pH = -log[H+] pH = -log(0.5) pH = 0.301 .pdf
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The document shows a calculation to determine the pH of a solution with a hydrogen ion concentration of 0.5 M. Taking the negative logarithm of 0.5 M hydrogen ions gives a pH value of 0.301 for the solution.
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The least soluble gas is N2 because it is inert and cannot form any .pdf
The least soluble gas is N2 because it is inert and cannot form any strong bonds with water.
The other compounds will react with water and dissolve:
CO2 + H2O => H2CO3
SO3 + H2O => H2SO4
NH3 + H2O => NH4+ + OH-
HCl + H2O => H3O+ + OH-
Solution
The least soluble gas is N2 because it is inert and cannot form any strong bonds with water.
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NH3 + H2O => NH4+ + OH-
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The below stated are the Challenges and business requirements faced by the hospital
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Solution
:
Inorder traversal: It will consider left child root, and right child.
Inorder Sequence: DHBEIAJFCGK
Explanation:
Root node is A.
left child of A is B.
It has children. So move B into stack.
stack |B|
Travel to the left child of B.It is D.
D has children. So move D into stack.
stack |D|B|
Travel to the left child of D. There is no left
child for D.
So traverse Back. pop the stack and print.
Sequence: D
stack |B|
move to the right child of D. It is H. it is a leaf.
So print it.
Sequence: DH
So traverse Back. pop the stack and print.
Sequence: DHB
stack empty
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The least soluble gas is N2 because it is inert and cannot form any strong bonds with water.
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CO2 + H2O => H2CO3
SO3 + H2O => H2SO4
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Solution
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Inorder traversal: It will consider left child root, and right child.
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Explanation:
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left child of A is B.
It has children. So move B into stack.
stack |B|
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D has children. So move D into stack.
stack |D|B|
Travel to the left child of D. There is no left
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So traverse Back. pop the stack and print.
Sequence: D
stack |B|
move to the right child of D. It is H. it is a leaf.
So print it.
Sequence: DH
So traverse Back. pop the stack and print.
Sequence: DHB
stack empty
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Form1.csusing System; using System.Collections.Generic; using .pdf
Form1.cs
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;
namespace LotteryGUI
{
public partial class Form1 : Form
{
private int[] iGuesses = new int[3]; // declared at form level
private int[] iSortedGuesses = new int[3];
public Form1()
{
InitializeComponent();
}
private void textBox1_Leave(object sender, EventArgs e)
{
if (textBox1.Text == \"\")
{
MessageBox.Show(\"You haven\'t chosen your first number yet\");
textBox1.Focus();
return;
}
int iGuess;
int.TryParse(textBox1.Text, out iGuess);
if (iGuess < 1 || iGuess > 4)
{
MessageBox.Show(\"You have entered an incorrect number please try again\");
textBox1.Focus();
return;
}
iGuesses[0] = iGuess;
iSortedGuesses[0] = iGuess;
}
private void textBox2_Leave(object sender, EventArgs e)
{
if (textBox2.Text == \"\")
{
MessageBox.Show(\"You haven\'t chosen your second number yet\");
textBox2.Focus();
return;
}
int iGuess;
int.TryParse(textBox2.Text, out iGuess);
if (iGuess < 1 || iGuess > 4)
{
MessageBox.Show(\"You have entered an incorrect number please try again\");
textBox2.Focus();
return;
}
iGuesses[1] = iGuess;
iSortedGuesses[1] = iGuess;
}
private void textBox3_Leave(object sender, EventArgs e)
{
if (textBox3.Text == \"\")
{
MessageBox.Show(\"You haven\'t chosen your third number yet\");
textBox3.Focus();
return;
}
int iGuess;
int.TryParse(textBox3.Text, out iGuess);
if (iGuess < 1 || iGuess > 4)
{
MessageBox.Show(\"You have entered an incorrect number please try again\");
textBox3.Focus();
return;
}
iGuesses[2] = iGuess;
iSortedGuesses[2] = iGuess;
}
private void button1_Click(object sender, EventArgs e)
{
if (textBox1.Text == \"\" || textBox2.Text == \"\" || textBox3.Text == \"\")
{
MessageBox.Show(\"You haven\'t yet chosen three numbers\");
return;
}
// declaring variables integer type
int iNum1;
int iNum2;
int iNum3;
// declaring the constants
const int iMATCHONE = 10;
const int iMATCHTWO = 100;
const int iMATCHTHREE = 1000;
const int iMATCHFOUR = 10000;
// declaring array
int[] iSortedNums = new int[3];
// generating random numbers
Random randomnumber = new Random();
iNum1 = randomnumber.Next(1, 5); // number between 1 and 4
iNum2 = randomnumber.Next(1, 5);
iNum3 = randomnumber.Next(1, 5);
iSortedNums[0] = iNum1;
iSortedNums[1] = iNum2;
iSortedNums[2] = iNum3;
Array.Sort(iSortedNums); // sort random numbers
Array.Sort(iSortedGuesses); // sort the guesses
label4.Text = String.Format(\"The random numbers are : \" + iNum1 + \", \" + iNum2
+ \", \" + iNum3);
label5.Text = String.Format(\"The numbers you chose are : \" + iGuesses[0] + \", \" +
iGuesses[1] + \", \" + iGuesses[2]);
if (iGuesses[0] == iNum1 && iGuesses[1] == iNum2 && iGuesses[2] == iNum3)
{
MessageBox.Show(\"you won $\" + iMATCHFOUR);
}
else if (iSortedGuesses[0] == iSortedNums[0] && iSortedGuesses[1] ==
iSortedNums[1]
&& iSortedGuesses[2] == iSortedNums[2])
{
Message.
