$ 90.08WorkingStep-1Monthly Payment CalculationMonhly Payment=L.pdf
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$ 90.08 Working:Step-1:Monthly Payment CalculationMonhly Payment=Loan Amount/Cumulative discount factor=$ 21,618/ 137.199=$ 157.57Working:Cumulative discount factor=(1-(1+i)^- n)/iWhere,=(1-(1+0.00417)^-204)/0.00417i=5%/12= 0.00417= 137.199n=17*12=204Step-2:Repayment schedule of first monthMonthBeginning Loan balanceInterest for the monthMonthly PaymentPrincipal RepaymentEnding Loan Balanceab=a*5%*1/12cd=c-ba-d1$ 21,618.00$ 90.08$ 157.57$ 67.49$ 21,550.51 Solution $ 90.08 Working:Step-1:Monthly Payment CalculationMonhly Payment=Loan Amount/Cumulative discount factor=$ 21,618/ 137.199=$ 157.57Working:Cumulative discount factor=(1-(1+i)^- n)/iWhere,=(1-(1+0.00417)^-204)/0.00417i=5%/12= 0.00417= 137.199n=17*12=204Step-2:Repayment schedule of first monthMonthBeginning Loan balanceInterest for the monthMonthly PaymentPrincipal RepaymentEnding Loan Balanceab=a*5%*1/12cd=c-ba-d1$ 21,618.00$ 90.08$ 157.57$ 67.49$ 21,550.51.
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1). The inheritance of recessive genes (disease causing) can remain .pdf
1). The inheritance of recessive genes (disease causing) can remain prevalent if presence of two
copies of an allele is disadvantageous and presence of one copy of the allele may be
advantageous.
For example in case of sickle cell anaemia, two alleles may cause sickle cell anaemia but
carrying one allele is advantageous in regions where malaria is common because this single copy
of allele offers resistance against malaria. Thus, many people still carry the allele of sickle cell
anaemia, as natural selection does not remove it; this is called “balanced polymorphism.”
Thus, people carrying one copy of sickle cell disease causing allele migrated to the regions of
Africa and develops resistance to malaria parasite. They passed this protective allele to their
offspring. However, two carriers with protective allele might have a child with sickle cell
disease.
Today, the prevalence of malaria parasite is decreased well by artificial control methods. This
change in climate caused decreased number of sickle cell allele carriers.
2). Yes, the individuals homozygous for the mutation are susceptible for the sickle cell anemia,
whereas the heterozygous individuals are resistant to malaria.
3). Increased incidence of malaria causes the persistence of sickle cell allele as it increases the
fitness of an individual.
Solution
1). The inheritance of recessive genes (disease causing) can remain prevalent if presence of two
copies of an allele is disadvantageous and presence of one copy of the allele may be
advantageous.
For example in case of sickle cell anaemia, two alleles may cause sickle cell anaemia but
carrying one allele is advantageous in regions where malaria is common because this single copy
of allele offers resistance against malaria. Thus, many people still carry the allele of sickle cell
anaemia, as natural selection does not remove it; this is called “balanced polymorphism.”
Thus, people carrying one copy of sickle cell disease causing allele migrated to the regions of
Africa and develops resistance to malaria parasite. They passed this protective allele to their
offspring. However, two carriers with protective allele might have a child with sickle cell
disease.
Today, the prevalence of malaria parasite is decreased well by artificial control methods. This
change in climate caused decreased number of sickle cell allele carriers.
2). Yes, the individuals homozygous for the mutation are susceptible for the sickle cell anemia,
whereas the heterozygous individuals are resistant to malaria.
3). Increased incidence of malaria causes the persistence of sickle cell allele as it increases the
fitness of an individual..
