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- You have to overcome the ionic forces of the CsI, the hydrogen bon.pdf
- You have to overcome the ionic forces of the CsI, the hydrogen bonding of the HF. In return
you get ion-dipole forces of the Cs(+) and I(-) in the solution
- Dispersion Forces and Metallic bonds
Solution
- You have to overcome the ionic forces of the CsI, the hydrogen bonding of the HF. In return
you get ion-dipole forces of the Cs(+) and I(-) in the solution
- Dispersion Forces and Metallic bonds.
#includeiostream struct node { char value; struct no.pdf
#include
struct node
{
char value;
struct node *next;
};
class StringOfNode
{
struct node * head;
public:
StringOfNode()
{
head=NULL;
}
int size()
{
struct node *p=head;
int count=1;
while(p)
{
count++;
p=p->next;
}
return count;
}
struct node * begin()
{
return head;
}
char &operator[](int index)
{
int i=0;
struct node *p=head;
while(index!=i)
{
p=p->next;
i++;
}
return p->value;
}
void insertFront(char valuee)
{
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(head==NULL)
head=temp;
else
{
temp->next=head;
head=temp;
}
}
void insert(char valuee, int pos)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
int i=0;
if(p==NULL)
{
head=temp;
}
else
{
while(i!=pos-1)
{
p=p->next;
i++;
}
temp->next=p->next;
p->next=temp;
}
}
void insertBack(char valuee)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(p==NULL)
{
head=temp;
}
else
{
while(p->next!=NULL)
p=p->next;
p->next=temp;
}
}
void print()
{
struct node *p=head;
while(p)
{
std::cout<value<<\"\ \";
p=p->next;
}
}
StringOfNode operator+(const StringOfNode& b)
{
StringOfNode *S;
struct node *p=head;
while(p->next!=NULL)
p=p->next;
p->next=S->begin();
}
};
int main()
{
StringOfNode *p1=new StringOfNode;
p1->insertBack(\'b\');
p1->insertBack(\'c\');
p1->insertBack(\'e\');
p1->insert(\'d\',2);
p1->insertFront(\'a\');
p1->print();
system(\"pause\");
return 0;
}
Solution
#include
struct node
{
char value;
struct node *next;
};
class StringOfNode
{
struct node * head;
public:
StringOfNode()
{
head=NULL;
}
int size()
{
struct node *p=head;
int count=1;
while(p)
{
count++;
p=p->next;
}
return count;
}
struct node * begin()
{
return head;
}
char &operator[](int index)
{
int i=0;
struct node *p=head;
while(index!=i)
{
p=p->next;
i++;
}
return p->value;
}
void insertFront(char valuee)
{
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(head==NULL)
head=temp;
else
{
temp->next=head;
head=temp;
}
}
void insert(char valuee, int pos)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
int i=0;
if(p==NULL)
{
head=temp;
}
else
{
while(i!=pos-1)
{
p=p->next;
i++;
}
temp->next=p->next;
p->next=temp;
}
}
void insertBack(char valuee)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(p==NULL)
{
head=temp;
}
else
{
while(p->next!=NULL)
p=p->next;
p->next=temp;
}
}
void print()
{
struct node *p=head;
while(p)
{
std::cout<value<<\"\ \";
p=p->next;
}
}
StringOfNode operator+(const StringOfNode& b)
{
StringOfNode *S;
struct node *p=head;
while(p->next!=NULL)
p=p->next;
p->next=S->begin();
}
};
int main()
{
StringOfNode *p1=new StringOfNode;
p1->insertBack(\'b\');
p1->insertBack(\'c\');
p1->insertBack(\'e\');
p1->insert(\'d\',2.
Two sp3 orbitals are filled by lone electron pair.pdf
Two sp3 orbitals are filled by lone electron pairs, so they cannot bond to another
atom.
Solution
Two sp3 orbitals are filled by lone electron pairs, so they cannot bond to another
atom..
X-Intercept is the value of x where cross x axis. Another name is .pdf
X-Intercept is the value of x where cross x axis.
Another name is the zeros of the function, when y is zero, the graph cross the x-axis
The answer is x-intercepts, Nataliya already answer the question. I just add some explanations to
it.
Solution
X-Intercept is the value of x where cross x axis.
Another name is the zeros of the function, when y is zero, the graph cross the x-axis
The answer is x-intercepts, Nataliya already answer the question. I just add some explanations to
it..
whether or not the viruses is not used to classify viruses.Vir.pdf
\"whether or not the viruses\" is not used to classify viruses.
Viruses are classified based on the type of genetic material. They have either DNA or RNA but
not both. If DNA is present, it may be single stranded or double stranded. In the same way RNA
may also be single stranded or double stranded. Based on the presence or absence of envelope
they are again classified into enveloped and non enveloped viruses. HIV is RNA virus and
enveloped. Herepes virus is DNA virus and enveloped. Polio virus is RNA virus and non
enveloped.
Solution
\"whether or not the viruses\" is not used to classify viruses.
Viruses are classified based on the type of genetic material. They have either DNA or RNA but
not both. If DNA is present, it may be single stranded or double stranded. In the same way RNA
may also be single stranded or double stranded. Based on the presence or absence of envelope
they are again classified into enveloped and non enveloped viruses. HIV is RNA virus and
enveloped. Herepes virus is DNA virus and enveloped. Polio virus is RNA virus and non
enveloped..
Vi may be a powerful text editor enclosed with most UNIX systems, ev.pdf
Vi may be a powerful text editor enclosed with most UNIX systems, even embedded ones.
generally you’ll ought toedit a document on a system that doesn’t embrace a friendlier text
editor, therefore knowing Vi is important.
Unlike Nano, Associate in Nursing easy-to-use terminal text editor, Vi doesn’t hold your hand
and supplya listingof keyboard shortcuts on the screen. It’s a modal text editor, Associate in
Nursingd it\'seach an insert and command mode.
Vi may be a terminal application, therefore you’ll ought tobegin it from a terminal window. Use
the vi /path/to/file command to open Associate in Nursing existing file with Vi. The vi
/path/to/file command additionally works if the file doesn’t exist yet; Vi canproducea brand new
file and write it to the desired location once you save.
Command Mode
This is what you’ll see once you open a get in vi. it\'ssuch as youwillsimplybeginwriting,
however you can’t. Vimay be a modal text editor, and it opens in command mode. making an
attempt to kind at this screen canlead tosudden behavior.
While in command mode, you\'ll move the pointer around with the arrow keys. Press the x key
to delete the characterbeneath the pointer. There area unita spread of alternative delete
commands — as an example, writingDD(press the d key twice) deletes a whole line of text.
You can choose, copy, cut and paste text in command mode. Position the pointer at the left or
right aspect of the text you would liketo repeat and press the v key. Move your pointerto pick out
text, so press y to repeatthe chosen text or x to chop it. Position your pointer at the required
location and press the p key to stick the text youtraced or cut.
Insert Mode
Aside from command mode, the opposite mode you wishto understandconcerning is insert
mode, thatpermitsyou to insert text in Vi. getting into insert mode is simple once you recognize it
exists — simply press the i key once when you’ve positioned the pointer in command mode.
beginwriting and Vi can insert the characters youkind into the file instead ofmaking an attempt to
interpret them as commands.
Once you’re worn out insert mode, press the escape key to come to command mode.
Saving and Quitting
You can save and quit vi from command mode. First, guarantee you’re in command mode by
pressing the escape key (pressing the escape key once morewill nothing if you’re already in
command mode.)
Type :wq and press enter to put in writing the file to disk and quit vi. you\'lladditionally split
this command up — as an example, kind :w and press enter to put in writing the file to disk
while not quitting or kind :q to quit vi while not saving the file.
Vi won’t allow you to quit if you’ve changed the file since you last saved, howeveryou\'llkind
:q! and press enter to ignore this warning.
Solution
Vi may be a powerful text editor enclosed with most UNIX systems, even embedded ones.
generally you’ll ought toedit a document on a system that doesn’t embrace a friendlier text
editor, therefore knowing Vi is .
