The probability of a persons being left-handed is .12. In the 1992.pdf
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The probability of a person\'s being left-handed is .12. In the 1992 U.S. presidential election all three major candidates were left-handed. What\'s the probability of selecting three unrelated people who are all left-handed? Solution Unrelated people have independent left-handedness. Thus, P(all three left handed) = 0.12*0.12*0.12 = 0.001728 [answer].
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![The probability of a person's being left-handed is .12. In the 1992 U.S. presidential election
all three major candidates were left-handed. What's the probability of selecting three
unrelated people who are all left-handed?
Solution
Unrelated people have independent left-handedness.
Thus,
P(all three left handed) = 0.12*0.12*0.12 = 0.001728 [answer]](https://image.slidesharecdn.com/theprobabilityofapersonsbeingleft-handedis-230703110145-7787d2e6/85/The-probability-of-a-persons-being-left-handed-is-12-In-the-1992-pdf-1-320.jpg)
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briefly write about the Aridity of Subtropics Causes, locations, etc.
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The Aridity of Subtropics: the differential air temperature at some specific latitudes makes
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Compare and contrast the Mann-Whitney (aka Wilcoxon Rank Sum) Test w.pdf
Compare and contrast the Mann-Whitney (aka Wilcoxon Rank Sum) Test with the Median test.
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Wilcoxon test) are rank sum tests and not median tests. Basically, the Mann-Whitney-Wilcoxon
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Determine whether the following statements about the accumulated eang.pdf
Determine whether the following statements about the accumulated eangstax are true or false a.
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the appropriate answer below the motives for the corporation\'s unreasonable accumulation of
earnings Select true/false Each corporate member of a controlled group can claim a separate
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only from the date the IRS or the courts determine that the tax is owed The accumulated earnings
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from any drop-down list and then continue to the next question
Solution
A. True - It is necessary for IRS to prove that the earnings has been accumulated by corporations
for evading tax after analysing and audting the tax return of corporation properly.
B. false - net capital gain after deducting related federal income taxes is a considered as a
negative adjustment done to get accumulated taxable income
C. false - A single AEC statutory exemption must be allocated among the members of a
controlled group.
D. False - there is no reduction in taxabl income due to dividiend paid deduction
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Discuss the changing demographics of Entrepreneurs and give an example each.
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Data StructuresPlease I need help completing this c++ program..pdf
Data Structures
Please I need help completing this c++ program. it was giving to complete it with no other
discriptions.
#include
using namespace std;
template
struct node
{
datatype info;
node *left, *right;
};
template
class tree
{
public:
tree();
~tree();
bool isempty();
void inserttree(datatype data);
void print();
private:
node *root;
};
template
tree::tree()
{
root = NULL;
}
template
void tree ::inserttree(datatype data)
{
node *temp;
temp = new node ;
temp->left = NULL;
temp->right = NULL;
temp->info = data;
if (root == NULL)
root = temp;
else
before = NULL;
after = root;
while (after != NULL)
{
b = a;
root
...;
else
}
if(temp->info< before->info)
before->left = temp;
else
before->right = temp;
//before = after;
//if (temp->info < after->info)
//after = after->left;
//else
//after = after->right;
//if(temp->info< before->info)
//before->left = temp;
//else
//before->right = temp;
}
template
void tree ::print(tree *temp)
{
if (temp != NULL)
{
print(temp->left);
cout << temp->info << endl;
print(temp->right);
}
p();
print(root);
}
int main()
{
tree t;
//put this in a loop
t.inserttree(90);
t.inserttree(80);
t.inserttree(108);
t.p();
}
Solution
// HEADER FILE FOR BINARY TREE//
// Define a structure to be used as the tree node
struct TreeNode
{
int Key;
float fValue;
int iValue;
char cArray[7];
TreeNode *left;
TreeNode *right;
};
class Code202_Tree
{
private:
TreeNode *root;
public:
Code202_Tree();
~Code202_Tree();
bool isEmpty();
TreeNode *SearchTree(int Key);
bool Insert(TreeNode *newNode);
bool Insert(int Key, float f, int i, char *cA);
bool Delete(int Key);
void PrintOne(TreeNode *T);
void PrintTree();
private:
void ClearTree(TreeNode *T);
TreeNode *DupNode(TreeNode * T);
void PrintAll(TreeNode *T);
};
#endif
//IMPLEMENTATION .CPP FILE FOR BINARY TREE//
#include
#include
#include \"Code202_Tree.h\"
using namespace std;
//--------------------------------------------
// Function: Code202_Tree()
// Purpose: Class constructor.
//--------------------------------------------
Code202_Tree::Code202_Tree()
{
root = NULL;
return;
}
//--------------------------------------------
// Function: Code202_Tree()
// Purpose: Class destructor.
//--------------------------------------------
Code202_Tree::~Code202_Tree()
{
ClearTree(root);
return;
}
//--------------------------------------------
// Function: ClearTree()
// Purpose: Perform a recursive traversal of
// a tree destroying all nodes.
//--------------------------------------------
void Code202_Tree::ClearTree(TreeNode *T)
{
if(T==NULL) return; // Nothing to clear
if(T->left != NULL) ClearTree(T->left); // Clear left sub-tree
if(T->right != NULL) ClearTree(T->right); // Clear right sub-tree
delete T; // Destroy this node
return;
}
//--------------------------------------------
// Function: isEmpty()
// Purpose: Return TRUE if tree is empty.
//--------------------------------------------
bool Code202_Tree::isEmpty()
{
return(root==NULL);
}
//--------------------------------------------
// Function.
Create Student Project Database with insert, select, update, and del.pdf
Create Student Project Database with insert, select, update, and delete.
Part I
- Create Student Table (Student) 10 Students
STUDENT NO TEXT (3) PRIMARYKEY
STUDENT_NAME TEXT (10)
STUDENT_DOB DATE
STUDENT_DOJ DATE
- Create Project Table (Project) 3 projects, Name, C++/Java/VB
PRJ_NO TEXT (3) PRIMARY KEY
PRJ_NAME TEXT (15)
PRJ_PLATFORM TEXT (10)
- Create Student Project Table (StudentProject) Student#, Project#, Role(programmer, manager,
or analyst)
STUDENT_NO TEXT (3)
PRJ_NO TEXT (3)
DESIGNATION TEXT (10)
PRIMARYKEY (STUDENT_NO,PRJ_NO,DESIGNATION)
FOREIGN KEY(STUDENT_NO)
FOREIGN KEY(PRJ_NO)
Solution
CREATE TABLE Student
(
STUDENT_NO CHAR(3) NOT NULL PRIMARY KEY,
STUDENT_NAME CHAR(10),
STUDENT_DOB DATE,
STUDENT_DOJ DATE
)
CREATE TABLE Project
(
PRJ_NO CHAR(3) NOT NULL PRIMARY KEY,
PRJ_NAME CHAR(15),
PRJ_PLATFORM CHAR(10)
)
CREATE TABLE StudentProject
(
STUDENT_NO CHAR(3) REFERENCES Student,
PRJ_NO CHAR(3) REFERENCES Project,
DESIGNATION CHAR(10),
PRIMARY KEY(STUDENT_NO,PRJ_NO,DESIGNATION)
)
INSERT INTO Student VALUES (\'AB0\',\"Paul\",\'1992-05-12\',\'2016-02-12\');
INSERT INTO Student VALUES (\'AB1\',\"Sam\",\'1992-06-11\',\'2016-03-11\');
INSERT INTO Student VALUES (\'AB2\',\"Raul\",\'1992-03-11\',\'2016-04-05\');
INSERT INTO Student VALUES (\'AB3\',\"John\",\'1992-07-09\',\'2016-05-12\');
INSERT INTO Student VALUES (\'AB4\',\"Mike\",\'1992-02-06\',\'2016-06-02\');
INSERT INTO Student VALUES (\'AB5\',\"Ram\",\'1992-01-03\',\'2016-07-06\');
INSERT INTO Student VALUES (\'AB6\',\"Jin\",\'1992-05-12\',\'2016-08-03\');
INSERT INTO Student VALUES (\'AB7\',\"Frodo\",\'1992-08-02\',\'2016-09-12\');
INSERT INTO Student VALUES (\'AB8\',\"Bill\",\'1992-09-01\',\'2016-11-04\');
INSERT INTO Student VALUES (\'AB9\',\"Tom\",\'1992-06-12\',\'2016-12-12\');
SELECT * FROM Student;
UPDATE Student
SET STUDENT_DOB=\'1992-03-04\',STUDENT_DOJ=\'1991-03-04\'
WHERE STUDENT_NO=\'AB5\';
INSERT INTO Project VALUES (\'XY0\',\"C++\",\"Coding\");
INSERT INTO Project VALUES (\'XY1\',\"JAVA\",\"Theory\");
INSERT INTO Project VALUES (\'XY2\',\"VB\",\"Design\");
DELETE FROM Project WHERE PRJ_PLATFORM=\"Design\"; //It Deletes row number 3
from Project table
SELECT * FROM Project;
INSERT INTO StudentProject VALUES (\'AB0\',XY0,\"Programmer\");
INSERT INTO StudentProject VALUES (\'AB1\',XY1,\"Manager\");
INSERT INTO StudentProject VALUES (\'AB2\',XY2,\"Analyst\");
INSERT INTO StudentProject VALUES (\'AB3\',XY0,\"Programmer\");
INSERT INTO StudentProject VALUES (\'AB4\',XY1,\"Manager\");
INSERT INTO StudentProject VALUES (\'AB5\',XY2,\"Analyst\");
INSERT INTO StudentProject VALUES (\'AB6\',XY0,\"Programmer\");
INSERT INTO StudentProject VALUES (\'AB7\',XY1,\"Manager\");
INSERT INTO StudentProject VALUES (\'AB8\',XY2,\"Analyst\");
INSERT INTO StudentProject VALUES (\'AB9\',XY0,\"Programmer\");
SELECT * FROM StudentProject;.
