C++ PROGRAMThis program builds on the code below#include iostr.pdf
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C++ PROGRAM This program builds on the code below: #include #include #include \"shirtType.h\" using namespace std; void getInput(shirtType s[],ifstream& infile, int& numShirts); int main() { int numShirts; shirtType shirts[500]; ifstream infile; infile.open(\"h:\\\\shirtlab.txt\"); getInput(shirts,infile,numShirts); for(int i =0; i>id; infile.ignore(100,\'\ \'); getline(infile,c); infile>>size>>p>>q; while(infile) { s[index].setAll(id, c, size, p, q); index++; infile>>id; infile.ignore(100,\'\ \'); getline(infile,c); infile>>size>>p>>q; } numShirts=index; } See also the description of the class shirtType below: #include #include using namespace std; class shirtType { private: int shirtID; string color; string size; double price; int quantity; public: void setAll(int i,string c,string s, double p, int q); void print() const; void newQuan(); shirtType(); shirtType(int,string,string,double,int); }; shirtType::shirtType(int i,string c,string s, double p, int q) { color = c; size = s; if(i>0) shirtID=i; else shirtID=0; if(p>0.0) price = p; else price=0.0; if (q>=0) quantity=q; else quantity = 0; } shirtType::shirtType() { shirtID=0; color=\"\"; size=\"\"; price=0.0; quantity=0; } void shirtType::setAll(int i,string c,string s, double p, int q) { color = c; size = s; if(i>0) shirtID=i; else shirtID=0; if(p>0.0) price = p; else price=0.0; if (q>=0) quantity=q; else quantity = 0; } void shirtType::newQuan() { cout<<\"Enter the new quantity of shirts: \"<>quantity; } void shirtType::print() const { cout<<\"Shirt ID: \"< Solution #include #include struct node { int data; struct node *next; }*head; void append(int nkkm) { struct node *temp,*right; temp= (struct node *)malloc(sizeof(struct node)); temp->data=nkkm; right=(struct node *)head; while(right->next != NULL) right=right->next; right->next =temp; right=temp; right->next=NULL; } void add( int nkkm ) { struct node *temp; temp=(struct node *)malloc(sizeof(struct node)); temp->data=nkkm; if (head== NULL) { head=temp; head->next=NULL; } else { temp->next=head; head=temp; } } void addafter(int nkkm, int loc) { int i; struct node *temp,*left,*right; right=head; for(i=1;inext; } temp=(struct node *)malloc(sizeof(struct node)); temp->data=nkkm; left->next=temp; left=temp; left->next=right; return; } void addaShirt(int nkkm) { int c=0; struct node *temp; temp=head; if(temp==NULL) { add(nkkm); } else { while(temp!=NULL) { if(temp->datanext; } if(c==0) add(nkkm); else if(cdata==nkkm) { if(temp==head) { head=temp->next; free(temp); return 1; } else { prev->next=temp->next; free(temp); return 1; } } else { prev=temp; temp= temp->next; } } return 0; } void display(struct node *r) { r=head; if(r==NULL) { return; } while(r!=NULL) { printf(\"%d \",r->data); r=r->next; } printf(\"\ \"); } int count() { struct node *n; int c=0; n=head; while(n!=NULL) { n=n->next; c++; } return c; } int main() { int i,nkkm; struct node *n; head=NULL; while(1) { printf(\"1.Add a shirt\ \"); printf(\"2.Sell a shirt\ \"); printf(\"3.Res.