#ifndef COURSES_H #define COURSES_H class Courses { privat.pdf
#ifndef COURSES_H
#define COURSES_H
class Courses
{
private:
struct CourseList
{
char courseNum[10];
char courseName[25];
char courseSch[20];
CourseList *next;
};
CourseList *head;
void displayArrayChar(char [], int);
public:
Courses(void)
{head=NULL;}
~Courses(void);
void appendCourse(char [],char [], char []);
void appendCourse(CourseList *);
void deleteCourse(char []);
bool searchforCourse(char []);
void displayAllCourses(void);
void editCourses(char [], char [], int );
void emptyCourses(void); //function called when a student is deleted
CourseList* getHead(void){return head;}
//friend void StudentInfo::insertStudent(char [], char [], int [], CourseList *);
};
#endif
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
//studentinfo.h
#ifndef STUDENTINFO_H
#define STUDENTINFO_H
#include
#include \"Courses.h\"
using namespace std;
class StudentInfo
{
private:
struct StudentList
{
char firstName[12];
char lastName[12];
int ssn[9];
Courses studentCourses;
struct StudentList *next;
};
StudentList *ssnhead;
int tempStoreSSN[9];
void displayArrayChar(char [], int);
void displayArrayInt(int []);
bool testIntArray(int [], int []);
public:
StudentInfo(void)
{ssnhead=NULL;}
~StudentInfo(void);
void insertStudent(char [], char [], int [], Courses &);
void insertStudent(char fn[], char ln[], int s[]);
void deleteStudent(int []);
bool searchBylast(char []);
bool searchByssn(int []);
void displayAllStudents(void);
void editStudentInfo(int []);
void editStudentInfo(char [],int);
int getTempssn(int);
void storessnTemp(char [], int);
};
#endif
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
//courses.cpp
#include
#include
#include
#include \"courses.h\"
#include \"studentinfo.h\"
using namespace std;
Courses::~Courses(void)
{
CourseList *nodePtr, *nextNode;
nodePtr=head;
while (nodePtr !=NULL)
{
nextNode=nodePtr->next;
delete nodePtr;
nodePtr=nextNode;
}
head=NULL;
}
void Courses::appendCourse(CourseList *tempHead) //should only pass the head pointer of the
node
{
CourseList *nodePtr, *nodeTempPtr, *newNode;
nodePtr=head;
nodeTempPtr=nodePtr;
while(nodeTempPtr->next !=NULL)
{nodePtr=nodePtr->next;
}
nodeTempPtr = tempHead;
while(nodeTempPtr->next !=NULL)
{
newNode=new CourseList;
strcpy(newNode->courseNum, nodeTempPtr->courseNum);
strcpy(newNode->courseName, nodeTempPtr->courseName);
strcpy(newNode->courseSch, nodeTempPtr->courseSch);
newNode->next=NULL;
nodePtr->next=newNode;
nodePtr=nodePtr->next;
nodeTempPtr=nodeTempPtr->next;
}
}
void Courses::appendCourse(char cNum[], char cName[], char cSch[])
{
CourseList *newNode, *nodePtr;
newNode=new CourseList;
strcpy(newNode->courseNum, cNum);
strcpy(newNode->courseName, cName);
strcpy(newNode->courseSch, cSch);
newNode->next=NULL;
if(!head)
head=newNode;
else
{
nodePtr=head;
while(nodePtr->next)
{
nodePtr=nodePtr->next;
}
nodePtr->next=newNode;
}
}
void Courses::emptyCourses(void)
{
CourseList *nodePtr, *nextNode;
nodePtr=head;
while (nodePtr !=NULL)
{
nextNode=nodePtr->next;
delete nodePtr;
nodePtr=nextNode;
}
head=NULL;
}
void Co.
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More from anupambedcovers
1). The inheritance of recessive genes (disease causing) can remain .pdf
1). The inheritance of recessive genes (disease causing) can remain prevalent if presence of two
copies of an allele is disadvantageous and presence of one copy of the allele may be
advantageous.
For example in case of sickle cell anaemia, two alleles may cause sickle cell anaemia but
carrying one allele is advantageous in regions where malaria is common because this single copy
of allele offers resistance against malaria. Thus, many people still carry the allele of sickle cell
anaemia, as natural selection does not remove it; this is called “balanced polymorphism.”