Before 1980, most cell control and management software was written b.pdf
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Solution
False
The C programming language was devised in the early 1970s as a system implementation
language for the nascent Unix operating system. Derived from the typeless language BCPL, it
evolved a type structure; created on a tiny machine as a tool to improve a meager programming
environment, it has become one of the dominant languages of today
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operating system; the most creative period occurred during 1972. Another spate of changes
peaked between 1977 and 1979, when portability of the Unix system was being demonstrated. In
the middle of this second period, the first widely available description of the language appeared:
The C Programming Language,often called the `white book\' or `K&R\' [Kernighan 78]. Finally,
in the middle 1980s, the language was officially standardized by the ANSI X3J11 committee,
which made further changes. Until the early 1980s, although compilers existed for a variety of
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with Unix; more recently, its use has spread much more widely, and today it is among the
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Laboratories [Ritchie 78] [Ritchie 84]. The company was pulling out of the Multics project
[Organick 75], which had started as a joint venture of MIT, General Electric, and Bell Labs; by
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Solution
The changes made by Congress in the Taxpayer Relief Act of 1997 now allow a premature IRA
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Solution
The Aridity of Subtropics: the differential air temperature at some specific latitudes makes
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some warm air currents may trigger more air temperature
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Compare and contrast the Mann-Whitney (aka Wilcoxon Rank Sum) Test with the Median test.
Discuss their similarities/differences.
Solution
the Mann-Whitney and the equivalent Wilcoxon test (hereafter called the Mann-Whitney-
Wilcoxon test) are rank sum tests and not median tests. Basically, the Mann-Whitney-Wilcoxon
test ranks all of the observations from both groups and then sums the ranks from one of the
groups which is compared with the expected rank sum. It is possible, although not very common,
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Determine whether the following statements about the accumulated eang.pdf
Determine whether the following statements about the accumulated eangstax are true or false a.
Before the IRS can impose the accumulated earnings tax, it need only show that tax avoidance
was one of b. Long-term capital gains are included in the accumulated earnings tax base Select
the appropriate answer below the motives for the corporation\'s unreasonable accumulation of
earnings Select true/false Each corporate member of a controlled group can claim a separate
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for both cash and property distributions (other than nontaxableC- stock dividends) made by a
corporation. This deduction reduces both regular taxable income and accumulated taxable
income c. d. e. The accumulated earnings tax liability cannot be eliminated by paying a
deficiency dividend f. f Interest and penalties on the accumulated earnings tax deficiency accrue
only from the date the IRS or the courts determine that the tax is owed The accumulated earnings
tax is self-reported on Form 1120-AET that is filed along with the corporate tax return g. Choose
from any drop-down list and then continue to the next question
Solution
A. True - It is necessary for IRS to prove that the earnings has been accumulated by corporations
for evading tax after analysing and audting the tax return of corporation properly.
B. false - net capital gain after deducting related federal income taxes is a considered as a
negative adjustment done to get accumulated taxable income
C. false - A single AEC statutory exemption must be allocated among the members of a
controlled group.
D. False - there is no reduction in taxabl income due to dividiend paid deduction
E. True - There is not relation between payment of dificiency dividend and AET liability
F. false - interst and penalties on the AET come into effect from the date or day the return is due.
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Discuss the changing demographics of Entrepreneurs and give an examp.pdf
Discuss the changing demographics of Entrepreneurs and give an example each.
Solution
Entrepreneurs :
Defination : Entrepreneur is a person to starts up new business or locally sets up a new company
with a new product and enters into the market. Entrepreneurs are individuals that notice a need in
the market and take the opportunity to make profits satisfying that needs.Intiating a process that
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sesrvices.
Charatersticts of Entrepreneurs :
Attributes of a entrepreneurs
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Seven Sources of Entrepreneurship :
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Data StructuresPlease I need help completing this c++ program..pdf
Data Structures
Please I need help completing this c++ program. it was giving to complete it with no other
discriptions.
#include
using namespace std;
template
struct node
{
datatype info;
node *left, *right;
};
template
class tree
{
public:
tree();
~tree();
bool isempty();
void inserttree(datatype data);
void print();
private:
node *root;
};
template
tree::tree()
{
root = NULL;
}
template
void tree ::inserttree(datatype data)
{
node *temp;
temp = new node ;
temp->left = NULL;
temp->right = NULL;
temp->info = data;
if (root == NULL)
root = temp;
else
before = NULL;
after = root;
while (after != NULL)
{
b = a;
root
...;
else
}
if(temp->info< before->info)
before->left = temp;
else
before->right = temp;
//before = after;
//if (temp->info < after->info)
//after = after->left;
//else
//after = after->right;
//if(temp->info< before->info)
//before->left = temp;
//else
//before->right = temp;
}
template
void tree ::print(tree *temp)
{
if (temp != NULL)
{
print(temp->left);
cout << temp->info << endl;
print(temp->right);
}
p();
print(root);
}
int main()
{
tree t;
//put this in a loop
t.inserttree(90);
t.inserttree(80);
t.inserttree(108);
t.p();
}
Solution
// HEADER FILE FOR BINARY TREE//
// Define a structure to be used as the tree node
struct TreeNode
{
int Key;
float fValue;
int iValue;
char cArray[7];
TreeNode *left;
TreeNode *right;
};
class Code202_Tree
{
private:
TreeNode *root;
public:
Code202_Tree();
~Code202_Tree();
bool isEmpty();
TreeNode *SearchTree(int Key);
bool Insert(TreeNode *newNode);
bool Insert(int Key, float f, int i, char *cA);
bool Delete(int Key);
void PrintOne(TreeNode *T);
void PrintTree();
private:
void ClearTree(TreeNode *T);
TreeNode *DupNode(TreeNode * T);
void PrintAll(TreeNode *T);
};
#endif
//IMPLEMENTATION .CPP FILE FOR BINARY TREE//
#include
#include
#include \"Code202_Tree.h\"
using namespace std;
//--------------------------------------------
// Function: Code202_Tree()
// Purpose: Class constructor.
//--------------------------------------------
Code202_Tree::Code202_Tree()
{
root = NULL;
return;
}
//--------------------------------------------
// Function: Code202_Tree()
// Purpose: Class destructor.
//--------------------------------------------
Code202_Tree::~Code202_Tree()
{
ClearTree(root);
return;
}
//--------------------------------------------
// Function: ClearTree()
// Purpose: Perform a recursive traversal of
// a tree destroying all nodes.
//--------------------------------------------
void Code202_Tree::ClearTree(TreeNode *T)
{
if(T==NULL) return; // Nothing to clear
if(T->left != NULL) ClearTree(T->left); // Clear left sub-tree
if(T->right != NULL) ClearTree(T->right); // Clear right sub-tree
delete T; // Destroy this node
return;
}
//--------------------------------------------
// Function: isEmpty()
// Purpose: Return TRUE if tree is empty.
//--------------------------------------------
bool Code202_Tree::isEmpty()
{
return(root==NULL);
}
//--------------------------------------------
// Function.
Create Student Project Database with insert, select, update, and del.pdf
Create Student Project Database with insert, select, update, and delete.