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![C++ PROGRAM
This program builds on the code below:
#include
#include
#include "shirtType.h"
using namespace std;
void getInput(shirtType s[],ifstream& infile, int& numShirts);
int main()
{
int numShirts;
shirtType shirts[500];
ifstream infile;
infile.open("h:shirtlab.txt");
getInput(shirts,infile,numShirts);
for(int i =0; i>id;
infile.ignore(100,' ');
getline(infile,c);
infile>>size>>p>>q;
while(infile)
{
s[index].setAll(id, c, size, p, q);
index++;
infile>>id;
infile.ignore(100,' ');
getline(infile,c);
infile>>size>>p>>q;
}
numShirts=index;
}
See also the description of the class shirtType below:
#include
#include
using namespace std;
class shirtType](https://image.slidesharecdn.com/cprogramthisprogrambuildsonthecodebelowincludeiostr-230707211215-cb726261/85/C-PROGRAMThis-program-builds-on-the-code-below-include-iostr-pdf-1-320.jpg)
Recommended
#include iostream #include fstream #include shirtType.h.pdf
#include
#include
#include \"shirtType.h\"
using namespace std;
void getInput(shirtType s[],ifstream& infile, int& numShirts);
int main()
{
int numShirts;
shirtType shirts[500];
ifstream infile;
infile.open(\"h:\\\\shirtlab.txt\");
getInput(shirts,infile,numShirts);
for(int i =0; i>id;
infile.ignore(100,\'\ \');
getline(infile,c);
infile>>size>>p>>q;
while(infile)
{
s[index].setAll(id, c, size, p, q);
index++;
infile>>id;
infile.ignore(100,\'\ \');
getline(infile,c);
infile>>size>>p>>q;
}
numShirts=index;
}
See also the description of the class shirtType below:
#include
#include
using namespace std;
class shirtType
{
private:
int shirtID;
string color;
string size;
double price;
int quantity;
public:
void setAll(int i,string c,string s, double p, int q);
void print() const;
void newQuan();
shirtType();
shirtType(int,string,string,double,int);
};
shirtType::shirtType(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
shirtType::shirtType()
{
shirtID=0;
color=\"\";
size=\"\";
price=0.0;
quantity=0;
}
void shirtType::setAll(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
void shirtType::newQuan()
{
cout<<\"Enter the new quantity of shirts: \"<>quantity;
}
void shirtType::print() const
{
cout<<\"Shirt ID: \"<
Solution
//shirtType.h
#include
#include
using namespace std;
class shirtType
{
private:
int shirtID;
string color;
string size;
double price;
int quantity;
public:
void setAll(int i,string c,string s, double p, int q);
void print() const;
void newQuan();
int changeQuan();
int getShirtId(int);
shirtType();
int search(int);
shirtType(int,string,string,double,int);
};
int shirtType::search(int sellId)
{
if(shirtID == sellId)
return 1;
else
return 0;
}
int shirtType::getShirtId(int i)
{
return shirtID;
}
shirtType::shirtType(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
shirtType::shirtType()
{
shirtID=0;
color=\"\";
size=\"\";
price=0.0;
quantity=0;
}
void shirtType::setAll(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
void shirtType::newQuan()
{
int q;
cout<<\"Enter the new quantity of shirts: \"<>q;
quantity += q;
}
int shirtType::changeQuan()
{
if(quantity > 0 )
{
quantity -= 1;
return 1;
}
{
return 2;
}
}
void shirtType::print() const
{
cout<<\"Shirt ID: \"<
#include
#include
#include \"shirtType.h\"
#include
using namespace std;
void getInput(shirtType s[],ifstream& infile, int& numShirts);
void addShirt(shirtType s[], int& numShirts);
void sellShirt(shirtType s[], int& numShirts);
void addstockShirt(shirtType s[], int&.
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Unlike the Federal income tax, most states do not require some form of state income tax.
True
False
Solution
Correct answer is True
Unlike Federal income tax, most states do not require some form of state income tax - This
statement is true as fililng of Fedral income tax return is mandatory for almost everybody but
some US states like Alaska, Florida, Navada, South Dakota, Texas, Washington and Wayoming
do not require filing of state income tax..
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Recommended
#include iostream #include fstream #include shirtType.h.pdf
#include
#include
#include \"shirtType.h\"
using namespace std;
void getInput(shirtType s[],ifstream& infile, int& numShirts);
int main()
{
int numShirts;
shirtType shirts[500];
ifstream infile;
infile.open(\"h:\\\\shirtlab.txt\");
getInput(shirts,infile,numShirts);
for(int i =0; i>id;
infile.ignore(100,\'\ \');
getline(infile,c);
infile>>size>>p>>q;
while(infile)
{
s[index].setAll(id, c, size, p, q);
index++;
infile>>id;
infile.ignore(100,\'\ \');
getline(infile,c);
infile>>size>>p>>q;
}
numShirts=index;
}
See also the description of the class shirtType below:
#include
#include
using namespace std;
class shirtType
{
private:
int shirtID;
string color;
string size;
double price;
int quantity;
public:
void setAll(int i,string c,string s, double p, int q);
void print() const;
void newQuan();
shirtType();
shirtType(int,string,string,double,int);
};
shirtType::shirtType(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
shirtType::shirtType()
{
shirtID=0;
color=\"\";
size=\"\";
price=0.0;
quantity=0;
}
void shirtType::setAll(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
void shirtType::newQuan()
{
cout<<\"Enter the new quantity of shirts: \"<>quantity;
}
void shirtType::print() const
{
cout<<\"Shirt ID: \"<
Solution
//shirtType.h
#include
#include
using namespace std;
class shirtType
{
private:
int shirtID;
string color;
string size;
double price;
int quantity;
public:
void setAll(int i,string c,string s, double p, int q);
void print() const;
void newQuan();
int changeQuan();
int getShirtId(int);
shirtType();
int search(int);
shirtType(int,string,string,double,int);
};
int shirtType::search(int sellId)
{
if(shirtID == sellId)
return 1;
else
return 0;
}
int shirtType::getShirtId(int i)
{
return shirtID;
}
shirtType::shirtType(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
shirtType::shirtType()
{
shirtID=0;
color=\"\";
size=\"\";
price=0.0;
quantity=0;
}
void shirtType::setAll(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
void shirtType::newQuan()
{
int q;
cout<<\"Enter the new quantity of shirts: \"<>q;
quantity += q;
}
int shirtType::changeQuan()
{
if(quantity > 0 )
{
quantity -= 1;
return 1;
}
{
return 2;
}
}
void shirtType::print() const
{
cout<<\"Shirt ID: \"<
#include
#include
#include \"shirtType.h\"
#include
using namespace std;
void getInput(shirtType s[],ifstream& infile, int& numShirts);
void addShirt(shirtType s[], int& numShirts);
void sellShirt(shirtType s[], int& numShirts);
void addstockShirt(shirtType s[], int&.
Basic C Programming Lab Practice
The document contains the answers to 12 tasks/problems involving C programming concepts like arrays, matrices, strings, functions, recursion, structures, files etc. Each answer provides the full code to solve the problem. The tasks include programs to sort arrays, add matrices, count characters in strings, check palindromes, find prime numbers using functions, calculate factorials and Fibonacci terms recursively, define nested structures, and read/write numbers to files.
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The document contains 14 code snippets that provide solutions to common programming problems in C including: printing patterns like numbers, stars, triangles using loops; finding sum of digits, checking palindrome, Armstrong number; sorting arrays using bubble, selection sort; string operations like concatenation; matrix addition; factorial; checking even-odd; Floyd's triangle. Each code snippet includes the header files, function definitions and main function to accept input, call functions and print output.
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Cs pritical file
The document discusses C++ programs involving classes and objects. It includes the definition of classes such as Applicant, Housing, Tour, Account and subclasses Current and Savings. The Applicant class stores applicant details and grades, Housing stores housing property data, Tour calculates tour fares, and Account is the base class for banking accounts with subclasses Current and Savings that inherit and expand its functionality. Member functions are defined to input, output and manipulate object data for these classes.
Using standard libraries like stdio and sdtlib.h and using stats.h a.pdf
Unlike the Federal income tax, most states do not require some form of state income tax.
True
False
Solution
Correct answer is True
Unlike Federal income tax, most states do not require some form of state income tax - This
statement is true as fililng of Fedral income tax return is mandatory for almost everybody but
some US states like Alaska, Florida, Navada, South Dakota, Texas, Washington and Wayoming
do not require filing of state income tax..
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This document provides an overview of arrays and strings in C programming. It discusses initializing and declaring arrays of different types, including multidimensional arrays. It also covers passing arrays as arguments to functions. For strings, it explains that strings are arrays of characters that are null-terminated. It provides examples of declaring and initializing string variables, and using string input/output functions like scanf() and printf().
