with that of one culture. The gestures and cultural characteristics are one culture may
be complete inverse from that of the other, which may again pose as a barrier to tolerate other
cultures. Some cultures are liberal while some are way too conservative hence arises the issue of
tolerance to other cultures. Hence despite investing in multicultural teams training, some
alliances still fail. If the leaders are not intellectually brilliant, they would find it difficult to
manage these barriers because their own culture has not taught to see the other cultures in such
details hence this intellect needs to be cultivated as well.
Usually the Companies fail in persuading and controlling people when the Company diversifies
because of the faulty applications of the management theories due to:
Therefore, in order to mitigate the same, one needs to understand the behavior of the employees
working at some other international location and thereby attempting to create a more efficient
organization on the basis of such a study. This strategy could help in applying various scientific
theories to study these individuals working in a Corporate set-up in order to optimize the
performance of overall human resource even in a culturally diversified set up. Understanding the
work culture and the d.
Describe at least one reason why transitioning from PVST+ to Rapid P.pdf
Describe at least one reason why transitioning from PVST+ to Rapid PVST+ is seamless
Solution
PVST+ is the per VLAN spanning tree.It means we run one spanning tree instance per VLAN.
RSTP is the rapid spanning tree port
Rapid PVST+ is the rapid VLAN this has the features of the above two.this is rapid used for
VLAN hence there requires the transition..
CASE 2-1 BUILDING UP OUR ASSETS DHR CONSTRUCTIONIn August 2011, w.pdf
CASE 2-1 BUILDING UP OUR ASSETS: DHR CONSTRUCTION
In August 2011, when the Dow Jones Industrial Average dipped under 8,000 from a high of
almost 11,500, Richard Davis and Stephen Hodgetts, academics, friends, and coauthors, were
lamenting their ever-shrinking retirement funds. As Hodgetts was fond of saying, “America
believes in education: The average professor earns more money in a year than a professional
athlete earns in a whole week.”
After a long discussion, they decided that they needed to take direct control of their investments
and not be prisoners to the rise and fall of the stock market. Davis had done enough preliminary
research on the real estate market in their area and convinced Hodgetts that there was money to
be made in property ownership and renting starter homes (three bedrooms, two baths) to
individuals with poor credit who could not afford a home but who had excellent rental track
records. Davis and Hodgetts formed D&H Management and found six families in three months
who were happy to rent and eventually purchase homes in the $175K price range. The deal was
so attractive that D&H even had a waiting list of new tenants. The six homes, though, gobbled up
their initial investment and required additional capital.
The New Firm: DHR Construction, LLC
Their construction company started off as just a smallcapital-raising venture. Hodgetts and Davis
would finish off the basements of their rental homes, get the homes reappraised, and then
remortgage the properties, pulling out an additional $10K to $20K per home. These funds could
then be used as down payments for future rental homes.
In chatting about this matter with some of their renters, Davis and Hodgetts were approached by
one of them to perform all of the nonlicensed work (excluding electrical, plumbing, HVAC,
etc.). Davis explained to the renters (Alan and Wilma Bronson) that they would have to form
their own company and act as any other subcontractor. After completing a few basements, Alan
and Wilma enjoyed working on these basements so much that they approached Davis and
Hodgetts about figuring out a way that Davis and Hodgetts could keep them occupied all year
round. In essence, Alan would quit his job and work as a subcontractor for Davis and Hodgetts.
This was not possible, though, since there was not enough work to keep Alan and Wilma busy.
Yet a few days later the situation dramatically changed.
One of Davis’s students, David Russ, who was designing their basements, said that he thought
that Davis and Hodgetts could cut out the middleman in terms of the rental business if they built
their own homes. Davis thought that Russ was crazy at the time, but they talked after class and
Russ said that he would be happy to act as the general contractor and that he knew all of the
subcontractors who were needed in order to construct new homes. Alan and Wilma would do all
of the interior work, and Alan could hire some part-time workers to help himself out. Davis and
Hodgetts coul.
C++ Write a function that takes two numbers are parameters and retu.pdf
C++ Write a function that takes two numbers are parameters and returns their product. Create a
program that program the prompts for to numbers, then uses this function calcuate the product,
and then displays the product.
Solution
#include
using namespace std;
int product(int number1,int number2) //function to compute product of two numbers
{
return (number1 * number2); //return product
}
int main()
{
int num1,num2;
cout<<\"Enter two numbers\";
cin>>num1>>num2;
cout<<\"\ Product of \"<.
B.1 Reaction of a Hydrate Addition of Water (2) Appearance Heating (1.pdf
B.1 Reaction of a Hydrate Addition of Water (2) Appearance Heating (1) Initial Final Why did
water droplets form at the mouth of the test tube when heated? lue color ted ought to hae n the r
outh artho at tempt tube . 05 When the hydrate of Cuso, 5H0) was heated, did the reaction
proceed in the direction of the reactants or products? produc Copyright © 2014 Pearson
Education, Inc.
Solution
CuSO4.5H2O
5. when heated CuSO4.5H2O, the reaction proceeds to the right handside, that is towards the
product end. It looses it water of crystallization to form anhydrous CuSO4 salt. Color changed to
white.
6. When water was added, the reaction proceeded to the left handside, such that towards the
reactant end. The color changed to blue due to the formation of CuSO4.5H2O..
Assignment isPage 349-350 #4 and #5 Use the Linked List lab.pdf
Assignment is:
\"Page 349-350 #4 and #5 Use the \"Linked List lab\" you have been working on in class and add
the two functions the questions are asking you to develop: divideMid and divideAt. Be sure to
include comments Use meaningful identifier names (constants where appropriate) Turn in .cpp
file AND Turn in a \"print-screen\' of your output (press \"print-screen\' on keyboard, then
\'paste\' in MS-Word)\"
How do you solve QUESTION #4 in the book data structures using c++ by D.S. Malik in Visiual
Studios using the linked list below with what is being asked? Please need help
Linked list :
#include
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void deleteLast(nodeType*& last, nodeType* first);
int main()
{
nodeType *first, *last;
int num;
createList(first, last);
int choice;
while(true)
{
cout<<\"1. Insert Front.\ 2. Insert Last.\ 3. Delete Front.\ 4. Delete Last.\ 5. Print List.\ 6. Exit.\
\";
cout<<\"Enter your choice: \";
cin>>choice;
switch(choice)
{
case 1: insertFront(first); break;
case 2: insertBack(last); break;
case 3: deleteFirst(first); break;
case 4: deleteLast(last, first); break;
case 5: printList(first); break;
case 6: return 0;
default: cout<<\"Invalid menu option. Try again.\"<>number;
while (number != -999)
{
newNode = new nodeType; // create new node
newNode->info = number;
newNode->link = NULL;
if (first == NULL)
{
first = newNode;
last = newNode;
}
else
{
last->link = newNode;
last = newNode;
}
cout<<\"Enter an integer (-999 to stop): \";
cin>>number;
} // end of while-loop
} // end of build list function
void deleteFirst(nodeType*& first)
{
nodeType *temp;
temp= first;
first= temp->link;
delete temp;
return;
}
void deleteLast(nodeType*& last, nodeType* current)
{
nodeType *temp;
while(current->link != NULL)
{
temp=current;
current=current->link;
}
temp=last;
current->link=NULL;
delete temp;
last = current;
return;
}
void insertFront(nodeType*& front)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= front;
front= newNode;
return;
}
void insertBack(nodeType*& last)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= NULL;
last->link= newNode;
last = newNode;
return;
}
void printList(nodeType*& first)
{
cout<<\"Inside printList...printing linked list...\ \"<info << \" \";
current = current->link;
}
cout<
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void deleteLast(nodeType*& last, nodeType* first);
int main()
{
nodeType *firs.