Influenza A The recently raised awareness of th.pdf
Influenza A: The recently raised awareness of the threat of a new influenza
pandemic has stimulated interest in the detection of influenza A viruses in human as well as
animal secretions. Virus isolation alone is unsatisfactory for this purpose because of its inherent
limited sensitivity and the lack of host cells that are universally permissive to all influenza A
viruses. Previously described PCR methods are more sensitive but are targeted predominantly at
virus strains currently circulating in humans, since the sequences of the primer sets display
considerable numbers of mismatches to the sequences of animal influenza A viruses. Therefore,
a new set of primers, based on highly conserved regions of the matrix gene, was designed for
single-tube reverse transcription-PCR for the detection of influenza A viruses from multiple
species. This PCR proved to be fully reactive with a panel of 25 genetically diverse virus isolates
that were obtained from birds, humans, pigs, horses, and seals and that included all known
subtypes of influenza A virus. It was not reactive with the 11 other RNA viruses tested.
Comparative tests with throat swab samples from humans and fecal and cloacal swab samples
from birds confirmed that the new PCR is faster and up to 100-fold more sensitive than classical
virus isolation procedures. Various Methods of Detection: Specimens. Cloacal swab specimens
were collected from ducks (widgeon [Mareca penelope], gadwall [Mareca strepera], and mallard
[Anas plathyrhynchos]) at a marshaling lake in Lekkerkerk, The Netherlands, and droppings as
well as cloacal swab specimens were collected from geese (greylag goose [Anser anser], white-
fronted goose [Anser albifrons albifrons], barnacle goose [Branta leucopsis], and brent goose
[Branta bernicla]) in Groningen and Eemdijk, The Netherlands, between 1997 and 1999. Cloacal
swab specimens and droppings were collected from shorebirds at Öland, Sweden, in the spring of
1999. Cotton swabs were used for sampling and were subsequently stored in transport medium.
Throat swab specimens collected from humans were also stored in transport medium. The
samples were stored at 4°C for a few days, at 20°C for less than a week, or at 70°C for extended
periods of time. Transport medium consisted of Hanks balanced salt solution supplemented with
10% glycerol, 200 U of penicillin per ml, 200 µg of streptomycin per ml, 100 U of polymyxin B
sulfate per ml, 250 µg of gentamicin per ml, and 50 U of nystatin per ml (all from ICN,
Zoetermeer, The Netherlands). RNA isolation. RNA was isolated with a high pure RNA
isolation kit (Roche Molecular Biochemicals) according to the instructions from the
manufacturer, with minor modifications. A 0.2-ml sample was homogenized by vortexing and
was subsequently lysed with 0.4 ml of lysis-binding buffer to which poly(A) (Roche Molecular
Biochemicals) was added as a carrier to 1 µg/ml. After binding to the column, DNase I digestion,
and washing, the RNA was eluted in 50 µl o.
I believe its named Methyl iodide. You wouldn.pdf
I believe it\'s named Methyl iodide. You wouldn\'t number the C that the Iodine is
on because there is only one carbon. Also i think that because there is only one carbon the base
name is Iodide instead of the longest carbon chain. Hope this helps! Enjoy.
Solution
I believe it\'s named Methyl iodide. You wouldn\'t number the C that the Iodine is
on because there is only one carbon. Also i think that because there is only one carbon the base
name is Iodide instead of the longest carbon chain. Hope this helps! Enjoy..
Data warehousing has quickly evolved into a unique and popular busin.pdf
Data warehousing has quickly evolved into a unique and popular business application class.
Early builders of data warehouses already consider their systems to be key components of their
IT strategy and architecture. Numerous examples can be cited of highly successful data
warehouses developed and deployed for businesses of all sizes and all types. Hardware and
software vendors have quickly developed products and services that specifically target the data
warehousing market. This paper will introduce key concepts surrounding the data warehousing
systems.
What is a data warehouse? A simple answer could be that a data warehouse is managed data
situated after and outside the operational systems. A complete definition requires discussion of
many key attributes of a data warehouse system. Later in Section 2, we will identify these key
attributes and discuss the definition they provide for a data warehouse. Section 3 briefly reviews
the activity against a data warehouse system. Initially in Section 1, however, we will take a brief
tour of the traditions of managing data after it passes through the operational systems and the
types of analysis generated from this historical data.
Evolution of an application class
This section reviews the historical management of the analysis data and the factors that have led
to the evolution of the data warehousing application class.
Traditional approaches to historical data
In reviewing the development of data warehousing, we need to begin with a review of what had
been done with the data before of evolution of data warehouses. Let us first look at how the kind
of data that ends up in today\'s data warehouses had been managed historically.
Throughout the history of systems development, the primary emphasis had been given to the
operational systems and the data they process. It is not practical to keep data in the operational
systems indefinitely; and only as an afterthought was a structure designed for archiving the data
that the operational system has processed. The fundamental requirements of the operational and
analysis systems are different: the operational systems need performance, whereas the analysis
systems need flexibility and broad scope. It has rarely been acceptable to have business analysis
interfere with and degrade performance of the operational systems.
Data from legacy systems
In the 1970s virtually all business system development was done on the IBM mainframe
computers using tools such as Cobol, CICS, IMS, DB2, etc. The 1980s brought in the new mini-
computer platforms such as AS/400 and VAX/VMS. The late eighties and early nineties made
UNIX a popular server platform with the introduction of client/server architecture.
Despite all the changes in the platforms, architectures, tools, and technologies, a remarkably
large number of business applications continue to run in the mainframe environment of the
1970s. By some estimates, more than 70 percent of business data for large corporations still
resi.
balready postedSolutionbalready posted.pdf
This document does not contain any substantive content to summarize. It appears to be a placeholder or note indicating that a solution had already been posted. The document consists of the word "already" repeated twice but does not convey any meaningful information that can be summarized in 3 sentences or less.
Assume the graph is strongly connected.SolutionAssume the grap.pdf
The document discusses a graph that is assumed to be strongly connected in one sentence. It then states "Solution" but provides no further details about the solution.
Half life is the time in which the compound reduc.pdf
Half life is the time in which the compound reduces to half of its concentration. t1/2
=0.693/wavelength
Solution
Half life is the time in which the compound reduces to half of its concentration. t1/2
=0.693/wavelength.
from the given reaction, order NO3=1 overall orde.pdf
from the given reaction, order NO3=1 overall order=2 the reaction is bimolecular
Solution
from the given reaction, order NO3=1 overall order=2 the reaction is bimolecular.
a) What is the role of Trizol in RNA extractionTrizol reagent is .pdf
a) What is the role of Trizol in RNA extraction?