U.S management is trained to provide the executives their own career.pdf
U.S management is trained to provide the executives their own career path; it provides flexibility
to the top executive over the journey. The top executive is set forward to accommodate their own
journey with their own set of interest. The management is expecting that the top executives will
be of good track record and will somehow contribute to the growth of the company and will
contribute solidly to the team and work as a team. The top executive should deliver best quality
efforts to the other team members. Management is trying to provide a corporate culture or
training to the team members and ill deliver appropriate efforts to the company. The
management is also trained towards the culture of the company and is dedicated to provide
cultural formal training. The U.S. management is focused on dividing the executive on the basis
of common language globally and domestically. To ensure the same the management is trained
for providing language courses to the executives for better understanding and training to the
organization.
Solution
U.S management is trained to provide the executives their own career path; it provides flexibility
to the top executive over the journey. The top executive is set forward to accommodate their own
journey with their own set of interest. The management is expecting that the top executives will
be of good track record and will somehow contribute to the growth of the company and will
contribute solidly to the team and work as a team. The top executive should deliver best quality
efforts to the other team members. Management is trying to provide a corporate culture or
training to the team members and ill deliver appropriate efforts to the company. The
management is also trained towards the culture of the company and is dedicated to provide
cultural formal training. The U.S. management is focused on dividing the executive on the basis
of common language globally and domestically. To ensure the same the management is trained
for providing language courses to the executives for better understanding and training to the
organization..
There are many operating systemsReal-Time Operating SystemReal-t.pdf
There are many operating systems
Real-Time Operating System
Real-time applications usually are executed on top of a Real-time Operating System (RTOS).
Specific scheduling algorithms can be designed. When possible, static cyclic schedules are
calculated off-line.
Real-time systems are those systems in which the correctness of the system depends not only on
the logical result of computation, but also on the time at which the results are produced.
RTOS is therefore an operating system that supports real-time applications by providing
logically correct result within the deadline required. Basic Structure is similar to regular OS but,
in addition, it provides mechanisms to allow real time scheduling of tasks.
Though real-time operating systems may or may not increase the speed of execution, they can
provide much more precise and predictable timing characteristics than general-purpose OS.
A real-time system is defined as a data processing system in which the time interval required to
process and respond to inputs is so small that it controls the environment. The time taken by the
system to respond to an input and display of required updated information is termed as the
response time. So in this method, the response time is very less as compared to online
processing.
Real-time systems are used when there are rigid time requirements on the operation of a
processor or the flow of data and real-time systems can be used as a control device in a dedicated
application. A real-time operating system must have well-defined, fixed time constraints,
otherwise the system will fail. For example, Scientific experiments, medical imaging systems,
industrial control systems, weapon systems, robots, air traffic control systems, etc.
Design considerations
Designing a proper RTOS architecture needs some delicate decisions. The basic services like
process management, inter-process communication, interrupt handling, or process
synchronization have to be provided in an efficient manner making use of a very restricted
resource budget.
Multi-core architectures need special techniques for process management, memory management,
and synchronization. The upcoming Wireless Sensor Networks (WSN) generate special demands
for RTOS support leading to dedicated solutions. Another special area is given by multimedia
applications. Very high data rates have to be supported under (soft) RT constraints.
The key difference between general-computing operating systems and real-time operating
systems is the need for \" deterministic \" timing behavior in the real-time operating systems.
Formally, \"deterministic\" timing means that operating system services consume only known
and expected amounts of time. In theory, these service times could be expressed as mathematical
formulas. These formulas must be strictly algebraic and not include any random timing
components. Random elements in service times could cause random delays in application
software and could then make the application randomly .
Recommended
- You have to overcome the ionic forces of the CsI, the hydrogen bon.pdf
- You have to overcome the ionic forces of the CsI, the hydrogen bonding of the HF. In return
you get ion-dipole forces of the Cs(+) and I(-) in the solution
- Dispersion Forces and Metallic bonds
Solution
- You have to overcome the ionic forces of the CsI, the hydrogen bonding of the HF. In return
you get ion-dipole forces of the Cs(+) and I(-) in the solution
- Dispersion Forces and Metallic bonds.
#includeiostream struct node { char value; struct no.pdf
#include
struct node
{
char value;
struct node *next;
};
class StringOfNode
{
struct node * head;
public:
StringOfNode()
{
head=NULL;
}
int size()
{
struct node *p=head;
int count=1;
while(p)
{
count++;
p=p->next;
}
return count;
}
struct node * begin()
{
return head;
}
char &operator[](int index)
{
int i=0;
struct node *p=head;
while(index!=i)
{
p=p->next;
i++;
}
return p->value;
}
void insertFront(char valuee)
{
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(head==NULL)
head=temp;
else
{
temp->next=head;
head=temp;
}
}
void insert(char valuee, int pos)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
int i=0;
if(p==NULL)
{
head=temp;
}
else
{
while(i!=pos-1)
{
p=p->next;
i++;
}
temp->next=p->next;
p->next=temp;
}
}
void insertBack(char valuee)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(p==NULL)
{
head=temp;
}
else
{
while(p->next!=NULL)
p=p->next;
p->next=temp;
}
}
void print()
{
struct node *p=head;
while(p)
{
std::cout<value<<\"\ \";
p=p->next;
}
}
StringOfNode operator+(const StringOfNode& b)
{
StringOfNode *S;
struct node *p=head;
while(p->next!=NULL)
p=p->next;
p->next=S->begin();
}
};
int main()
{
StringOfNode *p1=new StringOfNode;
p1->insertBack(\'b\');
p1->insertBack(\'c\');
p1->insertBack(\'e\');
p1->insert(\'d\',2);
p1->insertFront(\'a\');
p1->print();
system(\"pause\");
return 0;
}
Solution
#include
struct node
{
char value;
struct node *next;
};
class StringOfNode
{
struct node * head;
public:
StringOfNode()
{
head=NULL;
}
int size()
{
struct node *p=head;
int count=1;
while(p)
{
count++;
p=p->next;
}
return count;
}
struct node * begin()
{
return head;
}
char &operator[](int index)
{
int i=0;
struct node *p=head;
while(index!=i)
{
p=p->next;
i++;
}
return p->value;
}
void insertFront(char valuee)
{
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(head==NULL)
head=temp;
else
{
temp->next=head;
head=temp;
}
}
void insert(char valuee, int pos)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
int i=0;
if(p==NULL)
{
head=temp;
}
else
{
while(i!=pos-1)
{
p=p->next;
i++;
}
temp->next=p->next;
p->next=temp;
}
}
void insertBack(char valuee)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(p==NULL)
{
head=temp;
}
else
{
while(p->next!=NULL)
p=p->next;
p->next=temp;
}
}
void print()
{
struct node *p=head;
while(p)
{
std::cout<value<<\"\ \";
p=p->next;
}
}
StringOfNode operator+(const StringOfNode& b)
{
StringOfNode *S;
struct node *p=head;
while(p->next!=NULL)
p=p->next;
p->next=S->begin();
}
};
int main()
{
StringOfNode *p1=new StringOfNode;
p1->insertBack(\'b\');
p1->insertBack(\'c\');
p1->insertBack(\'e\');
p1->insert(\'d\',2.
Two sp3 orbitals are filled by lone electron pair.pdf
Two sp3 orbitals are filled by lone electron pairs, so they cannot bond to another
atom.
Solution
Two sp3 orbitals are filled by lone electron pairs, so they cannot bond to another
atom..
X-Intercept is the value of x where cross x axis. Another name is .pdf
X-Intercept is the value of x where cross x axis.
Another name is the zeros of the function, when y is zero, the graph cross the x-axis
The answer is x-intercepts, Nataliya already answer the question. I just add some explanations to
it.
Solution
X-Intercept is the value of x where cross x axis.
Another name is the zeros of the function, when y is zero, the graph cross the x-axis
The answer is x-intercepts, Nataliya already answer the question. I just add some explanations to
it..
whether or not the viruses is not used to classify viruses.Vir.pdf
\"whether or not the viruses\" is not used to classify viruses.
Viruses are classified based on the type of genetic material. They have either DNA or RNA but
not both. If DNA is present, it may be single stranded or double stranded. In the same way RNA
may also be single stranded or double stranded. Based on the presence or absence of envelope
they are again classified into enveloped and non enveloped viruses. HIV is RNA virus and
enveloped. Herepes virus is DNA virus and enveloped. Polio virus is RNA virus and non
enveloped.
Solution
\"whether or not the viruses\" is not used to classify viruses.