You maintain several virtual machines (VMs) in an offline state. How.pdf
You maintain several virtual machines (VMs) in an offline state. How do you proceed to add a
particular role to one of those VMs?
Solution
we instll the role for VHD(Virtual Hard Disk) when it remained in offline. and here VHD is the
file format used as virtual hard disk drive. It can contain files and folders as Hard Disk Drive..
Which of the following statements is not TRUE1)For a 64-bit compute.pdf
Which of the following statements is not TRUE?1)For a 64-bit computer, its memory address
space is 264, not 642;2)A process is a program in execution;3)A child process has a different
UID (User Identification) as its parent;4)DMA (Direct Memory Access) controls the flow of
data between memory and some controller without constant CPU intervention.
Solution
Answer:
2 , 3 and 4 are true .
1 is false only
process becomes program when it is been put from start state for processing.
Every child has different UID with the parent.
And we know that DMA controls the data flow..
You are running an ELISA on a sample to test for the presence of .pdf
You are running an ELISA on a sample to test for the presence of antigens from the influenza A
virus. When it comes time to add the antibody, you realize that you only have antibodies for the
influenza B virus. Would these antibodies give you accurate results? Why or why not?
Solution
Influenza virus is divided into three main types 1. Influenza A 2. Influenza B 3. Influenza C
Influenza A virus is mostly observed in a wide variety of birds and mammalian species and
Influenza B is mostly confined to humans.
Influenza A and B have significant differences in antigens that are present in their membrane
proteins.
Antibodies are specific to antigens, as the antigens present on influenza A membrane are
different from influenza B, the addition of influenza B antibodies instead of influenza A
antibodies would result in inaccurate results..
You are to write an efficient program that will read a dictionary of.pdf
You are to write an efficient program that will read a dictionary of 100,000 words, and then
check a document of 2,000,000 words. The program should insert into an array the position of
each misspelled word. I have provided the driver program SpellRunner.cpp, and the necessary
class stubs in speller.h, and speller.cpp. You may add any classes and/or code you wish to
speller.cpp, and speller.h. You may also use, and alter any Weiss files. CreateDocs.out creates
files used for testing. Doc-3-4-1 and Dict-3-4-1 are two testing files for small file.
_____________________________________________________________________________
_________________
SpellRunner.cpp
#include
#include
#include
#include
#include
#include \"CPUTimer.h\"
#include \"mynew.h\"
#include \"speller.h\"
// uncomment line below for DOS/Windows OS.
// #define DOS
#ifdef DOS
#include
#else
#include
#endif
extern int maxRAM;
extern int currentRAM;
using namespace std;
char* readDictionary(char *filename, char **dictionary, int dictSize)
{
int fileSize;
char *s;
struct stat statbuf;
stat(filename, &statbuf);
fileSize = statbuf.st_size;
s = new char[fileSize + 1];
ifstream inf(filename);
inf.read(s, fileSize);
inf.close();
s[fileSize] = \'\\0\';
dictionary[0] = strtok(s, \"\ \");
for(int i = 1; i < dictSize; i++)
dictionary[i] = strtok(NULL, \"\ \ \");
return s;
} // readDictionary
char* readDocument(char **document, int dictSize, int docSize, int seed)
{
char *s, filename[80];
int fileSize;
struct stat statbuf;
sprintf(filename, \"Doc-%d-%d-%d.txt\", dictSize, docSize, seed);
stat(filename, &statbuf);
fileSize = statbuf.st_size;
s = new char[fileSize + 100];
ifstream inf(filename);
inf.read(s, fileSize);
inf.close();
s[fileSize] = \'\\0\';
document[0] = strtok(s, \"\ \");
for(int i = 1; i < docSize; i++)
document[i] = strtok(NULL, \"\ \ \");
return s;
} // readDocument()
void readWrongs(int *wrongs, int dictSize, int docSize, int seed,
int &wrongCount)
{
char filename[80];
wrongCount = 0;
sprintf(filename, \"Wrongs-%d-%d-%d.txt\", dictSize, docSize, seed);
ifstream inf(filename);
while(inf >> wrongs[wrongCount++]);
wrongCount--;
} // readWrongs()
void checkAnswers(int wrongs[], int wrongCount, int misspelled[],
int misspelledCount)
{
for(int i = 0; i < wrongCount && i < misspelledCount; i++)
if(wrongs[i] < misspelled[i])
{
cout << \"Speller missed misspelled word # \" << wrongs[i] << endl;
return;
}
else
if(wrongs[i] > misspelled[i])
{
cout << \"Speller thought correctly spelled word # \" << misspelled[i]
<< \" was wrong\ \";
return;
}
if(wrongCount != misspelledCount)
cout << \"Speller found \" << misspelledCount << \" misspelled words when \"
<< \" there were really \" << wrongCount << \" misspelled words.\ \";
} // checkAnswers
void cleanup(char **dictionary, char *docFilePtr, char **document, int *wrongs, int
*misspelled)
{
delete [] dictionary;
delete [] docFilePtr;
delete [] document;
delete [] wrongs;
delete [] misspelled;
} // cleanup()
int main(int argc, char* argv[])
{
char .
X = C B - B C D; Use Accumulator Register-Register (LoadSt.pdf
X = C* B - B * C * D; Use Accumulator Register-Register (Load/Store): How many addresses?
Memory-Memory: How many addresses?
Solution
Question 11:
Register-Register (Load/store):how many addresses
Three address instructions were used.
Question 12:
Memory-Memory: How many addresses
Two and three address instructions were used..