Part I
- Create Student Table (Student) 10 Students
STUDENT NO TEXT (3) PRIMARYKEY
STUDENT_NAME TEXT (10)
STUDENT_DOB DATE
STUDENT_DOJ DATE
- Create Project Table (Project) 3 projects, Name, C++/Java/VB
PRJ_NO TEXT (3) PRIMARY KEY
PRJ_NAME TEXT (15)
PRJ_PLATFORM TEXT (10)
- Create Student Project Table (StudentProject) Student#, Project#, Role(programmer, manager,
or analyst)
STUDENT_NO TEXT (3)
PRJ_NO TEXT (3)
DESIGNATION TEXT (10)
PRIMARYKEY (STUDENT_NO,PRJ_NO,DESIGNATION)
FOREIGN KEY(STUDENT_NO)
FOREIGN KEY(PRJ_NO)
Solution
CREATE TABLE Student
(
STUDENT_NO CHAR(3) NOT NULL PRIMARY KEY,
STUDENT_NAME CHAR(10),
STUDENT_DOB DATE,
STUDENT_DOJ DATE
)
CREATE TABLE Project
(
PRJ_NO CHAR(3) NOT NULL PRIMARY KEY,
PRJ_NAME CHAR(15),
PRJ_PLATFORM CHAR(10)
)
CREATE TABLE StudentProject
(
STUDENT_NO CHAR(3) REFERENCES Student,
PRJ_NO CHAR(3) REFERENCES Project,
DESIGNATION CHAR(10),
PRIMARY KEY(STUDENT_NO,PRJ_NO,DESIGNATION)
)
INSERT INTO Student VALUES (\'AB0\',\"Paul\",\'1992-05-12\',\'2016-02-12\');
INSERT INTO Student VALUES (\'AB1\',\"Sam\",\'1992-06-11\',\'2016-03-11\');
INSERT INTO Student VALUES (\'AB2\',\"Raul\",\'1992-03-11\',\'2016-04-05\');
INSERT INTO Student VALUES (\'AB3\',\"John\",\'1992-07-09\',\'2016-05-12\');
INSERT INTO Student VALUES (\'AB4\',\"Mike\",\'1992-02-06\',\'2016-06-02\');
INSERT INTO Student VALUES (\'AB5\',\"Ram\",\'1992-01-03\',\'2016-07-06\');
INSERT INTO Student VALUES (\'AB6\',\"Jin\",\'1992-05-12\',\'2016-08-03\');
INSERT INTO Student VALUES (\'AB7\',\"Frodo\",\'1992-08-02\',\'2016-09-12\');
INSERT INTO Student VALUES (\'AB8\',\"Bill\",\'1992-09-01\',\'2016-11-04\');
INSERT INTO Student VALUES (\'AB9\',\"Tom\",\'1992-06-12\',\'2016-12-12\');
SELECT * FROM Student;
UPDATE Student
SET STUDENT_DOB=\'1992-03-04\',STUDENT_DOJ=\'1991-03-04\'
WHERE STUDENT_NO=\'AB5\';
INSERT INTO Project VALUES (\'XY0\',\"C++\",\"Coding\");
INSERT INTO Project VALUES (\'XY1\',\"JAVA\",\"Theory\");
INSERT INTO Project VALUES (\'XY2\',\"VB\",\"Design\");
DELETE FROM Project WHERE PRJ_PLATFORM=\"Design\"; //It Deletes row number 3
from Project table
SELECT * FROM Project;
INSERT INTO StudentProject VALUES (\'AB0\',XY0,\"Programmer\");
INSERT INTO StudentProject VALUES (\'AB1\',XY1,\"Manager\");
INSERT INTO StudentProject VALUES (\'AB2\',XY2,\"Analyst\");
INSERT INTO StudentProject VALUES (\'AB3\',XY0,\"Programmer\");
INSERT INTO StudentProject VALUES (\'AB4\',XY1,\"Manager\");
INSERT INTO StudentProject VALUES (\'AB5\',XY2,\"Analyst\");
INSERT INTO StudentProject VALUES (\'AB6\',XY0,\"Programmer\");
INSERT INTO StudentProject VALUES (\'AB7\',XY1,\"Manager\");
INSERT INTO StudentProject VALUES (\'AB8\',XY2,\"Analyst\");
INSERT INTO StudentProject VALUES (\'AB9\',XY0,\"Programmer\");
SELECT * FROM StudentProject;.
Before 1980, most cell control and management software was written b.pdf
Before 1980, most cell control and management software was written by the user using the C
programming language. True False
Solution
False
The C programming language was devised in the early 1970s as a system implementation
language for the nascent Unix operating system. Derived from the typeless language BCPL, it
evolved a type structure; created on a tiny machine as a tool to improve a meager programming
environment, it has become one of the dominant languages of today
C came into being in the years 1969-1973, in parallel with the early development of the Unix
operating system; the most creative period occurred during 1972. Another spate of changes
peaked between 1977 and 1979, when portability of the Unix system was being demonstrated. In
the middle of this second period, the first widely available description of the language appeared:
The C Programming Language,often called the `white book\' or `K&R\' [Kernighan 78]. Finally,
in the middle 1980s, the language was officially standardized by the ANSI X3J11 committee,
which made further changes. Until the early 1980s, although compilers existed for a variety of
machine architectures and operating systems, the language was almost exclusively associated
with Unix; more recently, its use has spread much more widely, and today it is among the
languages most commonly used throughout the computer industry.
The late 1960s were a turbulent era for computer systems research at Bell Telephone
Laboratories [Ritchie 78] [Ritchie 84]. The company was pulling out of the Multics project
[Organick 75], which had started as a joint venture of MIT, General Electric, and Bell Labs; by
1969, Bell Labs management, and even the researchers, came to believe that the promises of
Multics could be fulfilled only too late and too expensively. Even before the GE-645 Multics
machine was removed from the premises, an informal group, led primarily by Ken Thompson,
had begun investigating alternatives..
at age 51 can you take early distributions from traditional IRA for .pdf
at age 51 can you take early distributions from traditional IRA for family members educational
expenses
Solution
The changes made by Congress in the Taxpayer Relief Act of 1997 now allow a premature IRA
withdrawal to be taken without penalty if used to pay qualified higher education expenses of the
taxpayer or the taxpayer\'s spouse, child, or grandchild. However, the withdrawls are subject to
regular income taxes ..
What is a view What is it used for Give examples when a view can b.pdf
What is a view? What is it used for? Give examples when a view can be used for your project.
Solution
View is way of portraying information in a database by arranging data items in certain order by
highlighting certain items. Often thought of as Virtual table, doesn\'t store data itself but pulls out
from one or more tables.
Views are used according to purpose so that we query on selective data not whole of it. Using
views saves us lot of querying resources.
For example, people query telephone directory to search for Street Address associated with
Phone Number. We can create View which stores 2 columns namely Phone number and Street
Address ignoring all other fields..
^^^Q3. I am trying to implement double linked list but I was faile.pdf
^^^
Q3. I am trying to implement double linked list but I was failed to write the code so anyone gives
the main
Code in the main function
THANK YOU FOR ADVANCE
#include
#include
#include
struct node
{
int info;
struct node *lptr,*rptr;
};
typedef struct node DL;
DL *delete( ) , *insert ( );
void display();
DL *delete(DL *start,int x)
{
DL *left,*right,*curr;
curr = start;
if( start == NULL)
{
printf(\"\ Doubly Linked List is Empty\");
return(start);
}
while( (curr != NULL) && (curr->info != x) )
curr = curr->rptr;
if(curr == NULL)
{
printf(\"%d does not Exist ... Invalid Deletion! \ \",x);
return(start);
}
left = curr->lptr;
right = curr->rptr;
if(left == NULL)
{
start = right;
right->lptr = NULL;
}
else
{
left->rptr = right;
if(right != NULL)
right->lptr = left;
}
printf(\"\ The item %d is Deleted\ \",curr->info);
free(curr);
return(start);
}
void display(DL *start)
{
DL *last,*temp = start;
if( (temp != NULL) )
{
printf(\"\ The Elements of Doubly Linked List from Left to Right \");
printf(\"\ ROOT-> \");
while(temp != NULL )
{
printf(\"%d -> \",temp->info);
last = temp;
temp = temp->rptr;
}
printf(\"NULL\");
printf(\"\ The Elements of Doubly Linked List from Left to Right\");
printf(\"\ ROOT-> \");
while(last != NULL )
{
printf(\"%d -> \",last->info);
last = last->lptr;
}
printf(\"ROOT\ \ \");
}
else
printf(\"\ Empty List\ \");
}
DL *insert(DL *start,int x)
{
DL *new,*left,*temp;
new=(DL *) malloc(sizeof(DL));
new->info = x;
new->lptr = NULL;
new->rptr = NULL;
if( start == NULL)
start = new;
else
{
temp = start;
while(temp->rptr != NULL)
temp = temp->rptr;
temp->rptr = new;
new->lptr = temp;
}
return(start);
}
Main()
{
------------------Write the code here------------------------
}
Solution
main function for the given program.
main()
{
DL *root = NULL;
int item;
char c;
clrscr();
do
{
printf(\"Insert/Display/Remove/Quit <> I/D/R/Q \");
printf(\"\ Enter Choice : \");
fflush(stdin);
c = getchar();
switch(c)
{
case \'i\':
case \'I\': printf(\"Enter The Element you want to be Inserted : \");
scanf(\"%d\",&item);
root=insert(root,item);
break;
case \'d\':
case \'D\': display(root);
break;
case \'r\':
case \'R\':
printf(\"Enter the Item which you want to Remove :\");
scanf(\"%d\",&item);
root = delete(root,item);
break;
case \'Q\':
case \'q\': return;
}
} while(1);
}.
What is the evolutionary basis for sexual reproductionSolution.pdf
What is the evolutionary basis for sexual reproduction?
Solution
There are both asexual and sexual reproduction in nearly all groups of animals and plants, in
neither kingdom sex developed as a more rapid means.
When sexual reproduction first originated is still a question. But it is believed that sex originated
in green algae. There are certain species like Ulothrix have different size of spores are
produced. These spores comes in pair and become one and gives valuable new forms. In some
groups of unicellular plants ,the Peridinea, the Conjugatae and the Diatomeae- there is no asexual
reproduction. These are good example of the actual gametic origin.
The whole process of evolution is understand by these lines that first there was a visible
morphological differentiation of gametes, the one becoming a large inactive cell stored with food
, the other becoming small and mobile. Then came the evolution of various sex organs, and
finally the alternation of generation.
There are 3 main categories of sexual life cycles: diploid-dominant, demonstrated by most
animals; haploid - dominant, demonstrated by all fungi and some algae; and the alternation of
generations , demonstrated by plants and some algae.