Viruses are classified based on the type of genetic material. They have either DNA or RNA but
not both. If DNA is present, it may be single stranded or double stranded. In the same way RNA
may also be single stranded or double stranded. Based on the presence or absence of envelope
they are again classified into enveloped and non enveloped viruses. HIV is RNA virus and
enveloped. Herepes virus is DNA virus and enveloped. Polio virus is RNA virus and non
enveloped..
Vi may be a powerful text editor enclosed with most UNIX systems, ev.pdf
Vi may be a powerful text editor enclosed with most UNIX systems, even embedded ones.
generally you’ll ought toedit a document on a system that doesn’t embrace a friendlier text
editor, therefore knowing Vi is important.
Unlike Nano, Associate in Nursing easy-to-use terminal text editor, Vi doesn’t hold your hand
and supplya listingof keyboard shortcuts on the screen. It’s a modal text editor, Associate in
Nursingd it\'seach an insert and command mode.
Vi may be a terminal application, therefore you’ll ought tobegin it from a terminal window. Use
the vi /path/to/file command to open Associate in Nursing existing file with Vi. The vi
/path/to/file command additionally works if the file doesn’t exist yet; Vi canproducea brand new
file and write it to the desired location once you save.
Command Mode
This is what you’ll see once you open a get in vi. it\'ssuch as youwillsimplybeginwriting,
however you can’t. Vimay be a modal text editor, and it opens in command mode. making an
attempt to kind at this screen canlead tosudden behavior.
While in command mode, you\'ll move the pointer around with the arrow keys. Press the x key
to delete the characterbeneath the pointer. There area unita spread of alternative delete
commands — as an example, writingDD(press the d key twice) deletes a whole line of text.
You can choose, copy, cut and paste text in command mode. Position the pointer at the left or
right aspect of the text you would liketo repeat and press the v key. Move your pointerto pick out
text, so press y to repeatthe chosen text or x to chop it. Position your pointer at the required
location and press the p key to stick the text youtraced or cut.
Insert Mode
Aside from command mode, the opposite mode you wishto understandconcerning is insert
mode, thatpermitsyou to insert text in Vi. getting into insert mode is simple once you recognize it
exists — simply press the i key once when you’ve positioned the pointer in command mode.
beginwriting and Vi can insert the characters youkind into the file instead ofmaking an attempt to
interpret them as commands.
Once you’re worn out insert mode, press the escape key to come to command mode.
Saving and Quitting
You can save and quit vi from command mode. First, guarantee you’re in command mode by
pressing the escape key (pressing the escape key once morewill nothing if you’re already in
command mode.)
Type :wq and press enter to put in writing the file to disk and quit vi. you\'lladditionally split
this command up — as an example, kind :w and press enter to put in writing the file to disk
while not quitting or kind :q to quit vi while not saving the file.
Vi won’t allow you to quit if you’ve changed the file since you last saved, howeveryou\'llkind
:q! and press enter to ignore this warning.
Solution
Vi may be a powerful text editor enclosed with most UNIX systems, even embedded ones.
generally you’ll ought toedit a document on a system that doesn’t embrace a friendlier text
editor, therefore knowing Vi is .
U.S management is trained to provide the executives their own career.pdf
U.S management is trained to provide the executives their own career path; it provides flexibility
to the top executive over the journey. The top executive is set forward to accommodate their own
journey with their own set of interest. The management is expecting that the top executives will
be of good track record and will somehow contribute to the growth of the company and will
contribute solidly to the team and work as a team. The top executive should deliver best quality
efforts to the other team members. Management is trying to provide a corporate culture or
training to the team members and ill deliver appropriate efforts to the company. The
management is also trained towards the culture of the company and is dedicated to provide
cultural formal training. The U.S. management is focused on dividing the executive on the basis
of common language globally and domestically. To ensure the same the management is trained
for providing language courses to the executives for better understanding and training to the
organization.
Solution
U.S management is trained to provide the executives their own career path; it provides flexibility
to the top executive over the journey. The top executive is set forward to accommodate their own
journey with their own set of interest. The management is expecting that the top executives will
be of good track record and will somehow contribute to the growth of the company and will
contribute solidly to the team and work as a team. The top executive should deliver best quality
efforts to the other team members. Management is trying to provide a corporate culture or
training to the team members and ill deliver appropriate efforts to the company. The
management is also trained towards the culture of the company and is dedicated to provide
cultural formal training. The U.S. management is focused on dividing the executive on the basis
of common language globally and domestically. To ensure the same the management is trained
for providing language courses to the executives for better understanding and training to the
organization..
There are many operating systemsReal-Time Operating SystemReal-t.pdf
There are many operating systems
Real-Time Operating System
Real-time applications usually are executed on top of a Real-time Operating System (RTOS).
Specific scheduling algorithms can be designed. When possible, static cyclic schedules are
calculated off-line.
Real-time systems are those systems in which the correctness of the system depends not only on
the logical result of computation, but also on the time at which the results are produced.
RTOS is therefore an operating system that supports real-time applications by providing
logically correct result within the deadline required. Basic Structure is similar to regular OS but,
in addition, it provides mechanisms to allow real time scheduling of tasks.
Though real-time operating systems may or may not increase the speed of execution, they can
provide much more precise and predictable timing characteristics than general-purpose OS.
A real-time system is defined as a data processing system in which the time interval required to
process and respond to inputs is so small that it controls the environment. The time taken by the
system to respond to an input and display of required updated information is termed as the
response time. So in this method, the response time is very less as compared to online
processing.
Real-time systems are used when there are rigid time requirements on the operation of a
processor or the flow of data and real-time systems can be used as a control device in a dedicated
application. A real-time operating system must have well-defined, fixed time constraints,
otherwise the system will fail. For example, Scientific experiments, medical imaging systems,
industrial control systems, weapon systems, robots, air traffic control systems, etc.
Design considerations
Designing a proper RTOS architecture needs some delicate decisions. The basic services like
process management, inter-process communication, interrupt handling, or process
synchronization have to be provided in an efficient manner making use of a very restricted
resource budget.
Multi-core architectures need special techniques for process management, memory management,
and synchronization. The upcoming Wireless Sensor Networks (WSN) generate special demands
for RTOS support leading to dedicated solutions. Another special area is given by multimedia
applications. Very high data rates have to be supported under (soft) RT constraints.
The key difference between general-computing operating systems and real-time operating
systems is the need for \" deterministic \" timing behavior in the real-time operating systems.
Formally, \"deterministic\" timing means that operating system services consume only known
and expected amounts of time. In theory, these service times could be expressed as mathematical
formulas. These formulas must be strictly algebraic and not include any random timing
components. Random elements in service times could cause random delays in application
software and could then make the application randomly .
The three major forms of business organizations are1. Sole Propri.pdf
The three major forms of business organizations are:
1. Sole Proprietorship
2. Partnership &
3. Corporation
Sole Proprietorship: Ownership : In this form of organisation the ownership lies with a single
person i.e. the proprietor of the buiness. Along with the ownership all the liabilities and
responsibilities are with the sole proprietor of the organisation
Partnership: Ownership: In a partnership form of organisation more than one person come
together to form a business and all of them bring in capital in a decided proportion and they own
such decided portion of the organisation. They share liabilities and responsibilities in the pre
determined proportion i.e. profit sharing ratio
It is managed by all the partners together or by any working partner as decided by all of the
partners.
Corporation : Many people contribute to the Capital of the organisation and the management lies
with the directors and the management ho are elected by the shareholders and are completely
answerable and responsible to all the stakeholders.
Solution
The three major forms of business organizations are:
1. Sole Proprietorship
2. Partnership &
3. Corporation
Sole Proprietorship: Ownership : In this form of organisation the ownership lies with a single
person i.e. the proprietor of the buiness. Along with the ownership all the liabilities and
responsibilities are with the sole proprietor of the organisation
Partnership: Ownership: In a partnership form of organisation more than one person come
together to form a business and all of them bring in capital in a decided proportion and they own
such decided portion of the organisation. They share liabilities and responsibilities in the pre
determined proportion i.e. profit sharing ratio
It is managed by all the partners together or by any working partner as decided by all of the
partners.