Tic-Tac-Toe Simulator [C# Visual Basic] Create an application th.pdf
Tic-Tac-Toe Simulator [C# Visual Basic]
Create an application that simulates a game of tic-tac-toe. Figure 7-45 shows an ex-
ample of the application’s form. The form shown in the figure uses eight large Label
controls to display the Xs and Os.
The application should use a two-dimensional int array to simulate the game board
in memory. When the user clicks the New Game button, the application should step
through the array, storing a random number in the range of 0 through 1 in each ele-
ment. The number 0 represents the letter O, and the number 1 represents the letter
X. The form should then be updated to display the game board. The application
should display a message indicating whether player X won, player Y won, or the
game was a tie. Tic-Tac-Toe X O O X ins New Game
Solution
Answer:-
Tic-Tac-Toe Game :
The application should begin with a form that contains the following elements:-
1) A label saying \"Welcome to Tic-Tac-Toe\"
2) Two option buttons enclosed in a frame. The captions for the option buttons should be \"X\"
and \"O\". The caption for the frame should be \"Select \'X\' or \'O\' and click OK\".
3) A command button with the caption \"OK\".
4) When the user clicks OK, exit this form and display the main form (the game board).
5) At the start of each game, the application should select who should go first at random, and
display an appropriate message to the user in a message box (either \"You go first this time.\" or
\"This time, I will go first.\")
6) The game board should be set up with an array of label controls indexed 0 through 8. When a
player clicks one of the available labels, an \"X\" or an \"O\" (depending on the user\'s initial
selection) should appear in that label. Then the computer should make its move.
7) When the program detects a win, a line should be drawn through the \"three in a row\". If the
player wins, the message \"YOU WIN !!!\" should flash across the game board; if the computer
wins, the message \"YOU LOSE!!!\" (or \"I WIN!!!\" or \"COMPUTER WINS!!!\") should flash
across the game board. If the game ends in a tie, the message \"IT\'S A TIE!!!\" should flash
across the game board. (Hint: To get the flashing message, use a label in conjunction with a
timer and toggle the label\'s Visible property on and off.)
8) The program should provide options to play a new game and to quit.
9) The program should provide keep four counts: games played, games won, games lost, and
games tied. These counts should be displayed on the form.
10) The overall appearance of the application should be tasteful and suggest \"fun\". Experiment
with various colors and fonts until you get it the way you want it.
The Computer\'s AI Algorithm
Your program should implement the following algorithm to try to make the computer win:
(1) Examine the board for a winning move, and if you find one, make that move – otherwise
proceed to step (2).
(2) Examine the board to see if your opponent has a winning move, and if so, select that square
to blo.
Which substance has ionic bondsLiIH2ONH3Br2Which sub.pdf
Which substance has ionic bonds
LiI
H2O
NH3
Br2
Which substance has ionic bonds
LiI
H2O
NH3
Br2
Which substance has ionic bonds
LiI
H2O
NH3
Br2
Solution
There are two different ways to determine whether the bond is ionic or not. They are
1. If a compound is made of a metal atom and a non-metal atom, then the bonding will be ionic
2. And it can also be decided from the electronegativity difference between the two atoms of a
molecule. If the electronegativity difference is over or close to 1.7 (this value varies in literature
from 1.6-2.1), then the bond between the atom might be ionic..
What is a culture A subculture Give an example of each.Solutio.pdf
What is a culture? A subculture? Give an example of each.
Solution
Culture is the shared beliefs, characteristics and knowledge of a group of people based on
ideological beliefs, religion, social status, education, age, ethnicity etc. Examples of culture are
Western culture, Eastern culture, African-American culture, Asian culture, middle class culture.
Subculture is a subset of culture and share a distinct identity. Examples of sub culture are: pop
culture, bikers, hippies..
The last major idea in population dynamics is the idea of a minimum .pdf
The last major idea in population dynamics is the idea of a minimum threshold for a species in a
habitat. For some species, survival is not possible if the population size gets too small because of
natural interdependence. (Keep in mind that this is over and above the minimum threshold of two
animals required for sexual reproduction.) A differential equation that models that situation is
dN/ dt = rN (1 N/ T )( 1 N/ K) where 0 < T < K. Based on your experiences so far, choose
appropriate values of r, T, and K and plot a slope field. (a) Convince yourself that your slope
field indeed models a population of animals that is subject to a minimum threshold in order to
survive. (b) Describe the stability of each equilibrium you see. (c) Investigate the affect of
harvesting on this population. Explain how harvesting affects the equilibria and include slope
fields to support your answer. (d) Draw a bifurcation diagram and point out the bifurcation point.
What does that mean in terms of harvesting?
Solution
Exponential development is the rate of growth of population in the presence of unlimited
resources. Exponential development is also know as Density independent growth.
dN/dt = rN
dN is the adjustment or change in population measure
\"r\" is rate of increment of population (births-passings)
Logistic growth is the rate of development of populace when resources are constrained. The
growth is named as density dependent growth.
We represent the logistic growth by Sigmoidal or S-formed growth bend or curve.
dN/dt = rN [1-N/K]; where K is the carrying capacity
Carrying capacity is basically the maximum size of the population supported by the environment.
The development in populace stops at K.
The natural or environmental components modify and make the K to change.
a) Natural growth equation more genereally represented as
P(t) = P0 * ert
where P is the populace at given time t and r is the rate of development of populace
Logistic Growth dependably will exist in between carrying capacity, K and it will indicate S
formed development curve while exponential development or growth seems to move higher.
This development pattern is found in populaces that have entry to constrained resources. The
loest populace develops exponentially at first. Be that as it may, when the constrained resources
begin demonstrating their effects the development or growth slows down and achieves a the
carrying capacity or limiting value.
b) Logistic growth demonstrates discrete and continuous development models.
The discrete model demonstrates that populace development is reliant on most extreme rate of
development, carrying capacity, and rate of per-capita increment of the populace. Logistic
growth is ceaseless when the individual imitates at the rate that decreases as a linear function of
size of populace.
Equation for persistent model is dN/dt = r*N (1-N/K)
In populaces that are lower than K will increment in size while that are higher than K will
diminish in size. K will stay stea.