Trizol reagent is used for RNA isolation from cells and tissues. This reagent is used as it helps in
maintaining RNA integrity during tissue homogenization, while at the same time disrupting and
breaking down cells and cell components. The RNA remains only in the aqueous phase(Amrita).
b) What was the purpose of performing a reverse transcription reaction. What were the products
of thesereaction?
Reverse transcription polymerase chain reaction (RT-PCR), a variant of polymerase chain
reaction (PCR), is a technique commonly used in molecular biology to detect RNA expression
incontrast the traditional PCR is used to exponentially amplify target DNA sequences, (Freeman,
1999). The product of this reaction will be RNA
Solution
a) What is the role of Trizol in RNA extraction?
Trizol reagent is used for RNA isolation from cells and tissues. This reagent is used as it helps in
maintaining RNA integrity during tissue homogenization, while at the same time disrupting and
breaking down cells and cell components. The RNA remains only in the aqueous phase(Amrita).
b) What was the purpose of performing a reverse transcription reaction. What were the products
of thesereaction?
Reverse transcription polymerase chain reaction (RT-PCR), a variant of polymerase chain
reaction (PCR), is a technique commonly used in molecular biology to detect RNA expression
incontrast the traditional PCR is used to exponentially amplify target DNA sequences, (Freeman,
1999). The product of this reaction will be RNA.
A) odd numbers caseSequential or linear search is a searching mec.pdf
A) odd numbers case:
Sequential or linear search is a searching mechanism where every element one by one is
searched until the result is found.
where l is list and x is our search item.
Minimum number of sequential search comparisons = 1
Maximum number of sequential search comparisons = N
Average number of sequential search comparisons for successful searches = ½ N
To find the middle element we can use (biggest - smallest + 1)/ 2.
2)EVEN number case:.
if the array has even number of elements then divide the list into two subparts of equal size name
array1 array 2, the element which is immediately after array 1 is considered as mid element. so
the complexity even in this case if n/2.
3)
If we find the element in 1st position then comparisons = 1
2nd positions then 2. 3rd postion 3..... nth position = n so the average no of comparisons in linear
search is (1+2+3+4+5+6.....+n) = n(n+1)/2
Solution
A) odd numbers case:
Sequential or linear search is a searching mechanism where every element one by one is
searched until the result is found.
where l is list and x is our search item.
Minimum number of sequential search comparisons = 1
Maximum number of sequential search comparisons = N
Average number of sequential search comparisons for successful searches = ½ N
To find the middle element we can use (biggest - smallest + 1)/ 2.
2)EVEN number case:.
if the array has even number of elements then divide the list into two subparts of equal size name
array1 array 2, the element which is immediately after array 1 is considered as mid element. so
the complexity even in this case if n/2.
3)
If we find the element in 1st position then comparisons = 1
2nd positions then 2. 3rd postion 3..... nth position = n so the average no of comparisons in linear
search is (1+2+3+4+5+6.....+n) = n(n+1)/2.
#include stdio.h#include stdlib.h#include string.h#inclu.pdf
#include
#include
#include
#include
/*Structure for storing words received from each thread*/
struct return_val{
char wordlist[100][100];
int count[100];
} *arr;
pthread_t *threads;
int lines;
/*Aggregates Frequency Sum of different threads*/
int aggregate(char *a)
{
int sum = 0;
int i=0;
while(1)
{
if(i==lines) break;
int count = 0;
while(1)
{
if(arr[i].count[count] == -1)
{
break;
}
if(strcmp(a,arr[i].wordlist[count]) == 0)
{
sum += arr[i].count[count];
arr[i].count[count] = 0;
}
count++;
}
i++;
}
return sum;
}
/*Function executed by each thread on separate line*/
void *word_count(void* num)
{
int *ln = num;
unsigned int line_number = *ln;
line_number++;
char cmd_p1[9] = \"sed -n \'\\0\";
char cmd_p2[2];
sprintf(cmd_p2,\"%d\",line_number);
char cmd_p3[14] = \"p\' \'file.txt\'\\0\";
char command[100];
command[0] = \'\\0\';
strcat(command,cmd_p1);
strcat(command,cmd_p2);
strcat(command,cmd_p3);
//usleep(line_number);
char cmd[100] = \" | tr [:space:] \'\\\ \' | grep -v \'^\\\\s*$\' | sort | uniq -c | sort\\0\";
strcat(command,cmd);
FILE *in;
in= popen(command, \"r\");
rewind(in);
char buff[50];
int counter = 0;
while(fgets(buff,sizeof(buff),in))
{
char c=\' \';
int i = 0;
int cnt = atoi(buff);
arr[line_number-1].count[counter] = cnt;
while(c!=\'\\0\')
{
c=buff[i];
buff[i]=buff[i+6];
i++;
}
int cnnt = 0;
while(c!=\' \')
{
c = buff[cnnt];
cnnt++;
}
i=0;
while(c!=\'\\0\')
{
c=buff[i];
buff[i]=buff[i+cnnt];
i++;
}
sprintf(arr[line_number-1].wordlist[counter],\"%s\",buff);
counter++;
}
arr[line_number-1].count[counter] = -1;
fclose(in);
return NULL;
}
int main(void)
{
FILE *fp;
fp = fopen(\"file.txt\",\"r\");
char a[200];
lines = 0;
void *status = NULL;
while(fgets(a,sizeof(a),fp) !=NULL)
{
lines++;
}
fclose(fp);
arr = (struct return_val*) calloc(lines, sizeof(struct return_val) * lines);
threads = malloc(sizeof(pthread_t)*lines);
int i;
for(i=0;i= 0)
{
if(arr[i].count[ctr] > 0)
{
strcpy(final[final_ctr],arr[i].wordlist[ctr]);
final_cnt[final_ctr] = arr[i].count[ctr];
arr[i].count[ctr] = 0;
int sum = aggregate(arr[i].wordlist[ctr]);
final_cnt[final_ctr] += sum;
printf(\"%d %s\",final_cnt[final_ctr],final[final_ctr]);
}
ctr++; final_ctr++;
}
i++;
if(i==lines)
{
break;
}
}
free(arr);
return 0;
}
#include
#include
#include
#include
/*Structure for storing words received from each thread*/
struct return_val{
char wordlist[100][100];
int count[100];
} *arr;
pthread_t *threads;
int lines;
/*Aggregates Frequency Sum of different threads*/
int aggregate(char *a)
{
int sum = 0;
int i=0;
while(1)
{
if(i==lines) break;
int count = 0;
while(1)
{
if(arr[i].count[count] == -1)
{
break;
}
if(strcmp(a,arr[i].wordlist[count]) == 0)
{
sum += arr[i].count[count];
arr[i].count[count] = 0;
}
count++;
}
i++;
}
return sum;
}
/*Function executed by each thread on separate line*/
void *word_count(void* num)
{
int *ln = num;
unsigned int line_number = *ln;
line_number++;
char cmd_p1[9] = \"sed -n \'\\0\";
char cmd_p2[2];
sprintf(cmd_p2,\"%d\",line_number);
char cmd_p3[14] .