Corporation : Many people contribute to the Capital of the organisation and the management lies
with the directors and the management ho are elected by the shareholders and are completely
answerable and responsible to all the stakeholders..
the OSI model is an idea. it is abstract it has no value without imp.pdf
the OSI model is an idea. it is abstract it has no value without implementation. TCP/IP is an
implementation. it was shaped by the concept of the OSI model. the OSI model is unchanging,
implementations like TCP/IP conform to the mode
Solution
the OSI model is an idea. it is abstract it has no value without implementation. TCP/IP is an
implementation. it was shaped by the concept of the OSI model. the OSI model is unchanging,
implementations like TCP/IP conform to the mode.
The EVA metric effectively measures the amount of shareholder wealth.pdf
The EVA metric effectively measures the amount of shareholder wealth that the firm\'s
management has added to the firm\'s value during a time period. If EVA is positive , then
management has increased value.
While there cannot be a single factor which can be attributed to the managers performing their
primary task i.e. MVA/EVA as they are basically interrelated. This is due to the fact that both
aim towards creation of value.
The only difference is in the calculation and the interpretation as defined by mathematical
paradigms.
EVA is the difference of NOPAT (Net Operating Profit After Taxes ) and Cost of Capital.This
indicates the wealth the company generated over and above the cost of capital.
THe MVA on the other hand is simply the difference between the current total market value and
the capital contributed by investors (both shareholders and bondholders).
Thus as the company performs well on the EVA metric and as it retained earnings rise so will
the business. And as the business rises so will the share price and hence the MVA. SO basically
it is all interrelated.
EBIT = 80M-52M = 28M
EVA = 28* (1-0.4) - 115*0.125 = 2.425 M
MVA of a company is equal to the net present value of all future EVAs.
Solution
The EVA metric effectively measures the amount of shareholder wealth that the firm\'s
management has added to the firm\'s value during a time period. If EVA is positive , then
management has increased value.
While there cannot be a single factor which can be attributed to the managers performing their
primary task i.e. MVA/EVA as they are basically interrelated. This is due to the fact that both
aim towards creation of value.
The only difference is in the calculation and the interpretation as defined by mathematical
paradigms.
EVA is the difference of NOPAT (Net Operating Profit After Taxes ) and Cost of Capital.This
indicates the wealth the company generated over and above the cost of capital.
THe MVA on the other hand is simply the difference between the current total market value and
the capital contributed by investors (both shareholders and bondholders).
Thus as the company performs well on the EVA metric and as it retained earnings rise so will
the business. And as the business rises so will the share price and hence the MVA. SO basically
it is all interrelated.
EBIT = 80M-52M = 28M
EVA = 28* (1-0.4) - 115*0.125 = 2.425 M
MVA of a company is equal to the net present value of all future EVAs..
Reflection about the centre of the pentagon is not its symmetric and.pdf
Reflection about the centre of the pentagon is not its symmetric and the symmetry group has
trivial centre
Solution
Reflection about the centre of the pentagon is not its symmetric and the symmetry group has
trivial centre.
Question not visible. Please state again.SolutionQuestion not .pdf
The document does not contain any visible questions or information to summarize. No context or content was provided in the document to generate a 3 sentence summary.
Per my quiz, it was also D) formation of the carbocatio or bromonium.pdf
Per my quiz, it was also D) formation of the carbocatio or bromonium ion and B) addition of the
halide anion
Solution
Per my quiz, it was also D) formation of the carbocatio or bromonium ion and B) addition of the
halide anion.
Part ATay-Sachs disease is an autosomal recessive disorder, so, o.pdf
Part A:
Tay-Sachs disease is an autosomal recessive disorder, so, only a homozygous recessive person
will have the disease. 1 in 300 individuals of neither descent will be carrier. A child can have the
disease, only if both his/her parents are carriers.
Let the genotype of Tay Sachs disease be \'tt\'. So, carrier will be \'Tt\'
According to HW equilibrium, 2pq= 1/300
Given data is p = 0.0167 q = 0.983 (This data is totally wrong). Please note that \'q\' is the
recessive Tay-Sachs allele and \'p\' is the dominant one. So, the frequency of recessive allele
cannot be more than the dominant one.
When two carriers cross, 25% chance is there of having the genotype \'tt\'.
So, chances of having a child with Tay-Sachs disease = 25% of the total children born or 25% of
(1/300 * 1/300)
Part B.
If one individual is of Ashkenazi descent, then chances of being a carrier is 1/30
If the other parent is normal, then chances of being a carrier is 1/300
When two carriers cross, chances of having homozygous recessive genotype = 25%
So, 25% chance of having an affected child is there i.e
25% of (1/30 * 1/300)
Part C:
Again 25% chance of having an affected child is there.
You can also write as: 25% of (1/30 * 1/30)
Part D:
Again 25% chance of having an affected child is there.
You can also write as: 25% of (1/30 * 1/30)
Solution
Part A:
Tay-Sachs disease is an autosomal recessive disorder, so, only a homozygous recessive person
will have the disease. 1 in 300 individuals of neither descent will be carrier. A child can have the
disease, only if both his/her parents are carriers.
Let the genotype of Tay Sachs disease be \'tt\'. So, carrier will be \'Tt\'
According to HW equilibrium, 2pq= 1/300
Given data is p = 0.0167 q = 0.983 (This data is totally wrong). Please note that \'q\' is the
recessive Tay-Sachs allele and \'p\' is the dominant one. So, the frequency of recessive allele
cannot be more than the dominant one.
When two carriers cross, 25% chance is there of having the genotype \'tt\'.
So, chances of having a child with Tay-Sachs disease = 25% of the total children born or 25% of
(1/300 * 1/300)
Part B.
If one individual is of Ashkenazi descent, then chances of being a carrier is 1/30
If the other parent is normal, then chances of being a carrier is 1/300
When two carriers cross, chances of having homozygous recessive genotype = 25%
So, 25% chance of having an affected child is there i.e
25% of (1/30 * 1/300)
Part C:
Again 25% chance of having an affected child is there.
You can also write as: 25% of (1/30 * 1/30)
Part D:
Again 25% chance of having an affected child is there.
You can also write as: 25% of (1/30 * 1/30).
null is a subset of every setTrueSolutionnull is a sub.pdf
The document discusses the empty set, also known as the null set. It states that the null set is a subset of every set, and verifies that this statement is true. The null set contains no elements, so it is contained within any set regardless of its elements.
LiOH Sol.pdf
This 3-word document appears to be referring to a solution containing lithium hydroxide (LiOH). Lithium hydroxide is listed three times, suggesting it is the primary component of the solution under discussion. The brief document does not provide much other context around the LiOH solution.
Information is a valuable asset that can make or break your business.pdf
Information is a valuable asset that can make or break your business. When properly managed it
allows you to operate with confidence. Information security management gives you the freedom
to grow, innovate and broaden your customer-base in the knowledge that all your confidential
information will remain that way.
In this information era, Security has become more difficult to define and enforce.
information security was limited to controlling physical access to oral or written
communications. The importance of information security led societies to develop innovative
ways of protecting their information. Recent innovations in information technology, like the
Internet, have made it possible to send vast quantities of data across the globe with ease.
However, the challenge of controlling and protecting that information has grown exponentially
now that data can be easily transmitted, stored, copied, manipulated, and destroyed.
Within a large organization information technology generally refers to laptop and desktop
computers, servers, routers, and switches that form a computer network, although information
technology also includes fax machines, phone and voice mail systems, cellular phones, and other
electronic systems. A growing reliance on computers to work and communicate has made the
control of computer networks an important part of information security. Unauthorized access to
paper documents or phone conversations is still an information security concern, but the real
challenge has become protecting the security of computer networks, especially when they are
connected to the Internet. Most large organizations have their own local computer network, or
intranet, that links their computers together to share resources and support the communications
of employees and others with a legitimate need for access. Almost all of these networks are
connected to the Internet and allow employees to go “online.”
Solution
Information is a valuable asset that can make or break your business. When properly managed it
allows you to operate with confidence. Information security management gives you the freedom
to grow, innovate and broaden your customer-base in the knowledge that all your confidential
information will remain that way.
In this information era, Security has become more difficult to define and enforce.
information security was limited to controlling physical access to oral or written
communications. The importance of information security led societies to develop innovative
ways of protecting their information. Recent innovations in information technology, like the
Internet, have made it possible to send vast quantities of data across the globe with ease.