The Red Queen Hyothesis-
First proposed by Leigh Van Valen in 1973.
According to this hypothesis are species co-evolved with other organisms like predators evolve
with their prey and parasites evolve with their hosts. Each tiny advantages gained by favorable
variations gives a secies an edge over close competitors, predators, parasites, or even prey.
Species that cannot keep up become extinct.
August Weismann proposed an explanation for the evolution of sex, where the advantages of sex
is the creation of variation among siblings.
A similar hypothesis proposed by Charles Darwin in his book Origin of species..
Why is it that the Grams stain is not appropriate for acid fast ba.pdf
Why is it that the Gram\'s stain is not appropriate for acid fast bacterium?
Solution
During staining in microbiology, microbes are coloured or dyed so that their structure can be
viewed against a lighter background. There are different staining techniques such as direct(Dye
binds to the cell surface and all the microbes are stained alike),differential(based on their specific
cell surface different microbes get stained differently) special (usually a particular organelle is
targeted)and negative staining(back ground is stained instead of microbes) Gram stain is a
differential stain where the bacteria with a thick peptidoglycan layer around the cells are stained
with violet colour and those with an additional lipopolysaccharide layer will be stained pink. To
understand this lets take a look at Gram staining procedure. First of all bacterial culture is
smeared onto a glass slide and is heat fixed so that during the washes in the procedure the
bacteria stay fixed onto the slide. Primary stain crystal violet is added onto the slide and the dye
seperates into ions which penetrate or enter the cell membranes of both Gram positive and
negative cells. Next iodine is added onto the slide which forms huge complexes with crystal
violet. Then a dye remover usually called as decolouriser such as alchohol or acetone is used to
remove the stain. In Gram negative bacteria the outer most membrane is lost during the treatment
with alchohol whereas in the case of gram positive bacteria the dye is trapped inside the cell
membrane. A secondary stain saffronin is used and which gives pink colour to Gram negative
bacteria. Now coming to acid-fast bacteria, they have a compound called mycolic acid in their
cell walls and are resistant to decolourisation after the acid treatment and alchohol and hence
would not lose colour and appear violet. As we can see these organism can be easily confused
with Gram positive bacteria which also do not lose stain not because they are resistant to acids
but they have the crystal violet dye trapped in their cell membrane..
Why does this version of the swap function fail to work Is there a f.pdf
Why does this version of the swap function fail to work? Is there a fix? (Choose 2) oid swap (int
& lhs, int & rhs) Select one or more: It fails because the programmer forgot to make the
parameters call-by-reference. Of course it works ! Just look at it It clearly swaps the two
parameters! It fails OK, and we can fix it we can just reverse the order of the lines. To fix this,
we must save the the value in a local variable before making the first assignment indicated, then
instead of the second Iine, as int local = lhs; Ihs = rhs; rhs = local; It fails because the first line
destroys the old value of ihs without saving it. Then both variables have the old value of rhs in
them.
Solution
Choice d, and e are valid options.
To swap the elements, we need to first copy the value of first variable to temporary variable, then
copy the second variable to first variable, finally copy the value in temporary variable to second
variable.
So,
int local = lhs;
lhs = rhs;
rhs = local;
is the perfect way to swap two values.
The last option is the actual problem with the given code.
When we do lhs = rhs, the old value of lhs is destroyed, and now, both lhs and rhs are holding
the same value, that is the value in rhs. And then, the second operation rhs = lhs, is of no use, as
both of them are holding the same value.
Therefore, options d, and e are correct..
Which of the following is NOT a reason why heat is a bad method to ov.pdf
Which of the following is NOT a reason why heat is a bad method to overcome activation
energy? A. Heat is not specific enough compared to enzymes B. Too much heat would denature
enzymes C. Heat would allow for many enzymes to be activated, not just the ones that are
needed D. Heat does not hold enough energy to overcome most activation energies
Solution
temperature is one the influence factor for enzyme acitvity. In the body of the biochemical
reaction takes place
for energy production. enzymes plays vital role by mediating the reactions. temperature influnece
enzyme acticity. most the enzymes in general active between optimum temperature( 40-45c). in
conditions where body temperature above 50 degree celcius enzymes gets denaturation which
leads to derangment in the tertiary structure. in the become inactive at higher temperature (above
70c) answer B too much heat would denature the enzymes.
What are the trends in browsersSolution1.chrome Google chrom.pdf
What are the trends in browsers?
Solution
1.chrome :Google chrome is a freeware web browser evolved by Google.a Webkit layout is used
upto 27th version with the pecularity of IOS is declared. it is initially liberated in the form of
beta version for the windows in the year 2008 september 2nd.
2.Firefox:Mozilla firefox is shortly called as firefox .it is an open source and also lavish in
nature.moreover it was advanced for windows,linux and osx.firefox is periperal in nature.in
firefox a Gecko layout engine is seen to provide web pages that applies present and predicted
web attainments.
3.Opera:It is evolved by a opera software.the trendy version is obtainable for windows,linux and
osx.Blink layout engine is seen in this opera web browser.versions of opera is obtained for
widgets operating the IOS,Bada and blackberry.
4.Safari:Safari browser is evolved by Apple inc.containing the os x and IOS operarting
systems.initially in the form of public beta in the year 2003 january 7 .it flatters as a Apple,s
default browser starting with MAC OS X v 10.3 \"panther\".the congential browser of IOS is
additionally called as safari.However it had a discrete Graphical user interface and utilizes a
discrete webkit edition..
The mass of a radioactive substance follows a continuous exponential.pdf
The mass of a radioactive substance follows a continuous exponential decay model, with a decay
rate parameter of 5.1% per day. Find the half-life of this substance (that is, the time it takes for
one-half the original amount in a given sample of this substance to decay).
Solution
Half life = log2/ decay constant = 0.693/ decay constant
Decay constant 5.1% = 0.051
So, Half life =0.693 /0.051 = 13.58 days.
The groin is also known as the _____ region. crural inguinal perin.pdf
The groin is also known as the _____ region. crural inguinal perineum popliteal The upper
midportion of the abdomen is called the _____ region. hypochondriac iliac hypogastric
epigastric The thoracic cavity lies where In relationship to the abdominopelvic cavity? dorsal
(posterior) ventral (anterior) superior inferior The urinary bladder is in the _____ region.
inguinal epigastric umbilical hypogastric All of the following are found in the thoracic cavity
EXCEPT the lungs. pleurae. meninges. heart. Which of the following organs is retroperitoneal
in location? stomach liver heart kidney
Solution
1.b) Groin is also calles as inguinal region which is between the abdomen and the thigh.
2.d)epigastric region-It is the upper central region of the abdomen.It is located between the costal
margins and the subcostal plane.
3.c)superior
4.d)hypogastric region-It is a region of the abdomen located below the umbilical region.
5.c)meninges
Thoracic cavity contains trachea,bronchi,lungs,oesophagus,heart and great blood vessels,thymus
gland,lymph nodes,and nerves. It also contains two smaller cavities-pleural cavities surround
each lung and pericardial cavity covers the heart.
6.d)kidney
Retroperitoneal organs include- the kidneys,adrenal glands,ureters,urinary bladder,part of the
oesophagus,rectum,ovaries,uterus,aorta and caudal vena cava..
short answer List and describe two advances during the Industrial Re.pdf
short answer List and describe two advances during the Industrial Revolution that changed how
businesses delivered goods and services during the 18th and 19th centuries. What management
tactics did John D. Rockefeller employ to gain control of the country\'s oil industry? Give an
example of Theory X management style. Give an example of Theory Y management style.
What prompted new challenges for management? What does a country develop during an
industrial revolution?
Solution
19. innovative technology, rapid growth
20. concentrative tactics
21. people are not interested to work- autocratic management style is fit
22. people are interested and self motivated. consultative model
23. high competiton
24. its production and output will increase, and it generates employment..
SheilaSalmonella agar is _ because _ breaks down amino acids that co.pdf
Sheila/Salmonella agar is _ because _ breaks down amino acids that contain _, such as _ and_.
This releases _ which reacts with _ to produce a _ color.
Solution
Shigella / salmonella agar is differential media because bacteria beaks down amino acids that
contains sulfurgroup such as methionine and cystiene.This releases H2S which reacts with
indicator ferric ammonium sulfate to produce a black colour.
Security Challenges Please respond to the followingDistributed .pdf
Security Challenges\" Please respond to the following:
Distributed applications and cloud computing have become a viable option within the LAN-to-
WAN Domain. Discuss the primary challenges related to maintaining the security of both
applications and data in such an environment.
Describe the controls that are needed to ensure the security of distributed applications and data.
Identify the auditing approach that you believe is best suited to this arrangement. Provide a
rationale for your response.
No plagiarism please.
Solution
Challenges in Distributed applications :
1) Computational problems:-
Those kind of problems which cannot be resolved within a specific amount of time on a single
computer.
2) Scalability:-
The system has to remain effective enough so that it can handle the increase in the number of
resources and users.
3) Security:-
In Distributed applications, the possible threats could be the information leakage, integrity
violation, DoS(Denial Of Services)
, illegitimate usage of the system.
4) Concurrency:-
The shared access to the resources should be handled properly.