#include ctype.h #include stdio.h #include string.hstati.pdf
#include
#include
#include
static int read_data(char *buffer, size_t buflen);
static int direction_searching(char arr[18][18], char *str, int len, int m, int n, int dx, int dy);
static int searchData_from(char arr[18][18], char *str, int len, int m, int n);
int main(void)
{
int m, n, len;
char arr[18][18], str[50];
printf(\"Welcome to word search puzzle CTRL-C TO QUIT\ \ \");
FILE *fileData = fopen(\"dataSearch.txt\", \"r\");
if (!fileData)
{
printf(\"Error occured in opening file\ \");
return 1;
}
for (n = 0; n < 18; n++)
{
for (m = 0; m < 18; m++)
{
fscanf(fileData, \" %c\", &arr[n][m]);
arr[n][m] = toupper(arr[n][m]);
}
}
fclose(fileData);
printf(\" \");
for (m = 0; m < 18; m++)
printf(\"%-2d\", m);
printf(\"\ ______________________________________\ \");
for (n = 0; n < 18; n++)
{
printf(\"%-2d|\", n);
for (m = 0; m < 18; m++)
printf(\"%c \", arr[n][m]);
printf(\"\ \");
}
while ((len = read_data(str, sizeof(str))) != EOF)
{
printf(\"Searching for: [%s]\ \", str);
int totalCount = 0;
for (n = 0; n < 18; n++)
{
for (m = 0; m < 18; m++)
{
if (arr[n][m] == (str[0]) && searchData_from(arr, str, len, m, n))
totalCount++;
}
}
printf(\"Found %s %d times\ \", str, totalCount);
}
printf(\"\ \");
return 0;
}
// reading data from file
static int read_data(char *buffer, size_t buflen)
{
printf(\"\ Please enter the word to be searched: \");
if (fgets(buffer, buflen, stdin) == 0)
return EOF;
size_t len = strlen(buffer);
if (buffer[len-1] == \'\ \')
buffer[--len] = \'\\0\';
if (len == 0)
return EOF;
for (size_t i = 0; i < len; i++)
buffer[i] = toupper(buffer[i]);
return len;
}
// searching data from array
static int searchData_from(char arr[18][18], char *str, int len, int m, int n)
{
struct yx { int dy; int dx; } puzzleDirection[] =
{
{ +1, 0 }, { -1, 0 }, { +1, +1 }, { -1, +1 },
{ 0, +1 }, { 0, -1 }, { -1, -1 }, { +1, -1 },
};
enum { num_directions = sizeof(puzzleDirection) / sizeof(puzzleDirection[0]) };
int totalCount = 0;
for (int i = 0; i < num_directions; i++)
{
if (direction_searching(arr, str, len, m, n, puzzleDirection[i].dx, puzzleDirection[i].dy))
totalCount++;
}
return totalCount;
}
// checking in given direction
static int direction_searching(char arr[18][18], char *str, int len, int m, int n, int dx, int dy)
{
for (int i = 1; i < len; i++)
{
int x = m + i * dx;
int y = n + i * dy;
if (x < 0 || x >= 18 || y < 0 || y >= 18)
return 0;
if (arr[y][x] != str[i])
return 0;
}
printf(\"word Found %s starting at (%d,%d) to (%d,%d)\ \",
str, n, m, n + (len - 1) * dy, m + (len - 1) * dx);
/* checking words*/
char *pad = \"\";
for (int i = 0; i < len; i++)
{
int x = m + i * dx;
int y = n + i * dy;
printf(\"%s%c (%d,%d)\", pad, arr[y][x], y, x);
pad = \", \";
}
putchar(\'\ \');
return 1;
}
Solution
#include
#include
#include
static int read_data(char *buffer, size_t buflen);
static int direction_searching(char arr[18][18], char *str, int len, int m, int n, int dx, int dy);
static int searchData_from(char arr[18][18], char *str, int len, int m, int n);
in.
10. False.Asexually produced organisms are similar as well as iden.pdf
10. False.
Asexually produced organisms are similar as well as identical to their parents. Because the lack
of linkage and crossing over.
11. False. Prokaryotes have single free floating nucleoide.
12. False. Human cells are Diploid.
13. False. During S phase.
14. True.
15. True.
16. False. Prophase 1 and Metaphase 1 crossing over occurs. not in meiosis 2.
17. False.
18. True.
19. False.
20. True.
Solution
10. False.
Asexually produced organisms are similar as well as identical to their parents. Because the lack
of linkage and crossing over.
11. False. Prokaryotes have single free floating nucleoide.
12. False. Human cells are Diploid.
13. False. During S phase.
14. True.
15. True.
16. False. Prophase 1 and Metaphase 1 crossing over occurs. not in meiosis 2.
17. False.
18. True.
19. False.
20. True..
1) is associated with basic survival instincts of nutrition, memory .pdf
1) is associated with basic survival instincts of nutrition, memory and reproduction.