However, the challenge of controlling and protecting that information has grown exponentially
now that data can be easily transmitted, stored, copied, manipulated, and destroyed.
Within a large organization information technology generally refers to laptop and desktop
computers, servers, routers, and.
Ho there is no relationship between the age of the individual and t.pdf
Ho: there is no relationship between the age of the individual and the individual
Solution
Ho: there is no relationship between the age of the individual and the individual.
H2SO3 - H2O = SO2The oxide is sulfur dioxide SO2SolutionH2.pdf
H2SO3 - H2O => SO2
The oxide is sulfur dioxide: SO2
Solution
H2SO3 - H2O => SO2
The oxide is sulfur dioxide: SO2.
First of all we shoud understand what is HIV AIDSHIV AIDS is .pdf
First of all we shoud understand what is HIV / AIDS?
HIV / AIDS is not a single disease it is combination of diseases. Because once HIV is infected it
will start reduce immune system i.e defence mechanisms so it is called as Aquired Immuno
Defficiency Syndrome (AIDS). So because the HIV infected patient needs nutrient food for
increase their life span.
If you are HIV-positive, good nutrition can have several benefits. It can:
Basic principle of nutrition :
So overall the HIV infected patient shuld have high nutritious food havig good protein,
carbohydrates,vitamins , minerals to get good immune system to defence by the desease.
Thanks all the best
Solution
First of all we shoud understand what is HIV / AIDS?
HIV / AIDS is not a single disease it is combination of diseases. Because once HIV is infected it
will start reduce immune system i.e defence mechanisms so it is called as Aquired Immuno
Defficiency Syndrome (AIDS). So because the HIV infected patient needs nutrient food for
increase their life span.
If you are HIV-positive, good nutrition can have several benefits. It can:
Basic principle of nutrition :
So overall the HIV infected patient shuld have high nutritious food havig good protein,
carbohydrates,vitamins , minerals to get good immune system to defence by the desease.
Thanks all the best.
diffusion rate = 1Mr1r2 = m2m1 = t2t110.12.16 = m2 2mola.pdf
diffusion rate = 1/M
r1/r2 = m2/m1 = t2/t1
10.1/2.16 = m2 /2
molar mass of the unknown gas , m2 = 43.728 g
Solution
diffusion rate = 1/M
r1/r2 = m2/m1 = t2/t1
10.1/2.16 = m2 /2
molar mass of the unknown gas , m2 = 43.728 g.
Dear,The answer is .2.SolutionDear,The answer is .2..pdf
The document is a short note stating that the answer to an unknown problem or question is 0.2. It provides no context or explanation for how the answer was determined. In two sentences, it simply states "The answer is .2."
d. Increases venous PO2.systemic venous blood is likely to be 25 .pdf
d. Increases venous PO2.
systemic venous blood is likely to be 25% saturated with oxygen.
PO2 Partial presure of oxygen.
Solution
d. Increases venous PO2.
systemic venous blood is likely to be 25% saturated with oxygen.
PO2 Partial presure of oxygen..
Considering the general solution of the Laplace Equation we getXn(.pdf
Considering the general solution of the Laplace Equation we get
Xn(x) = sin(nx)
Yn(y) = Ancosh(ny) + Bnsinh(ny)
Lets have Yn(0) = 1 and An = 1 for convienience.
As we know Yn() = 0 , hence solving for Bn
=> Bn = -cosh(n)/sinh(n)
=> Yn(y) = sinh(n - ny)/sinh(n)
Since f(x) = sin(4x)
=> u(x,y) = n=1 sin(4x) * ( sinh(n - ny)/sinh(n) )
Solution
Considering the general solution of the Laplace Equation we get
Xn(x) = sin(nx)
Yn(y) = Ancosh(ny) + Bnsinh(ny)
Lets have Yn(0) = 1 and An = 1 for convienience.
As we know Yn() = 0 , hence solving for Bn
=> Bn = -cosh(n)/sinh(n)
=> Yn(y) = sinh(n - ny)/sinh(n)
Since f(x) = sin(4x)
=> u(x,y) = n=1 sin(4x) * ( sinh(n - ny)/sinh(n) ).
Cranial nerves which do not innervate the structures of the head but.pdf
Cranial nerves which do not innervate the structures of the head but innervate the structures
originating from branchial arches are the following
Solution
Cranial nerves which do not innervate the structures of the head but innervate the structures
originating from branchial arches are the following.
Code Include libraries. import javax.swing.JOptionPane;.pdf
Code:
//Include libraries.
import javax.swing.JOptionPane;
//Define a class lPayroll
public class lPayroll
{
//Main function
public static void main( String args[] )
{
//Initialize variables
String lfirst = \"\";
String lLast = \"\";
String lssn = \"\";
String lsalaryString;
double lsalary = 0;
String lwageString;
double lwage = 0;
String lhoursString;
int lhours = 0;
String lsalesString;
double lsales = 0;
String lrateString;
double lrate = 0;
String lempTypeString = \"\";
int lempType = 0;
int lcount;
for(lcount = 0; lcount < 5; lcount++)
{
//Receive user input
lempTypeString = JOptionPane.showInputDialog(null, \"Enter employee type \ \ Enter
your choice.\ 1. Salaried\ 2. Hourly\ 3. Commissioned\ 4. Commission plus base\", \"Employee
Type\", JOptionPane.QUESTION_MESSAGE);
lempType = Integer.parseInt(lempTypeString);
switch(lempType)
{
case 1:
lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\",
JOptionPane.QUESTION_MESSAGE);
lLast = JOptionPane.showInputDialog(null, \"Enter Last name \", \"lLast Name\",
JOptionPane.QUESTION_MESSAGE);
lssn = JOptionPane.showInputDialog(null, \"Enter social security number\", \"lssn\",
JOptionPane.QUESTION_MESSAGE);
lsalaryString = JOptionPane.showInputDialog(null, \"Enter weekly salary\", \"lsalary\",
JOptionPane.QUESTION_MESSAGE);
lsalary = Double.parseDouble(lsalaryString);
break;
case 2:
lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\",
JOptionPane.QUESTION_MESSAGE);
lLast = JOptionPane.showInputDialog(null, \"Enter Last name\", \"lLast Name\",
JOptionPane.QUESTION_MESSAGE);
lssn = JOptionPane.showInputDialog(null, \"Enter social security number\", \"lssn\",
JOptionPane.QUESTION_MESSAGE);
lwageString = JOptionPane.showInputDialog(null, \"Enter hourly wage\", \"Hourly
lwage\", JOptionPane.QUESTION_MESSAGE);
lwage = Double.parseDouble(lwageString);
lhoursString = JOptionPane.showInputDialog(null, \"Enter count of hours per week of
work\", \"lhours Worked\", JOptionPane.QUESTION_MESSAGE);
lhours = Integer.parseInt(lhoursString);
break;
case 3:
lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\",
JOptionPane.QUESTION_MESSAGE);
lLast = JOptionPane.showInputDialog(null, \"Enter Last name\", \"lLast Name\",
JOptionPane.QUESTION_MESSAGE);
lssn = JOptionPane.showInputDialog(null, \"Enter social security number\", \"lssn\",
JOptionPane.QUESTION_MESSAGE);
lsalesString = JOptionPane.showInputDialog(null, \"Enter gross sales on weekly
basis\", \"Gross lsales\", JOptionPane.QUESTION_MESSAGE);
lsales = Double.parseDouble(lsalesString);
lrateString = JOptionPane.showInputDialog(null, \"Enter commission percent\",
\"Commsission lrate\", JOptionPane.QUESTION_MESSAGE);
lrate = Double.parseDouble(lrateString);
break;
case 4:
lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\",
JOptionPane.QUESTION_MESSAGE);
lLast = JOptionPane.showInputDialog(null, \"Enter Last name\", \"lLast Name\",
JOptionPane.QUESTION_ME.