Challenges in Cloud computing :
1) Security & Privacy:-
The fact that the organization\'s data will reside outside the organization\'s firewall really raises
a serious concern.
Various types of attacks including hacking may affect multiple clients even if only one site is
attacked.These risks has to be mitigated by using proper
security applications, encrypted file-systems, using a proper storage devices to track unsual
activity across the servers .
2) Loss of control over sensitive data:-
The sensitive data that are stored in the servers has to be properly controlled.
3) Reliability and Availability:-
The frequent outages has to be avoided here. It is very important to monitor the activity of
servers using monitoring tool.
The controls that are essential to prevent security threats in distributed applications and data
include:
1) The control should protect data and integrity of system in case of invalid operation on secured
data:-
In case, a user tries to enter invalid data into the system, the control should protect the system
from such operations.
2) Unauthorized invocations should be rejected by security control.
3) Efficient protection against unauthorized users:-
Unauthorized access request by non-system users should be rejected by the control.
The auditing approach should firstly start by ensuring data integrity that is associated with
application. Control should specify the operation that can be invoked only by authorized person
while rejecting access to unauthorized users when the data resources are accessed. Irrespective of
operation, control should
select only that user which can be allowed access while restricting others non-system users.
Rules should be strictly followed by control and on the basis of defined rules, users access
should be granted or denied..
Sapling Learning There are several methods by which genetic material .pdf
Sapling Learning There are several methods by which genetic material can be exchanged
between bacteria of the following terms describes insertion of synthetic into a Which O
transduction O transformation O conjugation O transfection o Previous 8 Give up & View
Solution
Check Answer 0 Next HExt Hint.
Question 5 Checkboxes,Radio Buttons, and JTextAreaWrite a program.pdf
Question 5: Checkboxes,Radio Buttons, and JTextArea
Write a program to allow the user to enter their name (using a text field), age (using a text field),
gender (using radio buttons), and whether they are a student (using a check box). Include a
submit button. When the user clicks submit, display a summary of the information in a text area.
Solution
/*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package example;
import javafx.application.Application;
import javafx.beans.value.ChangeListener;
import javafx.beans.value.ObservableValue;
import javafx.event.ActionEvent;
import javafx.event.EventHandler;
import javafx.geometry.Pos;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.control.CheckBox;
import javafx.scene.control.Label;
import javafx.scene.control.RadioButton;
import javafx.scene.control.TextArea;
import javafx.scene.control.TextField;
import javafx.scene.control.Toggle;
import javafx.scene.control.ToggleGroup;
import javafx.scene.layout.GridPane;
import javafx.scene.layout.StackPane;
import javafx.stage.Stage;
/**
*
* @author MARIYABABU
*/
public class Example extends Application {
@Override
public void start(Stage primaryStage) {
GridPane grid = new GridPane();
grid.setAlignment(Pos.CENTER);
grid.setHgap(10);
grid.setVgap(10);
Label name = new Label(\"your Name:\");
grid.add(name, 0, 0);
TextField tname = new TextField();
grid.add(tname, 1, 0);
Label age = new Label(\"Your age:\");
grid.add(age, 0, 1);
TextField tage = new TextField();
grid.add(tage, 1, 1);
Label gender = new Label(\"Your Gender:\");
grid.add(gender, 0, 2);
ToggleGroup pickGender = new ToggleGroup();
RadioButton pickMale = new RadioButton(\"Male\");
pickMale.setToggleGroup(pickGender);
pickMale.setSelected(true);
grid.add(pickMale,1,2);
RadioButton pickFemale = new RadioButton(\"Female\");
pickFemale.setToggleGroup(pickGender);
grid.add(pickFemale,2, 2);
Label std = new Label(\"Are you a Student?:\");
grid.add(std, 0, 3);
CheckBox cb1 = new CheckBox(\"Student\");
grid.add(cb1,1,3);
CheckBox cb2 = new CheckBox(\"Employee\");
grid.add(cb2,2,3);
Label summa = new Label(\"Summary\");
grid.add(summa, 0, 4);
TextArea summary = new TextArea();
grid.add(summary, 0, 5);
Button btn = new Button();
btn.setText(\"GET SUMMARY\");
btn.setOnAction(new EventHandler() {
@Override
public void handle(ActionEvent event) {
summary.setText(
\"Name :\"+tname.getText()+\"\ \"+\"age :\"+tage.getText()+\"\ \");
//summary.appendText(name.getText());
System.out.println(\"I am doing Well\");
}
});
//final ToggleGroup group = new ToggleGroup();
pickGender.selectedToggleProperty().addListener(new ChangeListener(){
@Override
public void changed(ObservableValue ov,
Toggle old_toggle, Toggle new_toggle) {
if (pickGender.getSelectedToggle() != null) {
summary.appendText(\"\ Gender :\"+pickGender.getSelectedToggle().getUserData().toStrin.
1. Find at least two (2) organizations that are available in the sta.pdf
1. Find at least two (2) organizations that are available in the state of Missouri that would address
the following issue.(AIDS) (You need two DIFFERENT organizations for the health issue.)
2., List the name of the organization or agency, the TYPE of organization as listed in your text.
(ie. Governmental,Quasi-governmental, Voluntary, or Service / Social Health Organizations) and
1 or 2 statements of how each organization addresses that Health issue.
Health issue: AIDS
Solution
The Bureau of HIV, STD, and Hepatitis, Missouri Department of Health and Senior Services is a
Governmental organization and has a broad array of programs and services to help meet the
needs of individuals and communities affected by HIV/AIDS. These programs focus on
preventing the spread of HIV, counseling and testing persons at high risk, and caring for persons
living with the disease. Through strong partnerships consisting of federal, state and local
government agencies, and community based organizations and volunteers, the state of Missouri
is actively engaged in ending the HIV epidemic.
AIDS Project of the Ozarks (APO) is a non-profit community-based organization (Voluntary, or
Service) serving a 29 county region in southwest Missouri. APO currently serves over 600
clients with HIV/AIDS and their families. Concerned families and friends established the
organization as a grass roots entity in 1983. Services include: medical care, case management,
education to the general public, and services to persons with HIV infection, their families and
significant others in a confidential, caring environment..
Pros and Cons of these APIs Strava pros cons Gruvr pros co.pdf
Pros and Cons of these APIs Strava: pros: cons: Gruvr: pros: cons: Hi5 pros: cons:
Solution
strava
pros:
1)rich dataset
2)stable interface
cons:
1)shorter rate limit
2)still continiouly changes are made to it.
gruvr
pros
:offers tons of geocoded data
we can come to know about ucoming concerts using bands
cons:
takes more time the development.
PLEASE show all worksteps and expalin fullySolutionnull h.pdf
**PLEASE show all work/steps and expalin fully**
Solution
null hypothesis : miu > 40
alternative hypothesis : miu < 40
type I error : reject Ho when Ho is true
type II error : accept Ho when Ho is false
type I error
reject that miu > 40 , when in fact is true
type II error
accept that miu > 40 , when in fact is false.
Plants, fungi, and animals all have EUKARYOTIC cells. This is due to.pdf
Plants, fungi, and animals all have EUKARYOTIC cells. This is due to:a. Homologyb.
Convergent evolutionc. Oxygen catastrophed. Allopatric speciatione. Both C and D
Solution
Plants, fungi, and animals all have EUKARYOTIC cells. This is due to homology. In this
context, homology is the existence of shared ancestry between organisms.
Hence, the correct option is a. homology..
3. Describe the five basic attacks on authentication systems.4. De.pdf
3. Describe the five basic attacks on authentication systems.
4. Describe how password hashing works
a. Provide screenshots of using md5sum in your group VM under your account
5. Why can’t an attacker masquerade by using a hashed password?
6. Explain how biometric systems are vulnerable to the five generic attacks on authentication
systems.
a. Clone or borrow credential
b. Sniff credential
c. Trial and error guessing
d. Denial of service
e. Retrieve from backup
Solution
1) and 6)
1.Clone or borrow the credential or token - copy of the credential itself and uses it to log in.
2.Sniff the credential - intercepts and copies it while being transmitted to the login process.
3.Trial-and-error guessing - simply guess several words or names.
4.Denial of service - attack either damages the system or blocks access to it by others.
5.Retrieve from backup - find the authentication database or other information on a backup copy
of the hard drive
2)
Hashing is the transformation of a string of characters into a usually shorter fixed-length value
or key that represents the original string. Hashing is used to index and retrieve items in a
database because it is faster to find the item using the shorter hashed key than to find it using the
original value. It is also used in many encryption algorithms
>As a simple example of the using of hashing in databases, a group of people could be arranged
in a database like this:
Abernathy, Sara Epperdingle, Roscoe Moore, Wilfred Smith, David (and many more sorted into
alphabetical order)
>Each of these names would be the key in the database for that person\'s data. A database search
mechanism would first have to start looking character-by-character across the name for matches
until it found the match (or ruled the other entries out). But if each of the names were hashed, it
might be possible (depending on the number of names in the database) to generate a unique four-
digit key for each name
For example:
7864 Abernathy, Sara 9802 Epperdingle, Roscoe 1990 Moore, Wilfred 8822 Smith, David (and
so forth)
>A search for any name would first consist of computing the hash value (using the same hash
function used to store the item) and then comparing for a match using that value. It would, in
general, be much faster to find a match across four digits, each having only 10 possibilities, than
across an unpredictable value length where each character had 26 possibilities.