2) Pons and cerebellum.
3) cerebrum
Solution
1) is associated with basic survival instincts of nutrition, memory and reproduction.
2) Pons and cerebellum.
3) cerebrum.
The image is quite fuzzy. I cant really read. S.pdf
The image is quite fuzzy. I can\'t really read. Sorry
Solution
The image is quite fuzzy. I can\'t really read. Sorry.
Benzaldehyde oxidizes in air to produce benzoic a.pdf
Benzaldehyde oxidizes in air to produce benzoic acid.
Solution
Benzaldehyde oxidizes in air to produce benzoic acid..
B. a condensed structural formula Solution .pdf
This 1 sentence document provides the answer choice "B. a condensed structural formula" without any additional context or explanation. A condensed structural formula is a simplified way of depicting the bonding in organic molecules.
ANATOMY AND BIOMECHANICS OF HIP JOINT.pdf
it describes the bony anatomy including the femoral head , acetabulum, labrum . also discusses the capsule , ligaments . muscle that act on the hip joint and the range of motion are outlined. factors affecting hip joint stability and weight transmission through the joint are summarized.
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Long term capital = Common stock + shareholders equity + Retained ea.pdf
Long term capital = Common stock + shareholders equity + Retained earnings
Common stock = 4.7
Shareholders equity = 14.0
Retained earnings = 10.5
Total = 29.2
Solution
Long term capital = Common stock + shareholders equity + Retained earnings
Common stock = 4.7
Shareholders equity = 14.0
Retained earnings = 10.5
Total = 29.2.
Limbs evolve so frequently that it is impossible to determine otherw.pdf
Limbs evolve so frequently that it is impossible to determine otherwise if the limbs are
homologous.
Solution
Limbs evolve so frequently that it is impossible to determine otherwise if the limbs are
homologous..
hmmSolutionhmm.pdf
This very short document does not contain enough substantive information to summarize in 3 sentences or less. It consists of only three very brief statements - "hmm", "Solution", and "hmm" - which do not provide enough context to extract any essential high-level information.
Form1.csusing System; using System.Collections.Generic; using .pdf
Form1.cs
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;
namespace LotteryGUI
{
public partial class Form1 : Form
{
private int[] iGuesses = new int[3]; // declared at form level
private int[] iSortedGuesses = new int[3];
public Form1()
{
InitializeComponent();
}
private void textBox1_Leave(object sender, EventArgs e)
{
if (textBox1.Text == \"\")
{
MessageBox.Show(\"You haven\'t chosen your first number yet\");
textBox1.Focus();
return;
}
int iGuess;
int.TryParse(textBox1.Text, out iGuess);
if (iGuess < 1 || iGuess > 4)
{
MessageBox.Show(\"You have entered an incorrect number please try again\");
textBox1.Focus();
return;
}
iGuesses[0] = iGuess;
iSortedGuesses[0] = iGuess;
}
private void textBox2_Leave(object sender, EventArgs e)
{
if (textBox2.Text == \"\")
{
MessageBox.Show(\"You haven\'t chosen your second number yet\");
textBox2.Focus();
return;
}
int iGuess;
int.TryParse(textBox2.Text, out iGuess);
if (iGuess < 1 || iGuess > 4)
{
MessageBox.Show(\"You have entered an incorrect number please try again\");
textBox2.Focus();
return;
}
iGuesses[1] = iGuess;
iSortedGuesses[1] = iGuess;
}
private void textBox3_Leave(object sender, EventArgs e)
{
if (textBox3.Text == \"\")
{
MessageBox.Show(\"You haven\'t chosen your third number yet\");
textBox3.Focus();
return;
}
int iGuess;
int.TryParse(textBox3.Text, out iGuess);
if (iGuess < 1 || iGuess > 4)
{
MessageBox.Show(\"You have entered an incorrect number please try again\");
textBox3.Focus();
return;
}
iGuesses[2] = iGuess;
iSortedGuesses[2] = iGuess;
}
private void button1_Click(object sender, EventArgs e)
{
if (textBox1.Text == \"\" || textBox2.Text == \"\" || textBox3.Text == \"\")
{
MessageBox.Show(\"You haven\'t yet chosen three numbers\");
return;
}
// declaring variables integer type
int iNum1;
int iNum2;
int iNum3;
// declaring the constants
const int iMATCHONE = 10;
const int iMATCHTWO = 100;
const int iMATCHTHREE = 1000;
const int iMATCHFOUR = 10000;
// declaring array
int[] iSortedNums = new int[3];
// generating random numbers
Random randomnumber = new Random();
iNum1 = randomnumber.Next(1, 5); // number between 1 and 4
iNum2 = randomnumber.Next(1, 5);
iNum3 = randomnumber.Next(1, 5);
iSortedNums[0] = iNum1;
iSortedNums[1] = iNum2;
iSortedNums[2] = iNum3;
Array.Sort(iSortedNums); // sort random numbers
Array.Sort(iSortedGuesses); // sort the guesses
label4.Text = String.Format(\"The random numbers are : \" + iNum1 + \", \" + iNum2
+ \", \" + iNum3);
label5.Text = String.Format(\"The numbers you chose are : \" + iGuesses[0] + \", \" +
iGuesses[1] + \", \" + iGuesses[2]);
if (iGuesses[0] == iNum1 && iGuesses[1] == iNum2 && iGuesses[2] == iNum3)
{
MessageBox.Show(\"you won $\" + iMATCHFOUR);
}
else if (iSortedGuesses[0] == iSortedNums[0] && iSortedGuesses[1] ==
iSortedNums[1]
&& iSortedGuesses[2] == iSortedNums[2])
{
Message.