Python ProgramDo not use import please. Problem 1 Rolodex (Note .pdf
Python Program
Do not use import please. Problem 1: Rolodex (Note: A Rolodex is a Contact List) Write a
program that manages a rolodex. This program prompts the user for the name of a rol file. This
program should be able to load any rol file and require that the file extension -i.e the last 4 letters
of the file be rol. The file rol is consider a valid file. If the file exists the program loads in the
contacts stored in the file. A Rolodex contains unique individuals, no duplicate names are
allowed. However, name lookups are case sensitive, so \"Apple\" and \"apple\" are considered
different names allowed However, name
Solution
Rolodex.java
public class Rolodex{
String name;
String phone;
String email;
String note;
Rolodex(String name,String phone,String email,String note){
this.name=name;
this.phone=phone;
this.email=email;
this.note=note;
}
public String tostring(){
return \"name:\"+name+\"\ phone:\"+phone +\"\ email:\"+email+ \"\ note:\"+note;
}
}
rolodexmain.java
import java.util.*;
public class rolodexmain {
static Scanner sc=new Scanner(System.in);
static Listlist=new ArrayList();
static Setsets=new HashSet();
public static void main(String[] args) {
// TODO Auto-generated method stub
while(true){
System.out.println(\"Rolodex comands \\t Insert(i),remove(R),
Lookbyname(F),LookbyPhone number(N),Lookby email(E)saves(S),save and
quit(Q),printall(P),quit and dont save(X)\");
char ch=sc.next().charAt(0);
switch(ch){
case \'I\':
insert(list);
break;
case \'R\':
deleted(list);
break;
case \'F\':
Lookbyname(list);
break;
case \'N\':
Lookbyphonenumber(list);
break;
case \'E\':
Lookbyemail(list);
break;
case \'P\':
printall(list);
break;
case \'X\':
System.exit(0);
case \'S\':
saves(list);
break;
case \'Q\':
savesandquit(list);
System.exit(0);
}
}
}
public static void insert(List list){
System.out.println(\"please enter contact name:\");
String name=sc.next();
System.out.println(\"please enter contact phone number:\");
String phone=sc.next();
System.out.println(\"please enter contact email id:\");
String email=sc.next();
System.out.println(\"please enter contact note:\");
String note=sc.next();
Rolodex r=new Rolodex(name,phone,email,note);
list.add(r);
}
public static void deleted(List list){
boolean flag=false;
System.out.println(\"enter name of contact to delete :\");
String dname=sc.next();
for(int i=0;i list){
System.out.println(\"enter name of contact which you want to look :\");
String lname=sc.next();
for(int i=0;i list){
System.out.println(\"please enter phone number which you want to look:\");
String lphone=sc.next();
for(int i=0;i list){
System.out.println(\"please enter phone number which you want to look:\");
String lemail=sc.next();
for(int i=0;i list){
for(int i=0;i list){
sets.addAll(list);
for (Rolodex o : sets) {
// Rolodex r=(Rolodex)sets;
System.out.println(o.tostring());
}
}
public static void savesandquit(List list){
sets.addAll(list);
for (Rolodex o : sets) {
// Rolodex r=(Rolodex)sets;
System.out.println(o.tostring());
}
}.
The partial pressure of oxygen in the alveolar gas is affected by cer.pdf
The partial pressure of oxygen in the alveolar gas is affected by certain situations. Classify each
of the following conditions as causing an increase, decrease, or no change in alveolar PO_2.
Solution
Increase in pO2 will occur due to
Increased alveolar ventilation and unchanged metabolism
Decreased metabolism and unchanged alveolar ventilation.
Decreased pO2 will occur due to
Decreased alveolar ventilation and unchanged metabolism
Increased metabolism and unchanged alveolar ventilation
Breathing air with low pO2
No change in pO2 will occur due to
Proportional increase in metabolism and alveolar ventilation
Remember that, metabolism requires oxygen so it lowers partial pressure of oxygen whereas
alveolar ventilation always increase pO2..
A student had a set of solutions labeled A-E which contained in no o.pdf
A student had a set of solutions labeled A-E which contained in no order aquoeous soultions of
KOH,H2SO4,Fe(NO3)2, HBr, and Mg(NO3)2. After the initial pH test they determined that A
was a base,B/C were neutral and D/E were acids. They then mixed B and C with D and then E
and got a precipitate with the BD and CE mixtures but not the BE or CD mixtures. They then
came to the following erroneous conclusions A:KOH, B:FE(NO3)2 C:Mg(NO3)2 D: HBr and E:
H2SO4 with the only correct identification being that solution A was KOH. What did they do
wrong and what test should they have done to confirm their results?
Solution
What wrong they did war that they have not done the qualitative analysis of all the A, B, C, D
and E to conclude what are the ions or species present. After that, qualitative analysis for the
precipitate is also important. only looking at the acidic or basic behaviour of their solution, it is
very difficult to confirm the species..
A small insect viewed through a convex lens is 1.2 cm from the lens a.pdf
A small insect viewed through a convex lens is 1.2 cm from the lens and appear twice its actual
size. What is the focal length of the lens?
Solution
m = hi/ho = -di /do
2 = - di / do
and object distance is given , do = 1.2 cm
di = - 2 x 1.2 = - 2.4 cm
Applying lens equation,
1/f = 1/di + 1/do
1/f = 1/(-2.4) + 1/1.2
f = 2.4 cm ......Ans.
Advanced Java[Extra Concepts, Not Difficult].docx
This is part 2 of my Java Learning Journey. This contains Hashing, ArrayList, LinkedList, Date and Time Classes, Calendar Class and more.
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Disney\'s Expedition Everest
One of the newest thrill rides to open in the Walt Disney World Resort may just be the most
impressive. As Disney approached its 50th anniversary, the company wanted to celebrate in a
truly special way. What was its idea? Create a park attraction that would, in many ways, serve as
the link between Disney\'s amazing past and its prom ising future. Disney showed that it was
ready to pull out all stops in order to get everything just right.
In 2006, The Walt Disney Company introduced Expedition Everest in Disney\'s Animal
Kingdom Park at Lake Buena Vista, Florida. Expedition Everest is more than just a roller
coaster. It is the embodiment of the Disney spirit: a ride that combines Disney\'s trademark
thrills, unexpected twists and turns, incredible attention to detail, and impressive project
management skills.
First, let\'s consider some of the technical details of
Expedition Everest:
With a peak of just under 200 feet, the ride is con tained within the tallest of 18 mountains
created by Disney\'s Imagineers at Disney parks worldwide.
The ride contains nearly a mile of track, with twists, tight turns, and sudden drops.
The Disney team created a Yeti: an enormous, fur-covered, Audio-Animatronics monster pow
ered by a set of hydraulic cylinders whose com bined thrust equals that of a Boeing 747 airliner.
Through a series of sketches, computer-animated drawings, sculptures, and tests that took more
than two years to perfect, Disney created and pro grammed its Abominable Snowman to stand
over 10 feet tall and serve as the focal point of the ride.
More than 900 bamboo plants, 10 species of trees, and 110 species of shrubs were planted to re-
create the feeling of the Himalayan lowlands surrounding Mount Everest.
More than 1,800 tons of steel were used to construct the mountain. The covering of the
framework was done using more than 3,000 prefabricated \"chips\" created from 25,000
individual computer-molded pieces of steel.