C code#include stdio.hint main(void) { FILE infp, o.pdf
C code:
#include
int main(void)
{
FILE *infp, *outfp;
int ID,day,hr,i=1;
int Temp_ID,Total_hr;
int Line_No=0;
if ((infp = fopen(\"emp.txt\", \"r\"))==NULL)
{
printf(\"Input file cannot be opened\ \");
return-1;
}
if ((outfp = fopen(\"out.txt\", \"w\"))==NULL)
{
printf(\"Output file cannot be opened\ \");
return-1;
}
while(!feof(infp))
{
Line_No=Line_No+1;
fscanf(infp,\"%d %d\",&ID,&hr);
if(Line_No==1)
{
Temp_ID=ID;day=1;Total_hr=hr;
}
else
{
if(Temp_ID==ID)
{
day=day+1;
Total_hr=Total_hr+hr;
}
else
{
fprintf(outfp,\"%d %d %d\ \",Temp_ID,day,Total_hr);
Temp_ID=ID;day=1;Total_hr=hr;
}
}
}
fprintf(outfp,\"%d %d %d\ \",Temp_ID,day-1,Total_hr-hr);
fclose(infp);
fclose(outfp);
}
emp.txt:
11 2
11 8
11 5
15 10
18 4
18 16
20 23
20 11
out.txt:
11 3 15
15 1 10
18 2 20
20 2 34
Solution
C code:
#include
int main(void)
{
FILE *infp, *outfp;
int ID,day,hr,i=1;
int Temp_ID,Total_hr;
int Line_No=0;
if ((infp = fopen(\"emp.txt\", \"r\"))==NULL)
{
printf(\"Input file cannot be opened\ \");
return-1;
}
if ((outfp = fopen(\"out.txt\", \"w\"))==NULL)
{
printf(\"Output file cannot be opened\ \");
return-1;
}
while(!feof(infp))
{
Line_No=Line_No+1;
fscanf(infp,\"%d %d\",&ID,&hr);
if(Line_No==1)
{
Temp_ID=ID;day=1;Total_hr=hr;
}
else
{
if(Temp_ID==ID)
{
day=day+1;
Total_hr=Total_hr+hr;
}
else
{
fprintf(outfp,\"%d %d %d\ \",Temp_ID,day,Total_hr);
Temp_ID=ID;day=1;Total_hr=hr;
}
}
}
fprintf(outfp,\"%d %d %d\ \",Temp_ID,day-1,Total_hr-hr);
fclose(infp);
fclose(outfp);
}
emp.txt:
11 2
11 8
11 5
15 10
18 4
18 16
20 23
20 11
out.txt:
11 3 15
15 1 10
18 2 20
20 2 34.
How to Add Chatter in the odoo 17 ERP Module
In Odoo, the chatter is like a chat tool that helps you work together on records. You can leave notes and track things, making it easier to talk with your team and partners. Inside chatter, all communication history, activity, and changes will be displayed.
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
Denis is a dynamic and results-driven Chief Information Officer (CIO) with a distinguished career spanning information systems analysis and technical project management. With a proven track record of spearheading the design and delivery of cutting-edge Information Management solutions, he has consistently elevated business operations, streamlined reporting functions, and maximized process efficiency.
Certified as an ISO/IEC 27001: Information Security Management Systems (ISMS) Lead Implementer, Data Protection Officer, and Cyber Risks Analyst, Denis brings a heightened focus on data security, privacy, and cyber resilience to every endeavor.
His expertise extends across a diverse spectrum of reporting, database, and web development applications, underpinned by an exceptional grasp of data storage and virtualization technologies. His proficiency in application testing, database administration, and data cleansing ensures seamless execution of complex projects.
What sets Denis apart is his comprehensive understanding of Business and Systems Analysis technologies, honed through involvement in all phases of the Software Development Lifecycle (SDLC). From meticulous requirements gathering to precise analysis, innovative design, rigorous development, thorough testing, and successful implementation, he has consistently delivered exceptional results.
Throughout his career, he has taken on multifaceted roles, from leading technical project management teams to owning solutions that drive operational excellence. His conscientious and proactive approach is unwavering, whether he is working independently or collaboratively within a team. His ability to connect with colleagues on a personal level underscores his commitment to fostering a harmonious and productive workplace environment.
Date: May 29, 2024
Tags: Information Security, ISO/IEC 27001, ISO/IEC 42001, Artificial Intelligence, GDPR
-------------------------------------------------------------------------------
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Training: ISO/IEC 27001 Information Security Management System - EN | PECB
ISO/IEC 42001 Artificial Intelligence Management System - EN | PECB
General Data Protection Regulation (GDPR) - Training Courses - EN | PECB
Webinars: https://pecb.com/webinars
Article: https://pecb.com/article
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The three major forms of business organizations are1. Sole Propri.pdf
The three major forms of business organizations are:
1. Sole Proprietorship
2. Partnership &
3. Corporation
Sole Proprietorship: Ownership : In this form of organisation the ownership lies with a single
person i.e. the proprietor of the buiness. Along with the ownership all the liabilities and
responsibilities are with the sole proprietor of the organisation
Partnership: Ownership: In a partnership form of organisation more than one person come
together to form a business and all of them bring in capital in a decided proportion and they own
such decided portion of the organisation. They share liabilities and responsibilities in the pre
determined proportion i.e. profit sharing ratio
It is managed by all the partners together or by any working partner as decided by all of the
partners.
Corporation : Many people contribute to the Capital of the organisation and the management lies
with the directors and the management ho are elected by the shareholders and are completely
answerable and responsible to all the stakeholders.
Solution
The three major forms of business organizations are:
1. Sole Proprietorship
2. Partnership &
3. Corporation
Sole Proprietorship: Ownership : In this form of organisation the ownership lies with a single
person i.e. the proprietor of the buiness. Along with the ownership all the liabilities and
responsibilities are with the sole proprietor of the organisation
Partnership: Ownership: In a partnership form of organisation more than one person come
together to form a business and all of them bring in capital in a decided proportion and they own
such decided portion of the organisation. They share liabilities and responsibilities in the pre
determined proportion i.e. profit sharing ratio
It is managed by all the partners together or by any working partner as decided by all of the
partners.
Corporation : Many people contribute to the Capital of the organisation and the management lies
with the directors and the management ho are elected by the shareholders and are completely
answerable and responsible to all the stakeholders..
the OSI model is an idea. it is abstract it has no value without imp.pdf
the OSI model is an idea. it is abstract it has no value without implementation. TCP/IP is an
implementation. it was shaped by the concept of the OSI model. the OSI model is unchanging,
implementations like TCP/IP conform to the mode
Solution
the OSI model is an idea. it is abstract it has no value without implementation. TCP/IP is an
implementation. it was shaped by the concept of the OSI model. the OSI model is unchanging,
implementations like TCP/IP conform to the mode.
The EVA metric effectively measures the amount of shareholder wealth.pdf
The EVA metric effectively measures the amount of shareholder wealth that the firm\'s
management has added to the firm\'s value during a time period. If EVA is positive , then
management has increased value.
While there cannot be a single factor which can be attributed to the managers performing their
primary task i.e. MVA/EVA as they are basically interrelated. This is due to the fact that both
aim towards creation of value.
The only difference is in the calculation and the interpretation as defined by mathematical
paradigms.
EVA is the difference of NOPAT (Net Operating Profit After Taxes ) and Cost of Capital.This
indicates the wealth the company generated over and above the cost of capital.
THe MVA on the other hand is simply the difference between the current total market value and
the capital contributed by investors (both shareholders and bondholders).
Thus as the company performs well on the EVA metric and as it retained earnings rise so will
the business. And as the business rises so will the share price and hence the MVA. SO basically
it is all interrelated.
EBIT = 80M-52M = 28M
EVA = 28* (1-0.4) - 115*0.125 = 2.425 M
MVA of a company is equal to the net present value of all future EVAs.
Solution
The EVA metric effectively measures the amount of shareholder wealth that the firm\'s
management has added to the firm\'s value during a time period. If EVA is positive , then
management has increased value.
While there cannot be a single factor which can be attributed to the managers performing their
primary task i.e. MVA/EVA as they are basically interrelated. This is due to the fact that both
aim towards creation of value.
The only difference is in the calculation and the interpretation as defined by mathematical
paradigms.
EVA is the difference of NOPAT (Net Operating Profit After Taxes ) and Cost of Capital.This
indicates the wealth the company generated over and above the cost of capital.
THe MVA on the other hand is simply the difference between the current total market value and
the capital contributed by investors (both shareholders and bondholders).