Influenza A The recently raised awareness of th.pdf
Influenza A: The recently raised awareness of the threat of a new influenza
pandemic has stimulated interest in the detection of influenza A viruses in human as well as
animal secretions. Virus isolation alone is unsatisfactory for this purpose because of its inherent
limited sensitivity and the lack of host cells that are universally permissive to all influenza A
viruses. Previously described PCR methods are more sensitive but are targeted predominantly at
virus strains currently circulating in humans, since the sequences of the primer sets display
considerable numbers of mismatches to the sequences of animal influenza A viruses. Therefore,
a new set of primers, based on highly conserved regions of the matrix gene, was designed for
single-tube reverse transcription-PCR for the detection of influenza A viruses from multiple
species. This PCR proved to be fully reactive with a panel of 25 genetically diverse virus isolates
that were obtained from birds, humans, pigs, horses, and seals and that included all known
subtypes of influenza A virus. It was not reactive with the 11 other RNA viruses tested.
Comparative tests with throat swab samples from humans and fecal and cloacal swab samples
from birds confirmed that the new PCR is faster and up to 100-fold more sensitive than classical
virus isolation procedures. Various Methods of Detection: Specimens. Cloacal swab specimens
were collected from ducks (widgeon [Mareca penelope], gadwall [Mareca strepera], and mallard
[Anas plathyrhynchos]) at a marshaling lake in Lekkerkerk, The Netherlands, and droppings as
well as cloacal swab specimens were collected from geese (greylag goose [Anser anser], white-
fronted goose [Anser albifrons albifrons], barnacle goose [Branta leucopsis], and brent goose
[Branta bernicla]) in Groningen and Eemdijk, The Netherlands, between 1997 and 1999. Cloacal
swab specimens and droppings were collected from shorebirds at Öland, Sweden, in the spring of
1999. Cotton swabs were used for sampling and were subsequently stored in transport medium.
Throat swab specimens collected from humans were also stored in transport medium. The
samples were stored at 4°C for a few days, at 20°C for less than a week, or at 70°C for extended
periods of time. Transport medium consisted of Hanks balanced salt solution supplemented with
10% glycerol, 200 U of penicillin per ml, 200 µg of streptomycin per ml, 100 U of polymyxin B
sulfate per ml, 250 µg of gentamicin per ml, and 50 U of nystatin per ml (all from ICN,
Zoetermeer, The Netherlands). RNA isolation. RNA was isolated with a high pure RNA
isolation kit (Roche Molecular Biochemicals) according to the instructions from the
manufacturer, with minor modifications. A 0.2-ml sample was homogenized by vortexing and
was subsequently lysed with 0.4 ml of lysis-binding buffer to which poly(A) (Roche Molecular
Biochemicals) was added as a carrier to 1 µg/ml. After binding to the column, DNase I digestion,
and washing, the RNA was eluted in 50 µl o.
I believe its named Methyl iodide. You wouldn.pdf
I believe it\'s named Methyl iodide. You wouldn\'t number the C that the Iodine is
on because there is only one carbon. Also i think that because there is only one carbon the base
name is Iodide instead of the longest carbon chain. Hope this helps! Enjoy.
Solution
I believe it\'s named Methyl iodide. You wouldn\'t number the C that the Iodine is
on because there is only one carbon. Also i think that because there is only one carbon the base
name is Iodide instead of the longest carbon chain. Hope this helps! Enjoy..
Data warehousing has quickly evolved into a unique and popular busin.pdf
Data warehousing has quickly evolved into a unique and popular business application class.
Early builders of data warehouses already consider their systems to be key components of their
IT strategy and architecture. Numerous examples can be cited of highly successful data
warehouses developed and deployed for businesses of all sizes and all types. Hardware and
software vendors have quickly developed products and services that specifically target the data
warehousing market. This paper will introduce key concepts surrounding the data warehousing
systems.
What is a data warehouse? A simple answer could be that a data warehouse is managed data
situated after and outside the operational systems. A complete definition requires discussion of
many key attributes of a data warehouse system. Later in Section 2, we will identify these key
attributes and discuss the definition they provide for a data warehouse. Section 3 briefly reviews
the activity against a data warehouse system. Initially in Section 1, however, we will take a brief
tour of the traditions of managing data after it passes through the operational systems and the
types of analysis generated from this historical data.
Evolution of an application class
This section reviews the historical management of the analysis data and the factors that have led
to the evolution of the data warehousing application class.
Traditional approaches to historical data
In reviewing the development of data warehousing, we need to begin with a review of what had
been done with the data before of evolution of data warehouses. Let us first look at how the kind
of data that ends up in today\'s data warehouses had been managed historically.
Throughout the history of systems development, the primary emphasis had been given to the
operational systems and the data they process. It is not practical to keep data in the operational
systems indefinitely; and only as an afterthought was a structure designed for archiving the data
that the operational system has processed. The fundamental requirements of the operational and
analysis systems are different: the operational systems need performance, whereas the analysis
systems need flexibility and broad scope. It has rarely been acceptable to have business analysis
interfere with and degrade performance of the operational systems.
Data from legacy systems
In the 1970s virtually all business system development was done on the IBM mainframe
computers using tools such as Cobol, CICS, IMS, DB2, etc. The 1980s brought in the new mini-
computer platforms such as AS/400 and VAX/VMS. The late eighties and early nineties made
UNIX a popular server platform with the introduction of client/server architecture.
Despite all the changes in the platforms, architectures, tools, and technologies, a remarkably
large number of business applications continue to run in the mainframe environment of the
1970s. By some estimates, more than 70 percent of business data for large corporations still
resi.
balready postedSolutionbalready posted.pdf
This document does not contain any substantive content to summarize. It appears to be a placeholder or note indicating that a solution had already been posted. The document consists of the word "already" repeated twice but does not convey any meaningful information that can be summarized in 3 sentences or less.
Assume the graph is strongly connected.SolutionAssume the grap.pdf
The document discusses a graph that is assumed to be strongly connected in one sentence. It then states "Solution" but provides no further details about the solution.
Half life is the time in which the compound reduc.pdf
Half life is the time in which the compound reduces to half of its concentration. t1/2
=0.693/wavelength
Solution
Half life is the time in which the compound reduces to half of its concentration. t1/2
=0.693/wavelength.
from the given reaction, order NO3=1 overall orde.pdf
from the given reaction, order NO3=1 overall order=2 the reaction is bimolecular
Solution
from the given reaction, order NO3=1 overall order=2 the reaction is bimolecular.
a) What is the role of Trizol in RNA extractionTrizol reagent is .pdf
a) What is the role of Trizol in RNA extraction?