>The hashing algorithm is called the hash function-- probably the term is derived from the idea
that the resulting hash value can be thought of as a \"mixed up\" version of the represented value..
Need key terms and multiple choice answered Designers created WTLS a.pdf
Need key terms and multiple choice answered Designers created WTLS as a method to ensure
Privacy of data being broadcast over WAP. Key Terms 2.4 GHz band (305) 5 GHz band (308)
bluejacking B06) bluebugging (06) confidentiality 302) direct-sequence spread spectrum (DSSS)
(308) Key Terms Quiz Use terms from the Key Terms list to complete the sentences that follow.
Don\'t use the same term more than once. Not all terms will be used. 1. An AP uses to advertise
its existence to potential wireless clients. 318 IEEE 802.11 (299) initialization vector (IV) 302)
multiplex frequency division (OFDM) 308) RC4 stream cipher (310) service set identifier
(SSID) (309) Wired Equivalent Privacy (wEP) 310) Wireless Application Protocol (wAPo 301
Wireless Transport Layer Security (wTLs) pozy 2. The is the part of the RC4 cipher that has a
weak implementation in WEP is a standard for Ethernet authentication to 4. WAP uses the
attempt to ensure confidentiality of data. Principles of Computer Security CompTIA Secuitye
Solution
1. An AP uses ___ to advertise its existence to potential wireless clients
-> Beacon Frames
2. The ___ is the part of the RC4 cipher that has a weak implementation in WEP
-> Initialization Vector (IV)
3. ___ is a standard for Ethernet authentication.
-> IEEE 802.1x
4. WAP uses the ___ protocol to attempt to ensure confidentiality of data.
-> Wireless Transport Layer Security (WTLS)
5. The 32-character identifier attached to the header of a packet used for authentication to an
802.11 access point is the ___
-> Service Set Identifier (SSID)
6. 802.11g uses frequencies in the ___.
-> 2.4 GHz Band
7. 802.11i updates the flawed security deployed in ___.
-> Wired Equivalent Privacy (WEP)
8. The standard for wireless local area networks is called ___.
-> IEEE 802.11
9. Cryptography in the WEP protocol is an attempt to guarantee ___.
-> Confidentiality
10. 802.11a uses frequencies in the ___.
-> 5 GHz band.
Azure Interview Questions and Answers PDF By ScholarHat
Azure Interview Questions and Answers PDF By ScholarHat
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What is a view What is it used for Give examples when a view can b.pdf
What is a view? What is it used for? Give examples when a view can be used for your project.
Solution
View is way of portraying information in a database by arranging data items in certain order by
highlighting certain items. Often thought of as Virtual table, doesn\'t store data itself but pulls out
from one or more tables.
Views are used according to purpose so that we query on selective data not whole of it. Using
views saves us lot of querying resources.
For example, people query telephone directory to search for Street Address associated with
Phone Number. We can create View which stores 2 columns namely Phone number and Street
Address ignoring all other fields..
^^^Q3. I am trying to implement double linked list but I was faile.pdf
^^^
Q3. I am trying to implement double linked list but I was failed to write the code so anyone gives
the main
Code in the main function
THANK YOU FOR ADVANCE
#include
#include
#include
struct node
{
int info;
struct node *lptr,*rptr;
};
typedef struct node DL;
DL *delete( ) , *insert ( );
void display();
DL *delete(DL *start,int x)
{
DL *left,*right,*curr;
curr = start;
if( start == NULL)
{
printf(\"\ Doubly Linked List is Empty\");
return(start);
}
while( (curr != NULL) && (curr->info != x) )
curr = curr->rptr;
if(curr == NULL)
{
printf(\"%d does not Exist ... Invalid Deletion! \ \",x);
return(start);
}
left = curr->lptr;
right = curr->rptr;
if(left == NULL)
{
start = right;
right->lptr = NULL;
}
else
{
left->rptr = right;
if(right != NULL)
right->lptr = left;
}
printf(\"\ The item %d is Deleted\ \",curr->info);
free(curr);
return(start);
}
void display(DL *start)
{
DL *last,*temp = start;
if( (temp != NULL) )
{
printf(\"\ The Elements of Doubly Linked List from Left to Right \");
printf(\"\ ROOT-> \");
while(temp != NULL )
{
printf(\"%d -> \",temp->info);
last = temp;
temp = temp->rptr;
}
printf(\"NULL\");
printf(\"\ The Elements of Doubly Linked List from Left to Right\");
printf(\"\ ROOT-> \");
while(last != NULL )
{
printf(\"%d -> \",last->info);
last = last->lptr;
}
printf(\"ROOT\ \ \");
}
else
printf(\"\ Empty List\ \");
}
DL *insert(DL *start,int x)
{
DL *new,*left,*temp;
new=(DL *) malloc(sizeof(DL));
new->info = x;
new->lptr = NULL;
new->rptr = NULL;
if( start == NULL)
start = new;
else
{
temp = start;
while(temp->rptr != NULL)
temp = temp->rptr;
temp->rptr = new;
new->lptr = temp;
}
return(start);
}
Main()
{
------------------Write the code here------------------------
}
Solution
main function for the given program.
main()
{
DL *root = NULL;
int item;
char c;
clrscr();
do
{
printf(\"Insert/Display/Remove/Quit <> I/D/R/Q \");
printf(\"\ Enter Choice : \");
fflush(stdin);
c = getchar();
switch(c)
{
case \'i\':
case \'I\': printf(\"Enter The Element you want to be Inserted : \");
scanf(\"%d\",&item);
root=insert(root,item);
break;
case \'d\':
case \'D\': display(root);
break;
case \'r\':
case \'R\':
printf(\"Enter the Item which you want to Remove :\");
scanf(\"%d\",&item);
root = delete(root,item);
break;
case \'Q\':
case \'q\': return;
}
} while(1);
}.
What is the evolutionary basis for sexual reproductionSolution.pdf
What is the evolutionary basis for sexual reproduction?
Solution
There are both asexual and sexual reproduction in nearly all groups of animals and plants, in
neither kingdom sex developed as a more rapid means.
When sexual reproduction first originated is still a question. But it is believed that sex originated
in green algae. There are certain species like Ulothrix have different size of spores are
produced. These spores comes in pair and become one and gives valuable new forms. In some
groups of unicellular plants ,the Peridinea, the Conjugatae and the Diatomeae- there is no asexual
reproduction. These are good example of the actual gametic origin.
The whole process of evolution is understand by these lines that first there was a visible
morphological differentiation of gametes, the one becoming a large inactive cell stored with food
, the other becoming small and mobile. Then came the evolution of various sex organs, and
finally the alternation of generation.
There are 3 main categories of sexual life cycles: diploid-dominant, demonstrated by most
animals; haploid - dominant, demonstrated by all fungi and some algae; and the alternation of
generations , demonstrated by plants and some algae.
The Red Queen Hyothesis-
First proposed by Leigh Van Valen in 1973.
According to this hypothesis are species co-evolved with other organisms like predators evolve
with their prey and parasites evolve with their hosts. Each tiny advantages gained by favorable
variations gives a secies an edge over close competitors, predators, parasites, or even prey.
Species that cannot keep up become extinct.
August Weismann proposed an explanation for the evolution of sex, where the advantages of sex
is the creation of variation among siblings.
A similar hypothesis proposed by Charles Darwin in his book Origin of species..
Why is it that the Grams stain is not appropriate for acid fast ba.pdf
Why is it that the Gram\'s stain is not appropriate for acid fast bacterium?
Solution
During staining in microbiology, microbes are coloured or dyed so that their structure can be
viewed against a lighter background. There are different staining techniques such as direct(Dye
binds to the cell surface and all the microbes are stained alike),differential(based on their specific
cell surface different microbes get stained differently) special (usually a particular organelle is
targeted)and negative staining(back ground is stained instead of microbes) Gram stain is a
differential stain where the bacteria with a thick peptidoglycan layer around the cells are stained
with violet colour and those with an additional lipopolysaccharide layer will be stained pink. To
understand this lets take a look at Gram staining procedure. First of all bacterial culture is
smeared onto a glass slide and is heat fixed so that during the washes in the procedure the
bacteria stay fixed onto the slide. Primary stain crystal violet is added onto the slide and the dye
seperates into ions which penetrate or enter the cell membranes of both Gram positive and
negative cells. Next iodine is added onto the slide which forms huge complexes with crystal
violet. Then a dye remover usually called as decolouriser such as alchohol or acetone is used to
remove the stain. In Gram negative bacteria the outer most membrane is lost during the treatment
with alchohol whereas in the case of gram positive bacteria the dye is trapped inside the cell
membrane. A secondary stain saffronin is used and which gives pink colour to Gram negative
bacteria. Now coming to acid-fast bacteria, they have a compound called mycolic acid in their
cell walls and are resistant to decolourisation after the acid treatment and alchohol and hence
would not lose colour and appear violet. As we can see these organism can be easily confused
with Gram positive bacteria which also do not lose stain not because they are resistant to acids
but they have the crystal violet dye trapped in their cell membrane..