Thus, people carrying one copy of sickle cell disease causing allele migrated to the regions of
Africa and develops resistance to malaria parasite. They passed this protective allele to their
offspring. However, two carriers with protective allele might have a child with sickle cell
disease.
Today, the prevalence of malaria parasite is decreased well by artificial control methods. This
change in climate caused decreased number of sickle cell allele carriers.
2). Yes, the individuals homozygous for the mutation are susceptible for the sickle cell anemia,
whereas the heterozygous individuals are resistant to malaria.
3). Increased incidence of malaria causes the persistence of sickle cell allele as it increases the
fitness of an individual.
Solution
1). The inheritance of recessive genes (disease causing) can remain prevalent if presence of two
copies of an allele is disadvantageous and presence of one copy of the allele may be
advantageous.
For example in case of sickle cell anaemia, two alleles may cause sickle cell anaemia but
carrying one allele is advantageous in regions where malaria is common because this single copy
of allele offers resistance against malaria. Thus, many people still carry the allele of sickle cell
anaemia, as natural selection does not remove it; this is called “balanced polymorphism.”
Thus, people carrying one copy of sickle cell disease causing allele migrated to the regions of
Africa and develops resistance to malaria parasite. They passed this protective allele to their
offspring. However, two carriers with protective allele might have a child with sickle cell
disease.
Today, the prevalence of malaria parasite is decreased well by artificial control methods. This
change in climate caused decreased number of sickle cell allele carriers.
2). Yes, the individuals homozygous for the mutation are susceptible for the sickle cell anemia,
whereas the heterozygous individuals are resistant to malaria.
3). Increased incidence of malaria causes the persistence of sickle cell allele as it increases the
fitness of an individual..
#ifndef COURSES_H #define COURSES_H class Courses { privat.pdf
#ifndef COURSES_H
#define COURSES_H
class Courses
{
private:
struct CourseList
{
char courseNum[10];
char courseName[25];
char courseSch[20];
CourseList *next;
};
CourseList *head;
void displayArrayChar(char [], int);
public:
Courses(void)
{head=NULL;}
~Courses(void);
void appendCourse(char [],char [], char []);
void appendCourse(CourseList *);
void deleteCourse(char []);
bool searchforCourse(char []);
void displayAllCourses(void);
void editCourses(char [], char [], int );
void emptyCourses(void); //function called when a student is deleted
CourseList* getHead(void){return head;}
//friend void StudentInfo::insertStudent(char [], char [], int [], CourseList *);
};
#endif
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
//studentinfo.h
#ifndef STUDENTINFO_H
#define STUDENTINFO_H
#include
#include \"Courses.h\"
using namespace std;
class StudentInfo
{
private:
struct StudentList
{
char firstName[12];
char lastName[12];
int ssn[9];
Courses studentCourses;
struct StudentList *next;
};
StudentList *ssnhead;
int tempStoreSSN[9];
void displayArrayChar(char [], int);
void displayArrayInt(int []);
bool testIntArray(int [], int []);
public:
StudentInfo(void)
{ssnhead=NULL;}
~StudentInfo(void);
void insertStudent(char [], char [], int [], Courses &);
void insertStudent(char fn[], char ln[], int s[]);
void deleteStudent(int []);
bool searchBylast(char []);
bool searchByssn(int []);
void displayAllStudents(void);
void editStudentInfo(int []);
void editStudentInfo(char [],int);
int getTempssn(int);
void storessnTemp(char [], int);
};
#endif
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
//courses.cpp
#include
#include
#include
#include \"courses.h\"
#include \"studentinfo.h\"