To create the proper color schemes, 2,000 gallons of stain and paint were used on rockwork and
throughout the village Disney designed to serve as a backdrop for the ride.
More than 2,000 handcrafted items from Asia are used as props, cabinetry, and architectural
ornamentation.
Building an attraction does not come easily or quickly for Disney\'s Imagineers. Expedition
Everest was several years in development as Disney sent teams, including Walt Disney
Imagineering\'s Creative Executive Joe Rohde, on repeated trips to the Himalayas in Nepal to
study the lands, architecture, colors, ecol ogy, and culture in order to create the most authentic
setting for the new attraction. Disney\'s efforts reflect a desire to do much more than provide a
world-class ride experience; they demonstrate the Imagineers\' eagerness to tell a story-a story
that combines the mythology of the Yeti figure with the unique history of the Nepalese living in
the shadow of the world\'s tallest mountain. Ultimately, the attraction, with .
Discuss ONE risk that a company faces when trying to diversify inte.pdf
Discuss ONE risk that a company faces when trying to diversify internationally. Provide
examples, and also discuss how this risk can be mitigated
Solution
One very significant risk that a Company faces when trying to diversify internationally is the
cultural risk because there are certain Dominant cultures that express the core values that are
shared by a majority of the organization’s members while there are also certain subcultures
which could be regarded as the minicultures within an organization, typically defined by
department designations and geographical separation. Hence in terms of implications, both these
cultures impact the tolerance level of the people belonging to different cultures in a major way.
Here, it is a rare ability to bridge such a perception gap which has already been rooted down into
the beliefs of the people hailing from different culture. Further, the National culture has all more
greater impact than the Organizational culture on the minds of the people. In certain culture,
there is a facilitation of commitment to something larger than self-interest that enhances the
stability of the social system. However, it may not get along well with that of the other cultures
where social commitment does not form an intrinsic part of the culture. This is where the
ideologies clashes between the cultures and there arises the challenge of tolerance. Here, if the
leaders are not intellectually brilliant, the diversity management issue will fail. Further, when the
Company goes international, the diversified team ethical issues are more critical to barriers to
change and diversity which also comes in the way of welcoming or accepting another culture in
harmony with that of one culture. The gestures and cultural characteristics are one culture may
be complete inverse from that of the other, which may again pose as a barrier to tolerate other
cultures. Some cultures are liberal while some are way too conservative hence arises the issue of
tolerance to other cultures. Hence despite investing in multicultural teams training, some
alliances still fail. If the leaders are not intellectually brilliant, they would find it difficult to
manage these barriers because their own culture has not taught to see the other cultures in such
details hence this intellect needs to be cultivated as well.
Usually the Companies fail in persuading and controlling people when the Company diversifies
because of the faulty applications of the management theories due to:
Therefore, in order to mitigate the same, one needs to understand the behavior of the employees
working at some other international location and thereby attempting to create a more efficient
organization on the basis of such a study. This strategy could help in applying various scientific
theories to study these individuals working in a Corporate set-up in order to optimize the
performance of overall human resource even in a culturally diversified set up. Understanding the
work culture and the d.
Describe at least one reason why transitioning from PVST+ to Rapid P.pdf
Describe at least one reason why transitioning from PVST+ to Rapid PVST+ is seamless
Solution
PVST+ is the per VLAN spanning tree.It means we run one spanning tree instance per VLAN.
RSTP is the rapid spanning tree port
Rapid PVST+ is the rapid VLAN this has the features of the above two.this is rapid used for
VLAN hence there requires the transition..
CASE 2-1 BUILDING UP OUR ASSETS DHR CONSTRUCTIONIn August 2011, w.pdf
CASE 2-1 BUILDING UP OUR ASSETS: DHR CONSTRUCTION
In August 2011, when the Dow Jones Industrial Average dipped under 8,000 from a high of
almost 11,500, Richard Davis and Stephen Hodgetts, academics, friends, and coauthors, were
lamenting their ever-shrinking retirement funds. As Hodgetts was fond of saying, “America
believes in education: The average professor earns more money in a year than a professional
athlete earns in a whole week.”
After a long discussion, they decided that they needed to take direct control of their investments
and not be prisoners to the rise and fall of the stock market. Davis had done enough preliminary
research on the real estate market in their area and convinced Hodgetts that there was money to
be made in property ownership and renting starter homes (three bedrooms, two baths) to
individuals with poor credit who could not afford a home but who had excellent rental track
records. Davis and Hodgetts formed D&H Management and found six families in three months
who were happy to rent and eventually purchase homes in the $175K price range. The deal was
so attractive that D&H even had a waiting list of new tenants. The six homes, though, gobbled up
their initial investment and required additional capital.
The New Firm: DHR Construction, LLC
Their construction company started off as just a smallcapital-raising venture. Hodgetts and Davis
would finish off the basements of their rental homes, get the homes reappraised, and then
remortgage the properties, pulling out an additional $10K to $20K per home. These funds could
then be used as down payments for future rental homes.
In chatting about this matter with some of their renters, Davis and Hodgetts were approached by
one of them to perform all of the nonlicensed work (excluding electrical, plumbing, HVAC,
etc.). Davis explained to the renters (Alan and Wilma Bronson) that they would have to form
their own company and act as any other subcontractor. After completing a few basements, Alan
and Wilma enjoyed working on these basements so much that they approached Davis and
Hodgetts about figuring out a way that Davis and Hodgetts could keep them occupied all year
round. In essence, Alan would quit his job and work as a subcontractor for Davis and Hodgetts.
This was not possible, though, since there was not enough work to keep Alan and Wilma busy.
Yet a few days later the situation dramatically changed.
One of Davis’s students, David Russ, who was designing their basements, said that he thought
that Davis and Hodgetts could cut out the middleman in terms of the rental business if they built
their own homes. Davis thought that Russ was crazy at the time, but they talked after class and
Russ said that he would be happy to act as the general contractor and that he knew all of the
subcontractors who were needed in order to construct new homes. Alan and Wilma would do all
of the interior work, and Alan could hire some part-time workers to help himself out. Davis and
Hodgetts coul.
C++ Write a function that takes two numbers are parameters and retu.pdf
C++ Write a function that takes two numbers are parameters and returns their product. Create a
program that program the prompts for to numbers, then uses this function calcuate the product,
and then displays the product.
Solution
#include
using namespace std;
int product(int number1,int number2) //function to compute product of two numbers
{
return (number1 * number2); //return product
}
int main()
{
int num1,num2;
cout<<\"Enter two numbers\";
cin>>num1>>num2;
cout<<\"\ Product of \"<.