Thus as the company performs well on the EVA metric and as it retained earnings rise so will
the business. And as the business rises so will the share price and hence the MVA. SO basically
it is all interrelated.
EBIT = 80M-52M = 28M
EVA = 28* (1-0.4) - 115*0.125 = 2.425 M
MVA of a company is equal to the net present value of all future EVAs..
Reflection about the centre of the pentagon is not its symmetric and.pdf
Reflection about the centre of the pentagon is not its symmetric and the symmetry group has
trivial centre
Solution
Reflection about the centre of the pentagon is not its symmetric and the symmetry group has
trivial centre.
Question not visible. Please state again.SolutionQuestion not .pdf
The document does not contain any visible questions or information to summarize. No context or content was provided in the document to generate a 3 sentence summary.
Per my quiz, it was also D) formation of the carbocatio or bromonium.pdf
Per my quiz, it was also D) formation of the carbocatio or bromonium ion and B) addition of the
halide anion
Solution
Per my quiz, it was also D) formation of the carbocatio or bromonium ion and B) addition of the
halide anion.
Part ATay-Sachs disease is an autosomal recessive disorder, so, o.pdf
Part A:
Tay-Sachs disease is an autosomal recessive disorder, so, only a homozygous recessive person
will have the disease. 1 in 300 individuals of neither descent will be carrier. A child can have the
disease, only if both his/her parents are carriers.
Let the genotype of Tay Sachs disease be \'tt\'. So, carrier will be \'Tt\'
According to HW equilibrium, 2pq= 1/300
Given data is p = 0.0167 q = 0.983 (This data is totally wrong). Please note that \'q\' is the
recessive Tay-Sachs allele and \'p\' is the dominant one. So, the frequency of recessive allele
cannot be more than the dominant one.
When two carriers cross, 25% chance is there of having the genotype \'tt\'.
So, chances of having a child with Tay-Sachs disease = 25% of the total children born or 25% of
(1/300 * 1/300)
Part B.
If one individual is of Ashkenazi descent, then chances of being a carrier is 1/30
If the other parent is normal, then chances of being a carrier is 1/300
When two carriers cross, chances of having homozygous recessive genotype = 25%
So, 25% chance of having an affected child is there i.e
25% of (1/30 * 1/300)
Part C:
Again 25% chance of having an affected child is there.
You can also write as: 25% of (1/30 * 1/30)
Part D:
Again 25% chance of having an affected child is there.
You can also write as: 25% of (1/30 * 1/30)
Solution
Part A:
Tay-Sachs disease is an autosomal recessive disorder, so, only a homozygous recessive person
will have the disease. 1 in 300 individuals of neither descent will be carrier. A child can have the
disease, only if both his/her parents are carriers.
Let the genotype of Tay Sachs disease be \'tt\'. So, carrier will be \'Tt\'
According to HW equilibrium, 2pq= 1/300
Given data is p = 0.0167 q = 0.983 (This data is totally wrong). Please note that \'q\' is the
recessive Tay-Sachs allele and \'p\' is the dominant one. So, the frequency of recessive allele
cannot be more than the dominant one.
When two carriers cross, 25% chance is there of having the genotype \'tt\'.
So, chances of having a child with Tay-Sachs disease = 25% of the total children born or 25% of
(1/300 * 1/300)
Part B.
If one individual is of Ashkenazi descent, then chances of being a carrier is 1/30
If the other parent is normal, then chances of being a carrier is 1/300
When two carriers cross, chances of having homozygous recessive genotype = 25%
So, 25% chance of having an affected child is there i.e
25% of (1/30 * 1/300)
Part C:
Again 25% chance of having an affected child is there.
You can also write as: 25% of (1/30 * 1/30)
Part D:
Again 25% chance of having an affected child is there.
You can also write as: 25% of (1/30 * 1/30).
null is a subset of every setTrueSolutionnull is a sub.pdf
The document discusses the empty set, also known as the null set. It states that the null set is a subset of every set, and verifies that this statement is true. The null set contains no elements, so it is contained within any set regardless of its elements.
LiOH Sol.pdf
This 3-word document appears to be referring to a solution containing lithium hydroxide (LiOH). Lithium hydroxide is listed three times, suggesting it is the primary component of the solution under discussion. The brief document does not provide much other context around the LiOH solution.
Information is a valuable asset that can make or break your business.pdf
Information is a valuable asset that can make or break your business. When properly managed it
allows you to operate with confidence. Information security management gives you the freedom
to grow, innovate and broaden your customer-base in the knowledge that all your confidential
information will remain that way.
In this information era, Security has become more difficult to define and enforce.
information security was limited to controlling physical access to oral or written
communications. The importance of information security led societies to develop innovative
ways of protecting their information. Recent innovations in information technology, like the
Internet, have made it possible to send vast quantities of data across the globe with ease.
However, the challenge of controlling and protecting that information has grown exponentially
now that data can be easily transmitted, stored, copied, manipulated, and destroyed.
Within a large organization information technology generally refers to laptop and desktop
computers, servers, routers, and switches that form a computer network, although information
technology also includes fax machines, phone and voice mail systems, cellular phones, and other
electronic systems. A growing reliance on computers to work and communicate has made the
control of computer networks an important part of information security. Unauthorized access to
paper documents or phone conversations is still an information security concern, but the real
challenge has become protecting the security of computer networks, especially when they are
connected to the Internet. Most large organizations have their own local computer network, or
intranet, that links their computers together to share resources and support the communications
of employees and others with a legitimate need for access. Almost all of these networks are
connected to the Internet and allow employees to go “online.”
Solution
Information is a valuable asset that can make or break your business. When properly managed it
allows you to operate with confidence. Information security management gives you the freedom
to grow, innovate and broaden your customer-base in the knowledge that all your confidential
information will remain that way.
In this information era, Security has become more difficult to define and enforce.
information security was limited to controlling physical access to oral or written
communications. The importance of information security led societies to develop innovative
ways of protecting their information. Recent innovations in information technology, like the
Internet, have made it possible to send vast quantities of data across the globe with ease.
However, the challenge of controlling and protecting that information has grown exponentially
now that data can be easily transmitted, stored, copied, manipulated, and destroyed.
Within a large organization information technology generally refers to laptop and desktop
computers, servers, routers, and.
Ho there is no relationship between the age of the individual and t.pdf
Ho: there is no relationship between the age of the individual and the individual
Solution
Ho: there is no relationship between the age of the individual and the individual.
H2SO3 - H2O = SO2The oxide is sulfur dioxide SO2SolutionH2.pdf
H2SO3 - H2O => SO2
The oxide is sulfur dioxide: SO2
Solution
H2SO3 - H2O => SO2
The oxide is sulfur dioxide: SO2.
First of all we shoud understand what is HIV AIDSHIV AIDS is .pdf
First of all we shoud understand what is HIV / AIDS?
HIV / AIDS is not a single disease it is combination of diseases. Because once HIV is infected it
will start reduce immune system i.e defence mechanisms so it is called as Aquired Immuno
Defficiency Syndrome (AIDS). So because the HIV infected patient needs nutrient food for
increase their life span.
If you are HIV-positive, good nutrition can have several benefits. It can:
Basic principle of nutrition :
So overall the HIV infected patient shuld have high nutritious food havig good protein,
carbohydrates,vitamins , minerals to get good immune system to defence by the desease.
Thanks all the best
Solution
First of all we shoud understand what is HIV / AIDS?
HIV / AIDS is not a single disease it is combination of diseases. Because once HIV is infected it
will start reduce immune system i.e defence mechanisms so it is called as Aquired Immuno
Defficiency Syndrome (AIDS). So because the HIV infected patient needs nutrient food for
increase their life span.
If you are HIV-positive, good nutrition can have several benefits. It can:
Basic principle of nutrition :
So overall the HIV infected patient shuld have high nutritious food havig good protein,
carbohydrates,vitamins , minerals to get good immune system to defence by the desease.
Thanks all the best.
diffusion rate = 1Mr1r2 = m2m1 = t2t110.12.16 = m2 2mola.pdf
diffusion rate = 1/M
r1/r2 = m2/m1 = t2/t1
10.1/2.16 = m2 /2
molar mass of the unknown gas , m2 = 43.728 g
Solution
diffusion rate = 1/M
r1/r2 = m2/m1 = t2/t1
10.1/2.16 = m2 /2
molar mass of the unknown gas , m2 = 43.728 g.