using namespace std;
Courses::~Courses(void)
{
CourseList *nodePtr, *nextNode;
nodePtr=head;
while (nodePtr !=NULL)
{
nextNode=nodePtr->next;
delete nodePtr;
nodePtr=nextNode;
}
head=NULL;
}
void Courses::appendCourse(CourseList *tempHead) //should only pass the head pointer of the
node
{
CourseList *nodePtr, *nodeTempPtr, *newNode;
nodePtr=head;
nodeTempPtr=nodePtr;
while(nodeTempPtr->next !=NULL)
{nodePtr=nodePtr->next;
}
nodeTempPtr = tempHead;
while(nodeTempPtr->next !=NULL)
{
newNode=new CourseList;
strcpy(newNode->courseNum, nodeTempPtr->courseNum);
strcpy(newNode->courseName, nodeTempPtr->courseName);
strcpy(newNode->courseSch, nodeTempPtr->courseSch);
newNode->next=NULL;
nodePtr->next=newNode;
nodePtr=nodePtr->next;
nodeTempPtr=nodeTempPtr->next;
}
}
void Courses::appendCourse(char cNum[], char cName[], char cSch[])
{
CourseList *newNode, *nodePtr;
newNode=new CourseList;
strcpy(newNode->courseNum, cNum);
strcpy(newNode->courseName, cName);
strcpy(newNode->courseSch, cSch);
newNode->next=NULL;
if(!head)
head=newNode;
else
{
nodePtr=head;
while(nodePtr->next)
{
nodePtr=nodePtr->next;
}
nodePtr->next=newNode;
}
}
void Courses::emptyCourses(void)
{
CourseList *nodePtr, *nextNode;
nodePtr=head;
while (nodePtr !=NULL)
{
nextNode=nodePtr->next;
delete nodePtr;
nodePtr=nextNode;
}
head=NULL;
}
void Co.
C++ Program to Implement Singly Linked List #includei.pdf
/*
* C++ Program to Implement Singly Linked List
*/
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*start;
/*
* Class Declaration
*/
class single_llist
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos();
void sort();
void search();
void update();
void reverse();
void display();
single_llist()
{
start = NULL;
}
};
/*
* Main :contains menu
*/
main()
{
int choice, nodes, element, position, i;
single_llist sl;
start = NULL;
while (1)
{
cout<<\"1.Insert Node at beginning\"<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<info = value;
temp->next = NULL;
return temp;
}
}
/*
* Inserting element in beginning
*/
void single_llist::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<>value;
struct node *temp, *s, *ptr;
temp = create_node(value);
cout<<\"Enter the postion at which node to be inserted: \";
cin>>pos;
int i;
s = start;
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos == 1)
{
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
ptr = start;
start = temp;
start->next = ptr;
}
}
else if (pos > 1 && pos <= counter)
{
s = start;
for (i = 1; i < pos; i++)
{
ptr = s;
s = s->next;
}
ptr->next = temp;
temp->next = s;
}
else
{
cout<<\"Positon out of range\"<next;s !=NULL;s = s->next)
{
if (ptr->info > s->info)
{
value = ptr->info;
ptr->info = s->info;
s->info = value;
}
}
ptr = ptr->next;
}
}
/*
* Delete element at a given position
*/
void single_llist::delete_pos()
{
int pos, i, counter = 0;
if (start == NULL)
{
cout<<\"List is empty\"<>pos;
struct node *s, *ptr;
s = start;
if (pos == 1)
{
start = s->next;
}
else
{
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos > 0 && pos <= counter)
{
s = start;
for (i = 1;i < pos;i++)
{
ptr = s;
s = s->next;
}
ptr->next = s->next;
}
else
{
cout<<\"Position out of range\"<>pos;
cout<<\"Enter the new value: \";
cin>>value;
struct node *s, *ptr;
s = start;
if (pos == 1)
{
start->info = value;
}
else
{
for (i = 0;i < pos - 1;i++)
{
if (s == NULL)
{
cout<<\"There are less than \"<next;
}
s->info = value;
}
cout<<\"Node Updated\"<>value;
struct node *s;
s = start;
while (s != NULL)
{
pos++;
if (s->info == value)
{
flag = true;
cout<<\"Element \"<next;
}
if (!flag)
cout<<\"Element \"<next == NULL)
{
return;
}
ptr1 = start;
ptr2 = ptr1->next;
ptr3 = ptr2->next;
ptr1->next = NULL;
ptr2->next = ptr1;
while (ptr3 != NULL)
{
ptr1 = ptr2;
ptr2 = ptr3;
ptr3 = ptr3->next;
ptr2->next = ptr1;
}
start = ptr2;
}
/*
* Display Elements of a link list
*/
vo.