Why does this version of the swap function fail to work Is there a f.pdf
Why does this version of the swap function fail to work? Is there a fix? (Choose 2) oid swap (int
& lhs, int & rhs) Select one or more: It fails because the programmer forgot to make the
parameters call-by-reference. Of course it works ! Just look at it It clearly swaps the two
parameters! It fails OK, and we can fix it we can just reverse the order of the lines. To fix this,
we must save the the value in a local variable before making the first assignment indicated, then
instead of the second Iine, as int local = lhs; Ihs = rhs; rhs = local; It fails because the first line
destroys the old value of ihs without saving it. Then both variables have the old value of rhs in
them.
Solution
Choice d, and e are valid options.
To swap the elements, we need to first copy the value of first variable to temporary variable, then
copy the second variable to first variable, finally copy the value in temporary variable to second
variable.
So,
int local = lhs;
lhs = rhs;
rhs = local;
is the perfect way to swap two values.
The last option is the actual problem with the given code.
When we do lhs = rhs, the old value of lhs is destroyed, and now, both lhs and rhs are holding
the same value, that is the value in rhs. And then, the second operation rhs = lhs, is of no use, as
both of them are holding the same value.
Therefore, options d, and e are correct..
Which of the following is NOT a reason why heat is a bad method to ov.pdf
Which of the following is NOT a reason why heat is a bad method to overcome activation
energy? A. Heat is not specific enough compared to enzymes B. Too much heat would denature
enzymes C. Heat would allow for many enzymes to be activated, not just the ones that are
needed D. Heat does not hold enough energy to overcome most activation energies
Solution
temperature is one the influence factor for enzyme acitvity. In the body of the biochemical
reaction takes place
for energy production. enzymes plays vital role by mediating the reactions. temperature influnece
enzyme acticity. most the enzymes in general active between optimum temperature( 40-45c). in
conditions where body temperature above 50 degree celcius enzymes gets denaturation which
leads to derangment in the tertiary structure. in the become inactive at higher temperature (above
70c) answer B too much heat would denature the enzymes.
What are the trends in browsersSolution1.chrome Google chrom.pdf
What are the trends in browsers?
Solution
1.chrome :Google chrome is a freeware web browser evolved by Google.a Webkit layout is used
upto 27th version with the pecularity of IOS is declared. it is initially liberated in the form of
beta version for the windows in the year 2008 september 2nd.
2.Firefox:Mozilla firefox is shortly called as firefox .it is an open source and also lavish in
nature.moreover it was advanced for windows,linux and osx.firefox is periperal in nature.in
firefox a Gecko layout engine is seen to provide web pages that applies present and predicted
web attainments.
3.Opera:It is evolved by a opera software.the trendy version is obtainable for windows,linux and
osx.Blink layout engine is seen in this opera web browser.versions of opera is obtained for
widgets operating the IOS,Bada and blackberry.
4.Safari:Safari browser is evolved by Apple inc.containing the os x and IOS operarting
systems.initially in the form of public beta in the year 2003 january 7 .it flatters as a Apple,s
default browser starting with MAC OS X v 10.3 \"panther\".the congential browser of IOS is
additionally called as safari.However it had a discrete Graphical user interface and utilizes a
discrete webkit edition..
The mass of a radioactive substance follows a continuous exponential.pdf
The mass of a radioactive substance follows a continuous exponential decay model, with a decay
rate parameter of 5.1% per day. Find the half-life of this substance (that is, the time it takes for
one-half the original amount in a given sample of this substance to decay).
Solution
Half life = log2/ decay constant = 0.693/ decay constant
Decay constant 5.1% = 0.051
So, Half life =0.693 /0.051 = 13.58 days.
The groin is also known as the _____ region. crural inguinal perin.pdf
The groin is also known as the _____ region. crural inguinal perineum popliteal The upper
midportion of the abdomen is called the _____ region. hypochondriac iliac hypogastric
epigastric The thoracic cavity lies where In relationship to the abdominopelvic cavity? dorsal
(posterior) ventral (anterior) superior inferior The urinary bladder is in the _____ region.
inguinal epigastric umbilical hypogastric All of the following are found in the thoracic cavity
EXCEPT the lungs. pleurae. meninges. heart. Which of the following organs is retroperitoneal
in location? stomach liver heart kidney
Solution
1.b) Groin is also calles as inguinal region which is between the abdomen and the thigh.
2.d)epigastric region-It is the upper central region of the abdomen.It is located between the costal
margins and the subcostal plane.
3.c)superior
4.d)hypogastric region-It is a region of the abdomen located below the umbilical region.
5.c)meninges
Thoracic cavity contains trachea,bronchi,lungs,oesophagus,heart and great blood vessels,thymus
gland,lymph nodes,and nerves. It also contains two smaller cavities-pleural cavities surround
each lung and pericardial cavity covers the heart.
6.d)kidney
Retroperitoneal organs include- the kidneys,adrenal glands,ureters,urinary bladder,part of the
oesophagus,rectum,ovaries,uterus,aorta and caudal vena cava..
short answer List and describe two advances during the Industrial Re.pdf
short answer List and describe two advances during the Industrial Revolution that changed how
businesses delivered goods and services during the 18th and 19th centuries. What management
tactics did John D. Rockefeller employ to gain control of the country\'s oil industry? Give an
example of Theory X management style. Give an example of Theory Y management style.
What prompted new challenges for management? What does a country develop during an
industrial revolution?
Solution
19. innovative technology, rapid growth
20. concentrative tactics
21. people are not interested to work- autocratic management style is fit
22. people are interested and self motivated. consultative model
23. high competiton
24. its production and output will increase, and it generates employment..
SheilaSalmonella agar is _ because _ breaks down amino acids that co.pdf
Sheila/Salmonella agar is _ because _ breaks down amino acids that contain _, such as _ and_.
This releases _ which reacts with _ to produce a _ color.
Solution
Shigella / salmonella agar is differential media because bacteria beaks down amino acids that
contains sulfurgroup such as methionine and cystiene.This releases H2S which reacts with
indicator ferric ammonium sulfate to produce a black colour.
Security Challenges Please respond to the followingDistributed .pdf
Security Challenges\" Please respond to the following:
Distributed applications and cloud computing have become a viable option within the LAN-to-
WAN Domain. Discuss the primary challenges related to maintaining the security of both
applications and data in such an environment.
Describe the controls that are needed to ensure the security of distributed applications and data.
Identify the auditing approach that you believe is best suited to this arrangement. Provide a
rationale for your response.
No plagiarism please.
Solution
Challenges in Distributed applications :
1) Computational problems:-
Those kind of problems which cannot be resolved within a specific amount of time on a single
computer.
2) Scalability:-
The system has to remain effective enough so that it can handle the increase in the number of
resources and users.
3) Security:-
In Distributed applications, the possible threats could be the information leakage, integrity
violation, DoS(Denial Of Services)
, illegitimate usage of the system.
4) Concurrency:-
The shared access to the resources should be handled properly.
Challenges in Cloud computing :
1) Security & Privacy:-
The fact that the organization\'s data will reside outside the organization\'s firewall really raises
a serious concern.
Various types of attacks including hacking may affect multiple clients even if only one site is
attacked.These risks has to be mitigated by using proper
security applications, encrypted file-systems, using a proper storage devices to track unsual
activity across the servers .
2) Loss of control over sensitive data:-
The sensitive data that are stored in the servers has to be properly controlled.
3) Reliability and Availability:-
The frequent outages has to be avoided here. It is very important to monitor the activity of
servers using monitoring tool.
The controls that are essential to prevent security threats in distributed applications and data
include:
1) The control should protect data and integrity of system in case of invalid operation on secured
data:-
In case, a user tries to enter invalid data into the system, the control should protect the system
from such operations.
2) Unauthorized invocations should be rejected by security control.
3) Efficient protection against unauthorized users:-
Unauthorized access request by non-system users should be rejected by the control.
The auditing approach should firstly start by ensuring data integrity that is associated with
application. Control should specify the operation that can be invoked only by authorized person
while rejecting access to unauthorized users when the data resources are accessed. Irrespective of
operation, control should
select only that user which can be allowed access while restricting others non-system users.
Rules should be strictly followed by control and on the basis of defined rules, users access
should be granted or denied..
Sapling Learning There are several methods by which genetic material .pdf
Sapling Learning There are several methods by which genetic material can be exchanged
between bacteria of the following terms describes insertion of synthetic into a Which O
transduction O transformation O conjugation O transfection o Previous 8 Give up & View
Solution
Check Answer 0 Next HExt Hint.
Question 5 Checkboxes,Radio Buttons, and JTextAreaWrite a program.pdf
Question 5: Checkboxes,Radio Buttons, and JTextArea
Write a program to allow the user to enter their name (using a text field), age (using a text field),
gender (using radio buttons), and whether they are a student (using a check box). Include a
submit button. When the user clicks submit, display a summary of the information in a text area.