B.1 Reaction of a Hydrate Addition of Water (2) Appearance Heating (1.pdf
B.1 Reaction of a Hydrate Addition of Water (2) Appearance Heating (1) Initial Final Why did
water droplets form at the mouth of the test tube when heated? lue color ted ought to hae n the r
outh artho at tempt tube . 05 When the hydrate of Cuso, 5H0) was heated, did the reaction
proceed in the direction of the reactants or products? produc Copyright © 2014 Pearson
Education, Inc.
Solution
CuSO4.5H2O
5. when heated CuSO4.5H2O, the reaction proceeds to the right handside, that is towards the
product end. It looses it water of crystallization to form anhydrous CuSO4 salt. Color changed to
white.
6. When water was added, the reaction proceeded to the left handside, such that towards the
reactant end. The color changed to blue due to the formation of CuSO4.5H2O..
Assignment isPage 349-350 #4 and #5 Use the Linked List lab.pdf
Assignment is:
\"Page 349-350 #4 and #5 Use the \"Linked List lab\" you have been working on in class and add
the two functions the questions are asking you to develop: divideMid and divideAt. Be sure to
include comments Use meaningful identifier names (constants where appropriate) Turn in .cpp
file AND Turn in a \"print-screen\' of your output (press \"print-screen\' on keyboard, then
\'paste\' in MS-Word)\"
How do you solve QUESTION #4 in the book data structures using c++ by D.S. Malik in Visiual
Studios using the linked list below with what is being asked? Please need help
Linked list :
#include
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void deleteLast(nodeType*& last, nodeType* first);
int main()
{
nodeType *first, *last;
int num;
createList(first, last);
int choice;
while(true)
{
cout<<\"1. Insert Front.\ 2. Insert Last.\ 3. Delete Front.\ 4. Delete Last.\ 5. Print List.\ 6. Exit.\
\";
cout<<\"Enter your choice: \";
cin>>choice;
switch(choice)
{
case 1: insertFront(first); break;
case 2: insertBack(last); break;
case 3: deleteFirst(first); break;
case 4: deleteLast(last, first); break;
case 5: printList(first); break;
case 6: return 0;
default: cout<<\"Invalid menu option. Try again.\"<>number;
while (number != -999)
{
newNode = new nodeType; // create new node
newNode->info = number;
newNode->link = NULL;
if (first == NULL)
{
first = newNode;
last = newNode;
}
else
{
last->link = newNode;
last = newNode;
}
cout<<\"Enter an integer (-999 to stop): \";
cin>>number;
} // end of while-loop
} // end of build list function
void deleteFirst(nodeType*& first)
{
nodeType *temp;
temp= first;
first= temp->link;
delete temp;
return;
}
void deleteLast(nodeType*& last, nodeType* current)
{
nodeType *temp;
while(current->link != NULL)
{
temp=current;
current=current->link;
}
temp=last;
current->link=NULL;
delete temp;
last = current;
return;
}
void insertFront(nodeType*& front)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= front;
front= newNode;
return;
}
void insertBack(nodeType*& last)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= NULL;
last->link= newNode;
last = newNode;
return;
}
void printList(nodeType*& first)
{
cout<<\"Inside printList...printing linked list...\ \"<info << \" \";
current = current->link;
}
cout<
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void deleteLast(nodeType*& last, nodeType* first);
int main()
{
nodeType *firs.
You maintain several virtual machines (VMs) in an offline state. How.pdf
You maintain several virtual machines (VMs) in an offline state. How do you proceed to add a
particular role to one of those VMs?
Solution
we instll the role for VHD(Virtual Hard Disk) when it remained in offline. and here VHD is the
file format used as virtual hard disk drive. It can contain files and folders as Hard Disk Drive..
Which of the following statements is not TRUE1)For a 64-bit compute.pdf
Which of the following statements is not TRUE?1)For a 64-bit computer, its memory address
space is 264, not 642;2)A process is a program in execution;3)A child process has a different
UID (User Identification) as its parent;4)DMA (Direct Memory Access) controls the flow of
data between memory and some controller without constant CPU intervention.
Solution
Answer:
2 , 3 and 4 are true .
1 is false only
process becomes program when it is been put from start state for processing.
Every child has different UID with the parent.
And we know that DMA controls the data flow..
You are running an ELISA on a sample to test for the presence of .pdf
You are running an ELISA on a sample to test for the presence of antigens from the influenza A
virus. When it comes time to add the antibody, you realize that you only have antibodies for the
influenza B virus. Would these antibodies give you accurate results? Why or why not?
Solution
Influenza virus is divided into three main types 1. Influenza A 2. Influenza B 3. Influenza C
Influenza A virus is mostly observed in a wide variety of birds and mammalian species and
Influenza B is mostly confined to humans.
Influenza A and B have significant differences in antigens that are present in their membrane
proteins.
Antibodies are specific to antigens, as the antigens present on influenza A membrane are
different from influenza B, the addition of influenza B antibodies instead of influenza A
antibodies would result in inaccurate results..
You are to write an efficient program that will read a dictionary of.pdf
You are to write an efficient program that will read a dictionary of 100,000 words, and then
check a document of 2,000,000 words. The program should insert into an array the position of
each misspelled word. I have provided the driver program SpellRunner.cpp, and the necessary
class stubs in speller.h, and speller.cpp. You may add any classes and/or code you wish to
speller.cpp, and speller.h. You may also use, and alter any Weiss files. CreateDocs.out creates
files used for testing. Doc-3-4-1 and Dict-3-4-1 are two testing files for small file.
_____________________________________________________________________________
_________________
SpellRunner.cpp
#include
#include
#include
#include
#include
#include \"CPUTimer.h\"
#include \"mynew.h\"
#include \"speller.h\"
// uncomment line below for DOS/Windows OS.