Dear,The answer is .2.SolutionDear,The answer is .2..pdf
The document is a short note stating that the answer to an unknown problem or question is 0.2. It provides no context or explanation for how the answer was determined. In two sentences, it simply states "The answer is .2."
d. Increases venous PO2.systemic venous blood is likely to be 25 .pdf
d. Increases venous PO2.
systemic venous blood is likely to be 25% saturated with oxygen.
PO2 Partial presure of oxygen.
Solution
d. Increases venous PO2.
systemic venous blood is likely to be 25% saturated with oxygen.
PO2 Partial presure of oxygen..
Considering the general solution of the Laplace Equation we getXn(.pdf
Considering the general solution of the Laplace Equation we get
Xn(x) = sin(nx)
Yn(y) = Ancosh(ny) + Bnsinh(ny)
Lets have Yn(0) = 1 and An = 1 for convienience.
As we know Yn() = 0 , hence solving for Bn
=> Bn = -cosh(n)/sinh(n)
=> Yn(y) = sinh(n - ny)/sinh(n)
Since f(x) = sin(4x)
=> u(x,y) = n=1 sin(4x) * ( sinh(n - ny)/sinh(n) )
Solution
Considering the general solution of the Laplace Equation we get
Xn(x) = sin(nx)
Yn(y) = Ancosh(ny) + Bnsinh(ny)
Lets have Yn(0) = 1 and An = 1 for convienience.
As we know Yn() = 0 , hence solving for Bn
=> Bn = -cosh(n)/sinh(n)
=> Yn(y) = sinh(n - ny)/sinh(n)
Since f(x) = sin(4x)
=> u(x,y) = n=1 sin(4x) * ( sinh(n - ny)/sinh(n) ).
Cranial nerves which do not innervate the structures of the head but.pdf
Cranial nerves which do not innervate the structures of the head but innervate the structures
originating from branchial arches are the following
Solution
Cranial nerves which do not innervate the structures of the head but innervate the structures
originating from branchial arches are the following.
Code Include libraries. import javax.swing.JOptionPane;.pdf
Code:
//Include libraries.
import javax.swing.JOptionPane;
//Define a class lPayroll
public class lPayroll
{
//Main function
public static void main( String args[] )
{
//Initialize variables
String lfirst = \"\";
String lLast = \"\";
String lssn = \"\";
String lsalaryString;
double lsalary = 0;
String lwageString;
double lwage = 0;
String lhoursString;
int lhours = 0;
String lsalesString;
double lsales = 0;
String lrateString;
double lrate = 0;
String lempTypeString = \"\";
int lempType = 0;
int lcount;
for(lcount = 0; lcount < 5; lcount++)
{
//Receive user input
lempTypeString = JOptionPane.showInputDialog(null, \"Enter employee type \ \ Enter
your choice.\ 1. Salaried\ 2. Hourly\ 3. Commissioned\ 4. Commission plus base\", \"Employee
Type\", JOptionPane.QUESTION_MESSAGE);
lempType = Integer.parseInt(lempTypeString);
switch(lempType)
{
case 1:
lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\",
JOptionPane.QUESTION_MESSAGE);
lLast = JOptionPane.showInputDialog(null, \"Enter Last name \", \"lLast Name\",
JOptionPane.QUESTION_MESSAGE);
lssn = JOptionPane.showInputDialog(null, \"Enter social security number\", \"lssn\",
JOptionPane.QUESTION_MESSAGE);
lsalaryString = JOptionPane.showInputDialog(null, \"Enter weekly salary\", \"lsalary\",
JOptionPane.QUESTION_MESSAGE);
lsalary = Double.parseDouble(lsalaryString);
break;
case 2:
lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\",
JOptionPane.QUESTION_MESSAGE);
lLast = JOptionPane.showInputDialog(null, \"Enter Last name\", \"lLast Name\",
JOptionPane.QUESTION_MESSAGE);
lssn = JOptionPane.showInputDialog(null, \"Enter social security number\", \"lssn\",
JOptionPane.QUESTION_MESSAGE);
lwageString = JOptionPane.showInputDialog(null, \"Enter hourly wage\", \"Hourly
lwage\", JOptionPane.QUESTION_MESSAGE);
lwage = Double.parseDouble(lwageString);
lhoursString = JOptionPane.showInputDialog(null, \"Enter count of hours per week of
work\", \"lhours Worked\", JOptionPane.QUESTION_MESSAGE);
lhours = Integer.parseInt(lhoursString);
break;
case 3:
lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\",
JOptionPane.QUESTION_MESSAGE);
lLast = JOptionPane.showInputDialog(null, \"Enter Last name\", \"lLast Name\",
JOptionPane.QUESTION_MESSAGE);
lssn = JOptionPane.showInputDialog(null, \"Enter social security number\", \"lssn\",
JOptionPane.QUESTION_MESSAGE);
lsalesString = JOptionPane.showInputDialog(null, \"Enter gross sales on weekly
basis\", \"Gross lsales\", JOptionPane.QUESTION_MESSAGE);
lsales = Double.parseDouble(lsalesString);
lrateString = JOptionPane.showInputDialog(null, \"Enter commission percent\",
\"Commsission lrate\", JOptionPane.QUESTION_MESSAGE);
lrate = Double.parseDouble(lrateString);
break;
case 4:
lfirst = JOptionPane.showInputDialog(null, \"Enter first name\", \"lfirst Name\",
JOptionPane.QUESTION_MESSAGE);
lLast = JOptionPane.showInputDialog(null, \"Enter Last name\", \"lLast Name\",
JOptionPane.QUESTION_ME.
C code#include stdio.hint main(void) { FILE infp, o.pdf
C code:
#include
int main(void)
{
FILE *infp, *outfp;
int ID,day,hr,i=1;
int Temp_ID,Total_hr;
int Line_No=0;
if ((infp = fopen(\"emp.txt\", \"r\"))==NULL)
{
printf(\"Input file cannot be opened\ \");
return-1;
}
if ((outfp = fopen(\"out.txt\", \"w\"))==NULL)
{
printf(\"Output file cannot be opened\ \");
return-1;
}
while(!feof(infp))
{
Line_No=Line_No+1;
fscanf(infp,\"%d %d\",&ID,&hr);
if(Line_No==1)
{
Temp_ID=ID;day=1;Total_hr=hr;
}
else
{
if(Temp_ID==ID)
{
day=day+1;
Total_hr=Total_hr+hr;
}
else
{
fprintf(outfp,\"%d %d %d\ \",Temp_ID,day,Total_hr);
Temp_ID=ID;day=1;Total_hr=hr;
}
}
}
fprintf(outfp,\"%d %d %d\ \",Temp_ID,day-1,Total_hr-hr);
fclose(infp);
fclose(outfp);
}
emp.txt:
11 2
11 8
11 5
15 10
18 4
18 16
20 23
20 11
out.txt:
11 3 15
15 1 10
18 2 20
20 2 34
Solution
C code:
#include
int main(void)
{
FILE *infp, *outfp;
int ID,day,hr,i=1;
int Temp_ID,Total_hr;
int Line_No=0;
if ((infp = fopen(\"emp.txt\", \"r\"))==NULL)
{
printf(\"Input file cannot be opened\ \");
return-1;
}
if ((outfp = fopen(\"out.txt\", \"w\"))==NULL)
{
printf(\"Output file cannot be opened\ \");
return-1;
}
while(!feof(infp))
{
Line_No=Line_No+1;
fscanf(infp,\"%d %d\",&ID,&hr);
if(Line_No==1)
{
Temp_ID=ID;day=1;Total_hr=hr;
}
else
{
if(Temp_ID==ID)
{
day=day+1;
Total_hr=Total_hr+hr;
}
else
{
fprintf(outfp,\"%d %d %d\ \",Temp_ID,day,Total_hr);
Temp_ID=ID;day=1;Total_hr=hr;
}
}
}
fprintf(outfp,\"%d %d %d\ \",Temp_ID,day-1,Total_hr-hr);
fclose(infp);
fclose(outfp);
}
emp.txt:
11 2
11 8
11 5
15 10
18 4
18 16
20 23
20 11
out.txt:
11 3 15
15 1 10
18 2 20
20 2 34.
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