Solution
/*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package example;
import javafx.application.Application;
import javafx.beans.value.ChangeListener;
import javafx.beans.value.ObservableValue;
import javafx.event.ActionEvent;
import javafx.event.EventHandler;
import javafx.geometry.Pos;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.control.CheckBox;
import javafx.scene.control.Label;
import javafx.scene.control.RadioButton;
import javafx.scene.control.TextArea;
import javafx.scene.control.TextField;
import javafx.scene.control.Toggle;
import javafx.scene.control.ToggleGroup;
import javafx.scene.layout.GridPane;
import javafx.scene.layout.StackPane;
import javafx.stage.Stage;
/**
*
* @author MARIYABABU
*/
public class Example extends Application {
@Override
public void start(Stage primaryStage) {
GridPane grid = new GridPane();
grid.setAlignment(Pos.CENTER);
grid.setHgap(10);
grid.setVgap(10);
Label name = new Label(\"your Name:\");
grid.add(name, 0, 0);
TextField tname = new TextField();
grid.add(tname, 1, 0);
Label age = new Label(\"Your age:\");
grid.add(age, 0, 1);
TextField tage = new TextField();
grid.add(tage, 1, 1);
Label gender = new Label(\"Your Gender:\");
grid.add(gender, 0, 2);
ToggleGroup pickGender = new ToggleGroup();
RadioButton pickMale = new RadioButton(\"Male\");
pickMale.setToggleGroup(pickGender);
pickMale.setSelected(true);
grid.add(pickMale,1,2);
RadioButton pickFemale = new RadioButton(\"Female\");
pickFemale.setToggleGroup(pickGender);
grid.add(pickFemale,2, 2);
Label std = new Label(\"Are you a Student?:\");
grid.add(std, 0, 3);
CheckBox cb1 = new CheckBox(\"Student\");
grid.add(cb1,1,3);
CheckBox cb2 = new CheckBox(\"Employee\");
grid.add(cb2,2,3);
Label summa = new Label(\"Summary\");
grid.add(summa, 0, 4);
TextArea summary = new TextArea();
grid.add(summary, 0, 5);
Button btn = new Button();
btn.setText(\"GET SUMMARY\");
btn.setOnAction(new EventHandler() {
@Override
public void handle(ActionEvent event) {
summary.setText(
\"Name :\"+tname.getText()+\"\ \"+\"age :\"+tage.getText()+\"\ \");
//summary.appendText(name.getText());
System.out.println(\"I am doing Well\");
}
});
//final ToggleGroup group = new ToggleGroup();
pickGender.selectedToggleProperty().addListener(new ChangeListener(){
@Override
public void changed(ObservableValue ov,
Toggle old_toggle, Toggle new_toggle) {
if (pickGender.getSelectedToggle() != null) {
summary.appendText(\"\ Gender :\"+pickGender.getSelectedToggle().getUserData().toStrin.
1. Find at least two (2) organizations that are available in the sta.pdf
1. Find at least two (2) organizations that are available in the state of Missouri that would address
the following issue.(AIDS) (You need two DIFFERENT organizations for the health issue.)
2., List the name of the organization or agency, the TYPE of organization as listed in your text.
(ie. Governmental,Quasi-governmental, Voluntary, or Service / Social Health Organizations) and
1 or 2 statements of how each organization addresses that Health issue.
Health issue: AIDS
Solution
The Bureau of HIV, STD, and Hepatitis, Missouri Department of Health and Senior Services is a
Governmental organization and has a broad array of programs and services to help meet the
needs of individuals and communities affected by HIV/AIDS. These programs focus on
preventing the spread of HIV, counseling and testing persons at high risk, and caring for persons
living with the disease. Through strong partnerships consisting of federal, state and local
government agencies, and community based organizations and volunteers, the state of Missouri
is actively engaged in ending the HIV epidemic.
AIDS Project of the Ozarks (APO) is a non-profit community-based organization (Voluntary, or
Service) serving a 29 county region in southwest Missouri. APO currently serves over 600
clients with HIV/AIDS and their families. Concerned families and friends established the
organization as a grass roots entity in 1983. Services include: medical care, case management,
education to the general public, and services to persons with HIV infection, their families and
significant others in a confidential, caring environment..
Pros and Cons of these APIs Strava pros cons Gruvr pros co.pdf
Pros and Cons of these APIs Strava: pros: cons: Gruvr: pros: cons: Hi5 pros: cons:
Solution
strava
pros:
1)rich dataset
2)stable interface
cons:
1)shorter rate limit
2)still continiouly changes are made to it.
gruvr
pros
:offers tons of geocoded data
we can come to know about ucoming concerts using bands
cons:
takes more time the development.
PLEASE show all worksteps and expalin fullySolutionnull h.pdf
**PLEASE show all work/steps and expalin fully**
Solution
null hypothesis : miu > 40
alternative hypothesis : miu < 40
type I error : reject Ho when Ho is true
type II error : accept Ho when Ho is false
type I error
reject that miu > 40 , when in fact is true
type II error
accept that miu > 40 , when in fact is false.
Plants, fungi, and animals all have EUKARYOTIC cells. This is due to.pdf
Plants, fungi, and animals all have EUKARYOTIC cells. This is due to:a. Homologyb.
Convergent evolutionc. Oxygen catastrophed. Allopatric speciatione. Both C and D
Solution
Plants, fungi, and animals all have EUKARYOTIC cells. This is due to homology. In this
context, homology is the existence of shared ancestry between organisms.
Hence, the correct option is a. homology..
3. Describe the five basic attacks on authentication systems.4. De.pdf
3. Describe the five basic attacks on authentication systems.
4. Describe how password hashing works
a. Provide screenshots of using md5sum in your group VM under your account
5. Why can’t an attacker masquerade by using a hashed password?
6. Explain how biometric systems are vulnerable to the five generic attacks on authentication
systems.
a. Clone or borrow credential
b. Sniff credential
c. Trial and error guessing
d. Denial of service
e. Retrieve from backup
Solution
1) and 6)
1.Clone or borrow the credential or token - copy of the credential itself and uses it to log in.
2.Sniff the credential - intercepts and copies it while being transmitted to the login process.
3.Trial-and-error guessing - simply guess several words or names.
4.Denial of service - attack either damages the system or blocks access to it by others.
5.Retrieve from backup - find the authentication database or other information on a backup copy
of the hard drive
2)
Hashing is the transformation of a string of characters into a usually shorter fixed-length value
or key that represents the original string. Hashing is used to index and retrieve items in a
database because it is faster to find the item using the shorter hashed key than to find it using the
original value. It is also used in many encryption algorithms
>As a simple example of the using of hashing in databases, a group of people could be arranged
in a database like this:
Abernathy, Sara Epperdingle, Roscoe Moore, Wilfred Smith, David (and many more sorted into
alphabetical order)
>Each of these names would be the key in the database for that person\'s data. A database search
mechanism would first have to start looking character-by-character across the name for matches
until it found the match (or ruled the other entries out). But if each of the names were hashed, it
might be possible (depending on the number of names in the database) to generate a unique four-
digit key for each name
For example:
7864 Abernathy, Sara 9802 Epperdingle, Roscoe 1990 Moore, Wilfred 8822 Smith, David (and
so forth)
>A search for any name would first consist of computing the hash value (using the same hash
function used to store the item) and then comparing for a match using that value. It would, in
general, be much faster to find a match across four digits, each having only 10 possibilities, than
across an unpredictable value length where each character had 26 possibilities.
>The hashing algorithm is called the hash function-- probably the term is derived from the idea
that the resulting hash value can be thought of as a \"mixed up\" version of the represented value..
Need key terms and multiple choice answered Designers created WTLS a.pdf
Need key terms and multiple choice answered Designers created WTLS as a method to ensure
Privacy of data being broadcast over WAP. Key Terms 2.4 GHz band (305) 5 GHz band (308)
bluejacking B06) bluebugging (06) confidentiality 302) direct-sequence spread spectrum (DSSS)
(308) Key Terms Quiz Use terms from the Key Terms list to complete the sentences that follow.
Don\'t use the same term more than once. Not all terms will be used. 1. An AP uses to advertise
its existence to potential wireless clients. 318 IEEE 802.11 (299) initialization vector (IV) 302)
multiplex frequency division (OFDM) 308) RC4 stream cipher (310) service set identifier
(SSID) (309) Wired Equivalent Privacy (wEP) 310) Wireless Application Protocol (wAPo 301
Wireless Transport Layer Security (wTLs) pozy 2. The is the part of the RC4 cipher that has a
weak implementation in WEP is a standard for Ethernet authentication to 4. WAP uses the
attempt to ensure confidentiality of data. Principles of Computer Security CompTIA Secuitye
Solution
1. An AP uses ___ to advertise its existence to potential wireless clients
-> Beacon Frames
2. The ___ is the part of the RC4 cipher that has a weak implementation in WEP
-> Initialization Vector (IV)
3. ___ is a standard for Ethernet authentication.
-> IEEE 802.1x
4. WAP uses the ___ protocol to attempt to ensure confidentiality of data.
-> Wireless Transport Layer Security (WTLS)
5. The 32-character identifier attached to the header of a packet used for authentication to an
802.11 access point is the ___
-> Service Set Identifier (SSID)
6. 802.11g uses frequencies in the ___.
-> 2.4 GHz Band
7. 802.11i updates the flawed security deployed in ___.
-> Wired Equivalent Privacy (WEP)
8. The standard for wireless local area networks is called ___.
-> IEEE 802.11
9. Cryptography in the WEP protocol is an attempt to guarantee ___.
-> Confidentiality
10. 802.11a uses frequencies in the ___.
-> 5 GHz band.
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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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