// #define DOS
#ifdef DOS
#include
#else
#include
#endif
extern int maxRAM;
extern int currentRAM;
using namespace std;
char* readDictionary(char *filename, char **dictionary, int dictSize)
{
int fileSize;
char *s;
struct stat statbuf;
stat(filename, &statbuf);
fileSize = statbuf.st_size;
s = new char[fileSize + 1];
ifstream inf(filename);
inf.read(s, fileSize);
inf.close();
s[fileSize] = \'\\0\';
dictionary[0] = strtok(s, \"\ \");
for(int i = 1; i < dictSize; i++)
dictionary[i] = strtok(NULL, \"\ \ \");
return s;
} // readDictionary
char* readDocument(char **document, int dictSize, int docSize, int seed)
{
char *s, filename[80];
int fileSize;
struct stat statbuf;
sprintf(filename, \"Doc-%d-%d-%d.txt\", dictSize, docSize, seed);
stat(filename, &statbuf);
fileSize = statbuf.st_size;
s = new char[fileSize + 100];
ifstream inf(filename);
inf.read(s, fileSize);
inf.close();
s[fileSize] = \'\\0\';
document[0] = strtok(s, \"\ \");
for(int i = 1; i < docSize; i++)
document[i] = strtok(NULL, \"\ \ \");
return s;
} // readDocument()
void readWrongs(int *wrongs, int dictSize, int docSize, int seed,
int &wrongCount)
{
char filename[80];
wrongCount = 0;
sprintf(filename, \"Wrongs-%d-%d-%d.txt\", dictSize, docSize, seed);
ifstream inf(filename);
while(inf >> wrongs[wrongCount++]);
wrongCount--;
} // readWrongs()
void checkAnswers(int wrongs[], int wrongCount, int misspelled[],
int misspelledCount)
{
for(int i = 0; i < wrongCount && i < misspelledCount; i++)
if(wrongs[i] < misspelled[i])
{
cout << \"Speller missed misspelled word # \" << wrongs[i] << endl;
return;
}
else
if(wrongs[i] > misspelled[i])
{
cout << \"Speller thought correctly spelled word # \" << misspelled[i]
<< \" was wrong\ \";
return;
}
if(wrongCount != misspelledCount)
cout << \"Speller found \" << misspelledCount << \" misspelled words when \"
<< \" there were really \" << wrongCount << \" misspelled words.\ \";
} // checkAnswers
void cleanup(char **dictionary, char *docFilePtr, char **document, int *wrongs, int
*misspelled)
{
delete [] dictionary;
delete [] docFilePtr;
delete [] document;
delete [] wrongs;
delete [] misspelled;
} // cleanup()
int main(int argc, char* argv[])
{
char .
X = C B - B C D; Use Accumulator Register-Register (LoadSt.pdf
X = C* B - B * C * D; Use Accumulator Register-Register (Load/Store): How many addresses?
Memory-Memory: How many addresses?
Solution
Question 11:
Register-Register (Load/store):how many addresses
Three address instructions were used.
Question 12:
Memory-Memory: How many addresses
Two and three address instructions were used..
Tic-Tac-Toe Simulator [C# Visual Basic] Create an application th.pdf
Tic-Tac-Toe Simulator [C# Visual Basic]
Create an application that simulates a game of tic-tac-toe. Figure 7-45 shows an ex-
ample of the application’s form. The form shown in the figure uses eight large Label
controls to display the Xs and Os.
The application should use a two-dimensional int array to simulate the game board
in memory. When the user clicks the New Game button, the application should step
through the array, storing a random number in the range of 0 through 1 in each ele-
ment. The number 0 represents the letter O, and the number 1 represents the letter
X. The form should then be updated to display the game board. The application
should display a message indicating whether player X won, player Y won, or the
game was a tie. Tic-Tac-Toe X O O X ins New Game
Solution
Answer:-
Tic-Tac-Toe Game :
The application should begin with a form that contains the following elements:-
1) A label saying \"Welcome to Tic-Tac-Toe\"
2) Two option buttons enclosed in a frame. The captions for the option buttons should be \"X\"
and \"O\". The caption for the frame should be \"Select \'X\' or \'O\' and click OK\".
3) A command button with the caption \"OK\".
4) When the user clicks OK, exit this form and display the main form (the game board).
5) At the start of each game, the application should select who should go first at random, and
display an appropriate message to the user in a message box (either \"You go first this time.\" or
\"This time, I will go first.\")
6) The game board should be set up with an array of label controls indexed 0 through 8. When a
player clicks one of the available labels, an \"X\" or an \"O\" (depending on the user\'s initial
selection) should appear in that label. Then the computer should make its move.
7) When the program detects a win, a line should be drawn through the \"three in a row\". If the
player wins, the message \"YOU WIN !!!\" should flash across the game board; if the computer
wins, the message \"YOU LOSE!!!\" (or \"I WIN!!!\" or \"COMPUTER WINS!!!\") should flash
across the game board. If the game ends in a tie, the message \"IT\'S A TIE!!!\" should flash
across the game board. (Hint: To get the flashing message, use a label in conjunction with a
timer and toggle the label\'s Visible property on and off.)
8) The program should provide options to play a new game and to quit.
9) The program should provide keep four counts: games played, games won, games lost, and
games tied. These counts should be displayed on the form.
10) The overall appearance of the application should be tasteful and suggest \"fun\". Experiment
with various colors and fonts until you get it the way you want it.
The Computer\'s AI Algorithm
Your program should implement the following algorithm to try to make the computer win:
(1) Examine the board for a winning move, and if you find one, make that move – otherwise
proceed to step (2).
(2) Examine the board to see if your opponent has a winning move, and if so, select that square
to blo.
Which substance has ionic bondsLiIH2ONH3Br2Which sub.pdf
Which substance has ionic bonds
LiI
H2O
NH3
Br2
Which substance has ionic bonds
LiI
H2O
NH3
Br2
Which substance has ionic bonds
LiI
H2O
NH3
Br2
Solution
There are two different ways to determine whether the bond is ionic or not. They are
1. If a compound is made of a metal atom and a non-metal atom, then the bonding will be ionic
2. And it can also be decided from the electronegativity difference between the two atoms of a
molecule. If the electronegativity difference is over or close to 1.7 (this value varies in literature
from 1.6-2.1), then the bond between the atom might be ionic..
What is a culture A subculture Give an example of each.Solutio.pdf
What is a culture? A subculture? Give an example of each.
Solution
Culture is the shared beliefs, characteristics and knowledge of a group of people based on
ideological beliefs, religion, social status, education, age, ethnicity etc. Examples of culture are
Western culture, Eastern culture, African-American culture, Asian culture, middle class culture.
Subculture is a subset of culture and share a distinct identity. Examples of sub culture are: pop
culture, bikers, hippies..
The last major idea in population dynamics is the idea of a minimum .pdf
The last major idea in population dynamics is the idea of a minimum threshold for a species in a
habitat. For some species, survival is not possible if the population size gets too small because of
natural interdependence. (Keep in mind that this is over and above the minimum threshold of two
animals required for sexual reproduction.) A differential equation that models that situation is
dN/ dt = rN (1 N/ T )( 1 N/ K) where 0 < T < K. Based on your experiences so far, choose
appropriate values of r, T, and K and plot a slope field. (a) Convince yourself that your slope
field indeed models a population of animals that is subject to a minimum threshold in order to
survive. (b) Describe the stability of each equilibrium you see. (c) Investigate the affect of
harvesting on this population. Explain how harvesting affects the equilibria and include slope
fields to support your answer. (d) Draw a bifurcation diagram and point out the bifurcation point.
What does that mean in terms of harvesting?