Long term Expected return of the Stock return ty.pdf
Long term Expected return of the Stock return type Return Probability Probability
* return Booming 17% 0.57 9.69% recession 6% (1-0.57) 2.58% Expected Return of the
Stock 12.27%
Solution
Long term Expected return of the Stock return type Return Probability Probability
* return Booming 17% 0.57 9.69% recession 6% (1-0.57) 2.58% Expected Return of the
Stock 12.27%.
The English word chemistry has a few senses .pdf
The English word \'chemistry\' has a few senses: 1: the science of matter; the branch
of the natural sciences dealing with the composition of substances and their properties and
reactions; 2: the way two individuals relate to each other; 3: the chemical composition and
properties of a substance or object
Solution
The English word \'chemistry\' has a few senses: 1: the science of matter; the branch
of the natural sciences dealing with the composition of substances and their properties and
reactions; 2: the way two individuals relate to each other; 3: the chemical composition and
properties of a substance or object.
since it is (1)^infinity form ...... we get it b.pdf
since it is (1)^infinity form ...... we get it by lim n->infinity e ^ [ 1 - 2/5n - 1] *
(n+1) = e^(-2/5)...........
Solution
since it is (1)^infinity form ...... we get it by lim n->infinity e ^ [ 1 - 2/5n - 1] *
(n+1) = e^(-2/5)............
yes as NAI is ionic in nature water disolves ion.pdf
yes as NAI is ionic in nature water disolves ionic things in itself
Solution
yes as NAI is ionic in nature water disolves ionic things in itself.
Mg(0H)2 as there is ionic between Mg & OH and co.pdf
Mg(0H)2 as there is ionic between Mg & OH and covalent between O and H in OH
and same with K2C03 ionic between K and C03 and covalent between C and 0
Solution
Mg(0H)2 as there is ionic between Mg & OH and covalent between O and H in OH
and same with K2C03 ionic between K and C03 and covalent between C and 0.
It depends on how many unique protons are adjacen.pdf
It depends on how many unique protons are adjacent to the proton(s) that you are
looking at. First, if you have an isolated proton that is not adjacent to any other protons, it will
be a singlet. A good example is a tert-butyl group. The three CH3 appear equivalent and none of
them has directly adjacent protons. The signal appears as a huge singlet that integrates to nine
protons. Doublets arise when there is a single proton next to the proton you are observing. My
favorite example of this is 2-methyl cyclohexanone. If you were to look at the CH3 group, you
would see a doublet because there is one proton on the neighboring carbon. Now for some of the
more complicated stuff: Suppose you have an ethyl ether group in your molecule. If you look at
the signal for the CH3 at the end of the chain, you have a signal which would be split by the two
protons on the next carbon over. The rule is, you get n + 1 peaks, where n is the number of
protons that are causing the splitting. Since in the case of Et there are two equivalent protons on
the CH2, you end up with 2 + 1 = 3 peaks, a triplet. The peaks should appear in a 1:2:1 height
ratio. Now suppose we look at the CH2 protons, which have a different chemical shift. In this
case, the protons are adjacent to three equivalent protons (on the methyl group). As a result you
get 3 + 1 = 4 peaks, a quartet. In this case, the peaks will show up in a 1:3:3:1 height ratio.
