Create Student Project Database with insert, select, update, and del.pdf
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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;.
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/METHOD=SSTYPE(3)
/INTERCEPT=INCLUDE
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Notes
Output Created
23-Apr-2014 11:05:27
Comments
Input
Data
/Users/dan2088873/Downloads/Project 2 Part 1 (Lab 10) Data File.sav
Active Dataset
DataSet1
Filter
<none>
Weight
<none>
Split File
<none>
N of Rows in Working Data File
40
Missing Value Handling
Definition of Missing
User-defined missing values are treated as missing.
Cases Used
Statistics are based on all cases with valid data for all variables in the model.
Syntax
UNIANOVA class BY trouble
/METHOD=SSTYPE(3)
/INTERCEPT=INCLUDE
/POSTHOC=trouble(TUKEY QREGW C)
/EMMEANS=TABLES(trouble)
/PRINT=ETASQ HOMOGENEITY DESCRIPTIVE
/CRITERIA=ALPHA(.05)
/DESIGN=trouble.
Resources
Processor Time
00 00:00:00.006
Elapsed Time
00 00:00:00.000
[DataSet1] /Users/dan2088873/Downloads/Project 2 Part 1 (Lab 10) Data File.sav
Warnings
Post hoc tests are not performed for Number of times sent to office because at least one group has fewer than two cases.
Between-Subjects Factors
N
Number of times sent to office
3
1
4
4
5
6
6
5
7
9
8
3
9
5
10
1
11
2
12
1
13
1
15
1
17
1
Descriptive Statistics
Dependent Variable:Type of classroom management strategy
Number of times sent to office
Mean
Std. Deviation
N
3
3.00
.
1
4
1.50
1.000
4
5
2.00
.632
6
6
1.60
.548
5
7
1.78
.667
9
8
2.00
.000
3
9
2.00
.000
5
10
3.00
.
1
11
2.50
.707
2
12
3.00
.
1
13
3.00
.
1
15
3.00
.
1
17
3.00
.
1
Total
2.03
.698
40
Levene's Test of Equality of Error Variancesa
Dependent Variable:Type of classroom management strategy
F
df1
df2
Sig.
1.873
12
27
.086
Tests the null hypothesis that the error variance of the dependent variable is equal across groups.
a. Design: Intercept + trouble
Tests of Between-Subjects Effects
Dependent Variable:Type of classroom management strategy
Source
Type III Sum of Squares
df
Mean Square
F
Sig.
Partial Eta Squared
Corrected Model
8.719a
12
.727
1.913
.079
.460
Intercept
126.859
1
126.859
333.984
.000
.925
trouble
8.719
12
.727
1.913
.079
.460
Error
10.256
27
.380
Total
183.000
40
Corrected Total
18.975
39
a. R Squared = .460 (Adjusted R Squared = .219)
Estimated Marginal Means
Number of times sent to office
Dependent Variable:Type of classroom management strategy
Number of times sent to office
Mean
Std. Error
95% Confidence Interval
Lower Bound
Upper Bound
3
3.000
.616
1.735
4.265
4
1.500
.308
.868
2.132
5
2.000
.252
1.484
2.516
6
1.600
.276
1.034
2.166
7
1.778
.205
1.356
2.199
8
2.000
.356
1.270
2.730
9
2.000
.276
1.434
2.566
10
3.000
.616
1.735
4.265
11
2.500
.436
1.606
3.394
12
3.000
.616
1.735
4.265
13
3.000
.616
1.735
4.265
15
3.000
.616
1.735
4.265
17
3.000
.616
1.735
4.265
GET
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This document is a lab manual for the course "Analysis and Design of Algorithms" at Amity University Haryana. It contains 11 experiments on algorithms including binary search, merge sort, quicksort, knapsack problem, and graph algorithms like Kruskal, Prim's, BFS and DFS. The introduction provides an overview of algorithm design and analysis. The experiments provide code snippets in C/C++ to implement the given algorithms.
briefly write about the Aridity of Subtropics Causes, locations, etc.pdf
briefly write about the Aridity of Subtropics Causes, locations, etc.