Trizol reagent is used for RNA isolation from cells and tissues. This reagent is used as it helps in
maintaining RNA integrity during tissue homogenization, while at the same time disrupting and
breaking down cells and cell components. The RNA remains only in the aqueous phase(Amrita).
b) What was the purpose of performing a reverse transcription reaction. What were the products
of thesereaction?
Reverse transcription polymerase chain reaction (RT-PCR), a variant of polymerase chain
reaction (PCR), is a technique commonly used in molecular biology to detect RNA expression
incontrast the traditional PCR is used to exponentially amplify target DNA sequences, (Freeman,
1999). The product of this reaction will be RNA
Solution
a) What is the role of Trizol in RNA extraction?
Trizol reagent is used for RNA isolation from cells and tissues. This reagent is used as it helps in
maintaining RNA integrity during tissue homogenization, while at the same time disrupting and
breaking down cells and cell components. The RNA remains only in the aqueous phase(Amrita).
b) What was the purpose of performing a reverse transcription reaction. What were the products
of thesereaction?
Reverse transcription polymerase chain reaction (RT-PCR), a variant of polymerase chain
reaction (PCR), is a technique commonly used in molecular biology to detect RNA expression
incontrast the traditional PCR is used to exponentially amplify target DNA sequences, (Freeman,
1999). The product of this reaction will be RNA.
A) odd numbers caseSequential or linear search is a searching mec.pdf
A) odd numbers case:
Sequential or linear search is a searching mechanism where every element one by one is
searched until the result is found.
where l is list and x is our search item.
Minimum number of sequential search comparisons = 1
Maximum number of sequential search comparisons = N
Average number of sequential search comparisons for successful searches = ½ N
To find the middle element we can use (biggest - smallest + 1)/ 2.
2)EVEN number case:.
if the array has even number of elements then divide the list into two subparts of equal size name
array1 array 2, the element which is immediately after array 1 is considered as mid element. so
the complexity even in this case if n/2.
3)
If we find the element in 1st position then comparisons = 1
2nd positions then 2. 3rd postion 3..... nth position = n so the average no of comparisons in linear
search is (1+2+3+4+5+6.....+n) = n(n+1)/2
Solution
A) odd numbers case:
Sequential or linear search is a searching mechanism where every element one by one is
searched until the result is found.
where l is list and x is our search item.
Minimum number of sequential search comparisons = 1
Maximum number of sequential search comparisons = N
Average number of sequential search comparisons for successful searches = ½ N
To find the middle element we can use (biggest - smallest + 1)/ 2.
2)EVEN number case:.
if the array has even number of elements then divide the list into two subparts of equal size name
array1 array 2, the element which is immediately after array 1 is considered as mid element. so
the complexity even in this case if n/2.
3)
If we find the element in 1st position then comparisons = 1
2nd positions then 2. 3rd postion 3..... nth position = n so the average no of comparisons in linear
search is (1+2+3+4+5+6.....+n) = n(n+1)/2.
#include stdio.h#include stdlib.h#include string.h#inclu.pdf
#include
#include
#include
#include
/*Structure for storing words received from each thread*/
struct return_val{
char wordlist[100][100];
int count[100];
} *arr;
pthread_t *threads;
int lines;
/*Aggregates Frequency Sum of different threads*/
int aggregate(char *a)
{
int sum = 0;
int i=0;
while(1)
{
if(i==lines) break;
int count = 0;
while(1)
{
if(arr[i].count[count] == -1)
{
break;
}
if(strcmp(a,arr[i].wordlist[count]) == 0)
{
sum += arr[i].count[count];
arr[i].count[count] = 0;
}
count++;
}
i++;
}
return sum;
}
/*Function executed by each thread on separate line*/
void *word_count(void* num)
{
int *ln = num;
unsigned int line_number = *ln;
line_number++;
char cmd_p1[9] = \"sed -n \'\\0\";
char cmd_p2[2];
sprintf(cmd_p2,\"%d\",line_number);
char cmd_p3[14] = \"p\' \'file.txt\'\\0\";
char command[100];
command[0] = \'\\0\';
strcat(command,cmd_p1);
strcat(command,cmd_p2);
strcat(command,cmd_p3);
//usleep(line_number);
char cmd[100] = \" | tr [:space:] \'\\\ \' | grep -v \'^\\\\s*$\' | sort | uniq -c | sort\\0\";
strcat(command,cmd);
FILE *in;
in= popen(command, \"r\");
rewind(in);
char buff[50];
int counter = 0;
while(fgets(buff,sizeof(buff),in))
{
char c=\' \';
int i = 0;
int cnt = atoi(buff);
arr[line_number-1].count[counter] = cnt;
while(c!=\'\\0\')
{
c=buff[i];
buff[i]=buff[i+6];
i++;
}
int cnnt = 0;
while(c!=\' \')
{
c = buff[cnnt];
cnnt++;
}
i=0;
while(c!=\'\\0\')
{
c=buff[i];
buff[i]=buff[i+cnnt];
i++;
}
sprintf(arr[line_number-1].wordlist[counter],\"%s\",buff);
counter++;
}
arr[line_number-1].count[counter] = -1;
fclose(in);
return NULL;
}
int main(void)
{
FILE *fp;
fp = fopen(\"file.txt\",\"r\");
char a[200];
lines = 0;
void *status = NULL;
while(fgets(a,sizeof(a),fp) !=NULL)
{
lines++;
}
fclose(fp);
arr = (struct return_val*) calloc(lines, sizeof(struct return_val) * lines);
threads = malloc(sizeof(pthread_t)*lines);
int i;
for(i=0;i= 0)
{
if(arr[i].count[ctr] > 0)
{
strcpy(final[final_ctr],arr[i].wordlist[ctr]);
final_cnt[final_ctr] = arr[i].count[ctr];
arr[i].count[ctr] = 0;
int sum = aggregate(arr[i].wordlist[ctr]);
final_cnt[final_ctr] += sum;
printf(\"%d %s\",final_cnt[final_ctr],final[final_ctr]);
}
ctr++; final_ctr++;
}
i++;
if(i==lines)
{
break;
}
}
free(arr);
return 0;
}
#include
#include
#include
#include
/*Structure for storing words received from each thread*/
struct return_val{
char wordlist[100][100];
int count[100];
} *arr;
pthread_t *threads;
int lines;
/*Aggregates Frequency Sum of different threads*/
int aggregate(char *a)
{
int sum = 0;
int i=0;
while(1)
{
if(i==lines) break;
int count = 0;
while(1)
{
if(arr[i].count[count] == -1)
{
break;
}
if(strcmp(a,arr[i].wordlist[count]) == 0)
{
sum += arr[i].count[count];
arr[i].count[count] = 0;
}
count++;
}
i++;
}
return sum;
}
/*Function executed by each thread on separate line*/
void *word_count(void* num)
{
int *ln = num;
unsigned int line_number = *ln;
line_number++;
char cmd_p1[9] = \"sed -n \'\\0\";
char cmd_p2[2];
sprintf(cmd_p2,\"%d\",line_number);
char cmd_p3[14] .