Solution
It depends on how many unique protons are adjacent to the proton(s) that you are
looking at. First, if you have an isolated proton that is not adjacent to any other protons, it will
be a singlet. A good example is a tert-butyl group. The three CH3 appear equivalent and none of
them has directly adjacent protons. The signal appears as a huge singlet that integrates to nine
protons. Doublets arise when there is a single proton next to the proton you are observing. My
favorite example of this is 2-methyl cyclohexanone. If you were to look at the CH3 group, you
would see a doublet because there is one proton on the neighboring carbon. Now for some of the
more complicated stuff: Suppose you have an ethyl ether group in your molecule. If you look at
the signal for the CH3 at the end of the chain, you have a signal which would be split by the two
protons on the next carbon over. The rule is, you get n + 1 peaks, where n is the number of
protons that are causing the splitting. Since in the case of Et there are two equivalent protons on
the CH2, you end up with 2 + 1 = 3 peaks, a triplet. The peaks should appear in a 1:2:1 height
ratio. Now suppose we look at the CH2 protons, which have a different chemical shift. In this
case, the protons are adjacent to three equivalent protons (on the methyl group). As a result you
get 3 + 1 = 4 peaks, a quartet. In this case, the peaks will show up in a 1:3:3:1 height ratio..
x = 1087n = 1430p =xn = 0.76standard error, SE = sqrt(p(1.pdf
x = 1087
n = 1430
p =x/n = 0.76
standard error, SE = sqrt(p(1-p)/n)
SE = sqrt(0.76(1-0.76) / 1430)
= 0.0113
a.
alpha,a = 1-0.90 = 0.10
Za/2 = Z0.05= 1.645
confidence interval:
p +/- [Za/2 * SE]
0.76 +/- 0.0186
( 0.7416 , 0.7787 )
b.
alpha,a = 1-0.95 = 0.05
Za/2 = Z0.025= 1.96
confidence interval:
p +/- [Za/2 * SE]
0.76 +/- 0.0221
( 0.7380 , 0.7823 )
c.
as the confidence level increases, the confidence interval get wider
Solution
x = 1087
n = 1430
p =x/n = 0.76
standard error, SE = sqrt(p(1-p)/n)
SE = sqrt(0.76(1-0.76) / 1430)
= 0.0113
a.
alpha,a = 1-0.90 = 0.10
Za/2 = Z0.05= 1.645
confidence interval:
p +/- [Za/2 * SE]
0.76 +/- 0.0186
( 0.7416 , 0.7787 )
b.
alpha,a = 1-0.95 = 0.05
Za/2 = Z0.025= 1.96
confidence interval:
p +/- [Za/2 * SE]
0.76 +/- 0.0221
( 0.7380 , 0.7823 )
c.
as the confidence level increases, the confidence interval get wider.
Option B. ie More than one theseSolutionOption B. ie More than.pdf
Option B. ie More than one these
Solution
Option B. ie More than one these.
When two genes are located on the same chromosome they are called as.pdf
When two genes are located on the same chromosome they are called as linked. That means they
are inherited at the same time together thereby violating independent law of assortment. They
can be separated during meiosis by recombination process.
Solution
When two genes are located on the same chromosome they are called as linked. That means they
are inherited at the same time together thereby violating independent law of assortment. They
can be separated during meiosis by recombination process..
We have layered model used in network design since these layers grou.pdf
We have layered model used in network design since these layers group functions according to
the task that needs to be performed. Every function in this model is targeted to help a specific
layer perform its job.
5 layers in network model :
1) Application layer: It governs how to application work with each other.It provides applications
the ability to access the services of the other layers and defines the protocols that applications use
to exchange data.
Addressing : NA
Application layer Protocols: HTTP, FTP, SMTP, Telnet
2) Transport layer: It establishes the connection between applications running on different
hosts.It keeps track of the processes running in the applications above it by assigning port
numbers to them and uses the Network layer to access the TCP/IP network.
Addressing : Port address
Transport layer Protocols: TCP (Connection-oriented) , UDP (Connection-less)
3) Network layer: It governs the transmission of packets across an internet by sending them
through several routers along the route. It uses IP addresses to identify the packet’s source and
destination.