Solution
The Aridity of Subtropics: the differential air temperature at some specific latitudes makes
warmer weather than & some cold air currents may also trigger aridity & advection fog whereas
some warm air currents may trigger more air temperature
The main cause of aridity is due to the formation of desert weather through \"water deficiency\"
& dry climate. These causes are going to form sub-topical desert (BWh) and steppe (BSh) as
semi-arid locations in both tropical & temperate regions. The other causes are elevated
temperatures with cloudless sky and locations are Chile\'s Atacama Desert or Middle-Latitude
Desert.
Ocean conveyor, El Nino Southern Oscillation (ENSO): This effcet further increases the oceanic
atmospheric pressure in Pacific and finally warmer winds associated with much warmer water
leading global climate change..
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.
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,
for groups to have different rank sums and yet have equal or nearly equal medians..
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IDP_Project_2nd_Presentation_PPT Template.ppt
wdq qwdqwd q wert grwgg e rg ert g ert h rtheftyhj rfth sdth ewrt gh ew rger gwer h wret gh qerg e g qe r g qe rg eqrg qe rg eqrgqegqe qerg qer gqerg
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In the context of this assignment on Mongo, queries will be designed and executed on a mongo collection, simple operations on mongo will be executed with python while mapreduce jobs will also be designed and executed on a mongo collection.
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UNIANOVA class BY trouble METHOD=SSTYPE(3) INTERCEPT=INCLU.docx
UNIANOVA class BY trouble
/METHOD=SSTYPE(3)
/INTERCEPT=INCLUDE
/POSTHOC=trouble(TUKEY QREGW C)
/EMMEANS=TABLES(trouble)
/PRINT=ETASQ HOMOGENEITY DESCRIPTIVE
/CRITERIA=ALPHA(.05)
/DESIGN=trouble.
Univariate Analysis of Variance
Notes
Output Created
23-Apr-2014 11:05:27
Comments
Input
Data
/Users/dan2088873/Downloads/Project 2 Part 1 (Lab 10) Data File.sav
Active Dataset
DataSet1
Filter
<none>
Weight
<none>
Split File
<none>
N of Rows in Working Data File
40
Missing Value Handling
Definition of Missing
User-defined missing values are treated as missing.
Cases Used
Statistics are based on all cases with valid data for all variables in the model.
Syntax
UNIANOVA class BY trouble
/METHOD=SSTYPE(3)
/INTERCEPT=INCLUDE
/POSTHOC=trouble(TUKEY QREGW C)
/EMMEANS=TABLES(trouble)
/PRINT=ETASQ HOMOGENEITY DESCRIPTIVE
/CRITERIA=ALPHA(.05)
/DESIGN=trouble.
Resources
Processor Time
00 00:00:00.006
Elapsed Time
00 00:00:00.000
[DataSet1] /Users/dan2088873/Downloads/Project 2 Part 1 (Lab 10) Data File.sav
Warnings
Post hoc tests are not performed for Number of times sent to office because at least one group has fewer than two cases.
Between-Subjects Factors
N
Number of times sent to office
3
1
4
4
5
6
6
5
7
9
8
3
9
5
10
1
11
2
12
1
13
1
15
1
17
1
Descriptive Statistics
Dependent Variable:Type of classroom management strategy
Number of times sent to office
Mean
Std. Deviation
N
3
3.00
.
1
4
1.50
1.000
4
5
2.00
.632
6
6
1.60
.548
5
7
1.78
.667
9
8
2.00
.000
3
9
2.00
.000
5
10
3.00
.
1
11
2.50
.707
2
12
3.00
.
1
13
3.00
.
1
15
3.00
.
1
17
3.00
.
1
Total
2.03
.698
40
Levene's Test of Equality of Error Variancesa
Dependent Variable:Type of classroom management strategy
F
df1
df2
Sig.
1.873
12
27
.086
Tests the null hypothesis that the error variance of the dependent variable is equal across groups.
a. Design: Intercept + trouble
Tests of Between-Subjects Effects
Dependent Variable:Type of classroom management strategy
Source
Type III Sum of Squares
df
Mean Square
F
Sig.