Solution
Exponential development is the rate of growth of population in the presence of unlimited
resources. Exponential development is also know as Density independent growth.
dN/dt = rN
dN is the adjustment or change in population measure
\"r\" is rate of increment of population (births-passings)
Logistic growth is the rate of development of populace when resources are constrained. The
growth is named as density dependent growth.
We represent the logistic growth by Sigmoidal or S-formed growth bend or curve.
dN/dt = rN [1-N/K]; where K is the carrying capacity
Carrying capacity is basically the maximum size of the population supported by the environment.
The development in populace stops at K.
The natural or environmental components modify and make the K to change.
a) Natural growth equation more genereally represented as
P(t) = P0 * ert
where P is the populace at given time t and r is the rate of development of populace
Logistic Growth dependably will exist in between carrying capacity, K and it will indicate S
formed development curve while exponential development or growth seems to move higher.
This development pattern is found in populaces that have entry to constrained resources. The
loest populace develops exponentially at first. Be that as it may, when the constrained resources
begin demonstrating their effects the development or growth slows down and achieves a the
carrying capacity or limiting value.
b) Logistic growth demonstrates discrete and continuous development models.
The discrete model demonstrates that populace development is reliant on most extreme rate of
development, carrying capacity, and rate of per-capita increment of the populace. Logistic
growth is ceaseless when the individual imitates at the rate that decreases as a linear function of
size of populace.
Equation for persistent model is dN/dt = r*N (1-N/K)
In populaces that are lower than K will increment in size while that are higher than K will
diminish in size. K will stay stea.
Python ProgramDo not use import please. Problem 1 Rolodex (Note .pdf
Python Program
Do not use import please. Problem 1: Rolodex (Note: A Rolodex is a Contact List) Write a
program that manages a rolodex. This program prompts the user for the name of a rol file. This
program should be able to load any rol file and require that the file extension -i.e the last 4 letters
of the file be rol. The file rol is consider a valid file. If the file exists the program loads in the
contacts stored in the file. A Rolodex contains unique individuals, no duplicate names are
allowed. However, name lookups are case sensitive, so \"Apple\" and \"apple\" are considered
different names allowed However, name
Solution
Rolodex.java
public class Rolodex{
String name;
String phone;
String email;
String note;
Rolodex(String name,String phone,String email,String note){
this.name=name;
this.phone=phone;
this.email=email;
this.note=note;
}
public String tostring(){
return \"name:\"+name+\"\ phone:\"+phone +\"\ email:\"+email+ \"\ note:\"+note;
}
}
rolodexmain.java
import java.util.*;
public class rolodexmain {
static Scanner sc=new Scanner(System.in);
static Listlist=new ArrayList();
static Setsets=new HashSet();
public static void main(String[] args) {
// TODO Auto-generated method stub
while(true){
System.out.println(\"Rolodex comands \\t Insert(i),remove(R),
Lookbyname(F),LookbyPhone number(N),Lookby email(E)saves(S),save and
quit(Q),printall(P),quit and dont save(X)\");
char ch=sc.next().charAt(0);
switch(ch){
case \'I\':
insert(list);
break;
case \'R\':
deleted(list);
break;
case \'F\':
Lookbyname(list);
break;
case \'N\':
Lookbyphonenumber(list);
break;
case \'E\':
Lookbyemail(list);
break;
case \'P\':
printall(list);
break;
case \'X\':
System.exit(0);
case \'S\':
saves(list);
break;
case \'Q\':
savesandquit(list);
System.exit(0);
}
}
}
public static void insert(List list){
System.out.println(\"please enter contact name:\");
String name=sc.next();
System.out.println(\"please enter contact phone number:\");
String phone=sc.next();
System.out.println(\"please enter contact email id:\");
String email=sc.next();
System.out.println(\"please enter contact note:\");
String note=sc.next();
Rolodex r=new Rolodex(name,phone,email,note);
list.add(r);
}
public static void deleted(List list){
boolean flag=false;
System.out.println(\"enter name of contact to delete :\");
String dname=sc.next();
for(int i=0;i list){
System.out.println(\"enter name of contact which you want to look :\");
String lname=sc.next();
for(int i=0;i list){
System.out.println(\"please enter phone number which you want to look:\");
String lphone=sc.next();
for(int i=0;i list){
System.out.println(\"please enter phone number which you want to look:\");
String lemail=sc.next();
for(int i=0;i list){
for(int i=0;i list){
sets.addAll(list);
for (Rolodex o : sets) {
// Rolodex r=(Rolodex)sets;
System.out.println(o.tostring());
}
}
public static void savesandquit(List list){
sets.addAll(list);
for (Rolodex o : sets) {
// Rolodex r=(Rolodex)sets;
System.out.println(o.tostring());
}
}.
The partial pressure of oxygen in the alveolar gas is affected by cer.pdf
The partial pressure of oxygen in the alveolar gas is affected by certain situations. Classify each
of the following conditions as causing an increase, decrease, or no change in alveolar PO_2.
Solution
Increase in pO2 will occur due to
Increased alveolar ventilation and unchanged metabolism
Decreased metabolism and unchanged alveolar ventilation.
Decreased pO2 will occur due to
Decreased alveolar ventilation and unchanged metabolism
Increased metabolism and unchanged alveolar ventilation
Breathing air with low pO2
No change in pO2 will occur due to
Proportional increase in metabolism and alveolar ventilation
Remember that, metabolism requires oxygen so it lowers partial pressure of oxygen whereas
alveolar ventilation always increase pO2..
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A student had a set of solutions labeled A-E which contained in no order aquoeous soultions of
KOH,H2SO4,Fe(NO3)2, HBr, and Mg(NO3)2. After the initial pH test they determined that A
was a base,B/C were neutral and D/E were acids. They then mixed B and C with D and then E
and got a precipitate with the BD and CE mixtures but not the BE or CD mixtures. They then
came to the following erroneous conclusions A:KOH, B:FE(NO3)2 C:Mg(NO3)2 D: HBr and E:
H2SO4 with the only correct identification being that solution A was KOH. What did they do
wrong and what test should they have done to confirm their results?
Solution
What wrong they did war that they have not done the qualitative analysis of all the A, B, C, D
and E to conclude what are the ions or species present. After that, qualitative analysis for the
precipitate is also important. only looking at the acidic or basic behaviour of their solution, it is
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A small insect viewed through a convex lens is 1.2 cm from the lens and appear twice its actual
size. What is the focal length of the lens?
Solution
m = hi/ho = -di /do
2 = - di / do
and object distance is given , do = 1.2 cm
di = - 2 x 1.2 = - 2.4 cm
Applying lens equation,
1/f = 1/di + 1/do
1/f = 1/(-2.4) + 1/1.2
f = 2.4 cm ......Ans.
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