Addressing : IP Address
Network layer Protocols : IP, ARP, ICMP, IGMP
4) Data link layer: It governs transmission of frames across a single network.These frames
encapsulate the packets and use MAC addresses to identify the source and destination.
Addressing : MAC (Physical) address
Data link layer Protocols : WLAN protocols, Token rings
5) Physical layer: It governs transmission between adjacent devices connected by a transmission
medium.
Addressing : NA
Solution
We have layered model used in network design since these layers group functions according to
the task that needs to be performed. Every function in this model is targeted to help a specific
layer perform its job.
5 layers in network model :
1) Application layer: It governs how to application work with each other.It provides applications
the ability to access the services of the other layers and defines the protocols that applications use
to exchange data.
Addressing : NA
Application layer Protocols: HTTP, FTP, SMTP, Telnet
2) Transport layer: It establishes the connection between applications running on different
hosts.It keeps track of the processes running in the applications above it by assigning port
numbers to them and uses the Network layer to access the TCP/IP network.
Addressing : Port address
Transport layer Protocols: TCP (Connection-oriented) , UDP (Connection-less)
3) Network layer: It governs the transmission of packets across an internet by sending them
through several routers along the route. It uses IP addresses to identify the packet’s source and
destination.
Addressing : IP Address
Network layer Protocols : IP, ARP, ICMP, IGMP
4) Data link layer: It governs transmission of frames across a single network.These frames
encapsulate the packets and use MAC addresses to identify the source and destination.
Addressing : MAC (Physical) address
Data link layer Protocols : WLAN protocols, Token rings
5) .
The half-reactions areAnode Fe = [Fe]2+ + 2 e-Fe(s) release.pdf
The half-reactions are:
Anode: Fe => [Fe]2+ + 2 e-
Fe(s) releases electrons and is oxidized to Fe2+(aq) at the anode.
Cathode: [Ag]+ + e- => Ag
Ag+(aq) accepts electrons and is reduced to Ag(s) at the cathode
Solution
The half-reactions are:
Anode: Fe => [Fe]2+ + 2 e-
Fe(s) releases electrons and is oxidized to Fe2+(aq) at the anode.
Cathode: [Ag]+ + e- => Ag
Ag+(aq) accepts electrons and is reduced to Ag(s) at the cathode.
The answer is CH2Since moles = massmolar massMoles of C H = .pdf
The answer is: CH2
Since moles = mass/molar mass
Moles of C : H = 85.5/12 : 14.5/1
= 7.125 : 14.5
= 1 : 2
Thus the empircal formula is CH2
Solution
The answer is: CH2
Since moles = mass/molar mass
Moles of C : H = 85.5/12 : 14.5/1
= 7.125 : 14.5
= 1 : 2
Thus the empircal formula is CH2.
SolutionFICASocial Security withheld taxMedicre tax withheldRates.pdf
Solution
:FICASocial Security withheld taxMedicre tax withheldRates Applicable for
20157.65%6.20%1.45%FICA tax in amount2,420.381,961.62458.76Calculation of social
security tax withheld= 2,420.38 / 7.65 % * 6.20 %Calculation of medicre withheld = 2,420.38 /
7.65 % * 1.45%.
Since HCL is a strong acid we can assume that it will fulyy ionise.pdf
Since HCL is a strong acid we can assume that it will fulyy ionise
HCl -----> H+ + Cl-
0.2 0 0
0 0.2 0.2
H2S ------- > H+ + HS-
Initial 0.10 0.2 0
Final 0.10-x 0.2+x x
I am taking pKa = 6.9
=> Ka = 1.259 *10-7 = x*(0.2+x)/(0.10-x)
=> x2 + 0.2x - 1.259*10-8 = 0 => x = 6.295* 10-8
=> Concentration of HS- = [HS-] = 6 * 10-8 M
Solution
Since HCL is a strong acid we can assume that it will fulyy ionise
HCl -----> H+ + Cl-
0.2 0 0
0 0.2 0.2
H2S ------- > H+ + HS-
Initial 0.10 0.2 0
Final 0.10-x 0.2+x x
I am taking pKa = 6.9
=> Ka = 1.259 *10-7 = x*(0.2+x)/(0.10-x)
=> x2 + 0.2x - 1.259*10-8 = 0 => x = 6.295* 10-8
=> Concentration of HS- = [HS-] = 6 * 10-8 M.