Partial Eta Squared
Corrected Model
8.719a
12
.727
1.913
.079
.460
Intercept
126.859
1
126.859
333.984
.000
.925
trouble
8.719
12
.727
1.913
.079
.460
Error
10.256
27
.380
Total
183.000
40
Corrected Total
18.975
39
a. R Squared = .460 (Adjusted R Squared = .219)
Estimated Marginal Means
Number of times sent to office
Dependent Variable:Type of classroom management strategy
Number of times sent to office
Mean
Std. Error
95% Confidence Interval
Lower Bound
Upper Bound
3
3.000
.616
1.735
4.265
4
1.500
.308
.868
2.132
5
2.000
.252
1.484
2.516
6
1.600
.276
1.034
2.166
7
1.778
.205
1.356
2.199
8
2.000
.356
1.270
2.730
9
2.000
.276
1.434
2.566
10
3.000
.616
1.735
4.265
11
2.500
.436
1.606
3.394
12
3.000
.616
1.735
4.265
13
3.000
.616
1.735
4.265
15
3.000
.616
1.735
4.265
17
3.000
.616
1.735
4.265
GET
FILE='/Users/dan2088873/Downloads/Project 2 Part 2 (Lab 9) Data File.sav'.
Warning. Command name: GET FILE
SPSS S ...
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Running Head STUDENT RECORD KEEPING SYSTEM DATABASE PROJECT 1.docx
Running Head: STUDENT RECORD KEEPING SYSTEM DATABASE PROJECT 1
STUDENT RECORD KEEPING SYSTEM DATABASE PROJECT 15
Student Record Keeping System Database Project
Professor’s Name
Student’s Name
Course Title
Date
Project Description
In any learning environment like most learning institutions, their success in academic endeavors depends on the ability of the institution to acquire accurate as well as up to date information regarding its daily activities. This will specifically help the organizational management in effectively managing as well as the use of the available data in analyzing as well as guiding its activities. The major important data that a learning organization must take good care of is the information regarding the students. The students being the main customers or stakeholders of learning institutions, their data which ranges from the student’s background information, the course being undertaken by the student, the financial status of the student besides the beginning and ending years of their study must be well kept and in an organized manner. Therefore the main and specific objective of this project is to create or develop a student database system that will assist in storing the students’ records, hence reducing the duration spent on managerial issues (Al-Dmour, 2016).
According to the architecture of the database, it will mainly accept, process, besides generating the students' reports which will encompass the grades and transcripts respectively. The provision of better services to the registrar, provision of meaningful information, keeping consistency, timely information as well as efficiency in the conversion of paperwork to electronic form are also some of the areas where this “Students Record-Keeping System Database” will assist (Frantiska, 2018).
This “Students Record-Keeping System Database” will be developed by the use of basic technologies which are Microsoft access, the embedded visual basic, as well as the graffiti creator. It must be noted that this system will be free of errors with 99% efficiency, note unless the error is caused by the user like errors of omission. It will also be less time consuming as a result of the great care that will be taken when developing the system. It must be noted also that all the database system creation phases will be employed while developing the “Students Record-Keeping System Database”, and therefore, it is with no doubt that this database system will be a user-friendly one besides being strong (Al-Btoush, 2015). The system will also have some provisions for future development of the system if need be.
The main features of the systems will be; the ability to deal with all details of the students starting from the first day to the last day of the student’s his or her course. This can be specifically be also utilized for all the reporting purposes, tracking for attendance, progress in the course, the completed semester years, coming year curriculum details, t ...
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This document is a lab manual for the course "Analysis and Design of Algorithms" at Amity University Haryana. It contains 11 experiments on algorithms including binary search, merge sort, quicksort, knapsack problem, and graph algorithms like Kruskal, Prim's, BFS and DFS. The introduction provides an overview of algorithm design and analysis. The experiments provide code snippets in C/C++ to implement the given algorithms.
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briefly write about the Aridity of Subtropics Causes, locations, etc.pdf
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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.
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Entrepreneurs :
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the market and take the opportunity to make profits satisfying that needs.Intiating a process that
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Attributes of a entrepreneurs
Types of Entrepreneurs :
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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.
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
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.
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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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^^^
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 probability of a persons being left-handed is .12. In the 1992.pdf
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].
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..
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Create Student Project Database with insert, select, update, and del.pdf
- 1. 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) )
- 2. 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");
- 3. INSERT INTO StudentProject VALUES ('AB7',XY1,"Manager"); INSERT INTO StudentProject VALUES ('AB8',XY2,"Analyst"); INSERT INTO StudentProject VALUES ('AB9',XY0,"Programmer"); SELECT * FROM StudentProject;