#include ctype.h #include stdio.h #include string.hstati.pdf
#include
#include
#include
static int read_data(char *buffer, size_t buflen);
static int direction_searching(char arr[18][18], char *str, int len, int m, int n, int dx, int dy);
static int searchData_from(char arr[18][18], char *str, int len, int m, int n);
int main(void)
{
int m, n, len;
char arr[18][18], str[50];
printf(\"Welcome to word search puzzle CTRL-C TO QUIT\ \ \");
FILE *fileData = fopen(\"dataSearch.txt\", \"r\");
if (!fileData)
{
printf(\"Error occured in opening file\ \");
return 1;
}
for (n = 0; n < 18; n++)
{
for (m = 0; m < 18; m++)
{
fscanf(fileData, \" %c\", &arr[n][m]);
arr[n][m] = toupper(arr[n][m]);
}
}
fclose(fileData);
printf(\" \");
for (m = 0; m < 18; m++)
printf(\"%-2d\", m);
printf(\"\ ______________________________________\ \");
for (n = 0; n < 18; n++)
{
printf(\"%-2d|\", n);
for (m = 0; m < 18; m++)
printf(\"%c \", arr[n][m]);
printf(\"\ \");
}
while ((len = read_data(str, sizeof(str))) != EOF)
{
printf(\"Searching for: [%s]\ \", str);
int totalCount = 0;
for (n = 0; n < 18; n++)
{
for (m = 0; m < 18; m++)
{
if (arr[n][m] == (str[0]) && searchData_from(arr, str, len, m, n))
totalCount++;
}
}
printf(\"Found %s %d times\ \", str, totalCount);
}
printf(\"\ \");
return 0;
}
// reading data from file
static int read_data(char *buffer, size_t buflen)
{
printf(\"\ Please enter the word to be searched: \");
if (fgets(buffer, buflen, stdin) == 0)
return EOF;
size_t len = strlen(buffer);
if (buffer[len-1] == \'\ \')
buffer[--len] = \'\\0\';
if (len == 0)
return EOF;
for (size_t i = 0; i < len; i++)
buffer[i] = toupper(buffer[i]);
return len;
}
// searching data from array
static int searchData_from(char arr[18][18], char *str, int len, int m, int n)
{
struct yx { int dy; int dx; } puzzleDirection[] =
{
{ +1, 0 }, { -1, 0 }, { +1, +1 }, { -1, +1 },
{ 0, +1 }, { 0, -1 }, { -1, -1 }, { +1, -1 },
};
enum { num_directions = sizeof(puzzleDirection) / sizeof(puzzleDirection[0]) };
int totalCount = 0;
for (int i = 0; i < num_directions; i++)
{
if (direction_searching(arr, str, len, m, n, puzzleDirection[i].dx, puzzleDirection[i].dy))
totalCount++;
}
return totalCount;
}
// checking in given direction
static int direction_searching(char arr[18][18], char *str, int len, int m, int n, int dx, int dy)
{
for (int i = 1; i < len; i++)
{
int x = m + i * dx;
int y = n + i * dy;
if (x < 0 || x >= 18 || y < 0 || y >= 18)
return 0;
if (arr[y][x] != str[i])
return 0;
}
printf(\"word Found %s starting at (%d,%d) to (%d,%d)\ \",
str, n, m, n + (len - 1) * dy, m + (len - 1) * dx);
/* checking words*/
char *pad = \"\";
for (int i = 0; i < len; i++)
{
int x = m + i * dx;
int y = n + i * dy;
printf(\"%s%c (%d,%d)\", pad, arr[y][x], y, x);
pad = \", \";
}
putchar(\'\ \');
return 1;
}
Solution
#include
#include
#include
static int read_data(char *buffer, size_t buflen);
static int direction_searching(char arr[18][18], char *str, int len, int m, int n, int dx, int dy);
static int searchData_from(char arr[18][18], char *str, int len, int m, int n);
in.
10. False.Asexually produced organisms are similar as well as iden.pdf
10. False.
Asexually produced organisms are similar as well as identical to their parents. Because the lack
of linkage and crossing over.
11. False. Prokaryotes have single free floating nucleoide.
12. False. Human cells are Diploid.
13. False. During S phase.
14. True.
15. True.
16. False. Prophase 1 and Metaphase 1 crossing over occurs. not in meiosis 2.
17. False.
18. True.
19. False.
20. True.
Solution
10. False.
Asexually produced organisms are similar as well as identical to their parents. Because the lack
of linkage and crossing over.
11. False. Prokaryotes have single free floating nucleoide.
12. False. Human cells are Diploid.
13. False. During S phase.
14. True.
15. True.
16. False. Prophase 1 and Metaphase 1 crossing over occurs. not in meiosis 2.
17. False.
18. True.
19. False.
20. True..
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1) is associated with basic survival instincts of nutrition, memory and reproduction.
2) Pons and cerebellum.
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1) is associated with basic survival instincts of nutrition, memory and reproduction.
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3) cerebrum.
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