d) Peptide Note that forming a peptide bond resu.pdf
d) Peptide Note that forming a peptide bond results in elimination of a molecule of
water. since isoleucine, leucine and valine, these are hydrophobic and tend to orient towards the
interior of the folded protein molecule. Hence Peptide bonds are the ideal to form..... Hope u
understand.... :)
Solution
d) Peptide Note that forming a peptide bond results in elimination of a molecule of
water. since isoleucine, leucine and valine, these are hydrophobic and tend to orient towards the
interior of the folded protein molecule. Hence Peptide bonds are the ideal to form..... Hope u
understand.... :).
Security has been the number 1 issue for any IT industry and organiz.pdf
Security has been the number 1 issue for any IT industry and organizations have to adopt to the
new security technology catering to their needs from time to time. Though the DBMS itself has
security features, however multiple features like SQL injection protection,vulnerability
assessments,user control activity etc features are still left unattended.Let us now see what are the
emerging security technologies in the protection of databases.
Protecting sensitive and confidential data of the organization is getting more and more difficult
as new threats are discovered. Database is a ever growing thing and their is a chance for the
presence of loopholes.The main problem when it comes to databases is the DBA should have the
view of all the data as a central controller and he should scour the database for any
vulnerabilities.VMT ( vulnerability management technology) is very useful to view the entire
database landscape and help the DBA to ensure his data of any exploitations.
Once the landscape is clearly scene a series of scans can be administered to see where the
sensitive or high protection needed data is present and using DAM technology (Database
monitoring technology) it is easy to protect the sensitive data effectively and add patches
wherever required. The technology also generates warnings and alerts on violations as it
monitors the database in real time. Most useful feature of using DAM is to collect information
about the use of DBMS which can help in audit purposes.It also provides database firewalls
which is similar to web application firewalls which can discard malicious activity,filter the
queries and have blacklist and whitelist feature as well.
Yes, software security is what people are after.However hardware security is needed wherever
software security cant protect.Nowadays cloud computing has emerged as the most powerful
technology where every information we have will be stored in remote servers and can be
accessed by us whenever needed. But the question arises about the security of the data. So, one
wonderful solution for this involves a hardware security systems where the dedicated chips are
inserted in the servers which can be overridden by the third parties via malicious attacks.These
chips provide security and security verification and makes the cloud system safe and reliable.
It is an obvious fact that providing hardware security for the device gives more confidence that
having a software security solution which can be accessed remotely and malware can come and
reside anywhere, But in these hardware systems there are only designated tasks and limited space
and malware cant even hide there.
Hardware security chips also manage physical problems.For example if there are fluctuations in
temperature,problems in electrical flow and breach of the casting then these chips can erase the
sensitive data.In this way these chips can protect themselves. There is also a facilty of every chip
having a unique digital signature and the user whe.
P.E = - k e2 rnP.E in nth orbit is UnP.E is inversely propotio.pdf
P.E = - k e2 / rn
P.E in nth orbit is Un
P.E is inversely propotional to rn
but, radius rn n2
now, P.E 1/n2
so, (P.E)n / (P.E)n+1 = { ( n+1)/n }2
Un/ P.En+1 = (n+1)2 /n2
P.En+1 = Un * (n/n+1)2
Solution
P.E = - k e2 / rn
P.E in nth orbit is Un
P.E is inversely propotional to rn
but, radius rn n2
now, P.E 1/n2
so, (P.E)n / (P.E)n+1 = { ( n+1)/n }2
Un/ P.En+1 = (n+1)2 /n2
P.En+1 = Un * (n/n+1)2.
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