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C-Program: Custom Library, Header File, and Implementation Files
I have this main.c file:
#include
#include
#include \"stats.h\"
int readGrades(char fileName[], float grades[]);
//Include this function prototype
void WelcomeToTheProgram();
int main()
{
///Declare Variables
char fileName[30];
float grades[1000]; //to store grades
int i, ArrayCount[10] = {0}; //to store histogram counts
/* Rewrite this line */
int num_grades; //for # of grades in file
///Call Intro Comment
WelcomeToTheProgram();
///Enter Filename
printf(\"\ \ \ Enter the data filename: \");
scanf(\"%s\", &fileName);
//Get # of grades in file
num_grades = readGrades(fileName, grades); //get # of grades in file
///Print Results
printf(\"There are %d grades read\ \", num_grades);
printf(\"Mean = %f\ \", mean(grades, num_grades));
printf(\"Variance = %f \ \", variance(grades, num_grades));
printf(\"Maximum = %f \ \", maximum(grades, num_grades));
printf(\"Minumum = %f \ \", minimum(grades, num_grades));
//Call histogram
histogram(grades, ArrayCount, num_grades);
//Print the histogram
printf(\"Grade Histogram\ \");
for (i = 0; i < 10; i++)
{
if (i != 9)
{
printf(\"\ %d %% - %d %%: %d \", i * 10, ((i + 1) * 10) - 1, ArrayCount[i]);
}
else
{
printf(\"\ 90 %% - 100 %%: %d\ \", ArrayCount[i]);
}
}
return 0;
}
///Reads all grades from the file to array
int readGrades(char fileName[], float grades[])
{
FILE *fpin = fopen(fileName, \"r\");
int num_scores = 0;
float value;
if (fpin == NULL)
{
printf(\"Cannot open the file\ \");
exit(0);
}
while ((fscanf(fpin, \"%f\", &value)) != EOF)
{
grades[num_scores] = value;
num_scores++;
}
return num_scores;
}
void WelcomeToTheProgram()
{
printf(\"***Welcome to the Program Which Involves a Custom Library***\");
printf(\"\ The program finds the minimum value, maximum value, mean,\");
printf(\"variance, and a histogram of the grades of data by calling\");
printf(\"implementation files min(), max(), mean(), variance(), and histogram().\ \");
}
But, the only thing that prints is
Enter the data file name: grades.txt
There are 400 grades read.
Then the program just stops, it does not print the min, max, mean, variance, or histogram. Any
ideas? Am I using the custom library wrong, the implementation files (for min, max, mean,
variance, and histogram) all check out with no warnings or errors.
Here is stats.h:
#ifndef STATS_H
#define STATS_H
float minimum(float *grades[], int Size); // NOTE: You need to complete the prototypes
float maximum(float *grades[], int Size);
float mean(float *grades[], int Size);
float variance(float *grades[], int Size);
void histogram(float *grades[], int *ArrayCount, int Size);
#endif // STATS_H
And here is the grades file:
Solution
//stats.h
#ifndef STATS_H
#define STATS_H
float minimum(float grades[], int Size); // NOTE: You need to complete the prototypes
float maximum(float grades[], int Size);
float mean(float grades[], int Size);
float variance(float grades[], int Size);
void histogram(float grades[], int ArrayCount[], int S.
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1) Finding perfect, Armstrong, and prime numbers as well as reversing, summing digits, and checking palindromes of numbers.
2) Converting between decimal and binary, multiplying matrices, and calculating LCM and factorial.
3) Checking for leap years and strong, palindrome, and generic root of numbers.
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c++ program for Railway reservation
This document contains C++ code for a railway reservation system. It defines classes for login, passenger details, reservations, cancellations, and manages the user interface and database interactions. The main function displays a menu and calls additional functions for administrative tasks like adding/viewing train schedules or passenger details, and user tasks like reservation, cancellation or inquiries. Database files store details, reservations and user login information.
4. chapter iii
A pointer is a variable that stores the address of another variable. Pointers allow variables to reference memory locations and access the values stored there. An array is a collection of variables of the same type stored contiguously in memory. Strings in C are arrays of characters terminated by a null character. Pointers can also be used to point to strings by storing the base address of the character array. Multidimensional arrays and manipulating strings are also discussed.
#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] .
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In nonvascular plants the _ is usually no photosynthetic and is_ dependent on the garnet
zoophyte nutritionally, sporophyte sporophyte reproducing. gametophyte gametophyte
structurally, sporophyte gametophyte reproductively. sporophyte sporophyte. nutritional.
gametophyte
Solution
sporophyte,nutrionally,gametophyte
In non vascular plants the sporophyte is usually non-photosynthetic and is nutrionally dependant
on the gametophyte
Explination
In bryophytes like mosses gametophyte n is the dominant generation and sporophyte 2n lives on
the gametophyte .....
i need help to edit the two methods below so that i can have them wo.pdf
I have to make a program for my C class and im stuck on the part where i have to calculate the
variance with elements from my array. This is exactly what im suppose to do:
Reset the sum variable to 0
Using another for loop (which needs its own loop variable), run through all valid elements of the
array, starting from index 0. Inside the loop:
Use the sum variable to accumulate the numerator of the variance formula shown on page 1.
After the end of this loop, compute the variance by dividing the new sum by the
dataElementsToProcess parameter.
this is the entire function
bool GenerateStatistics(FILE *fp2, double myArray[], int dataElementsToProcess)
{
if (fp2 == NULL && myArray == NULL && dataElementsToProcess < 1)
{
return false;
}
else
{
int i = 0,o = 0;
int sum = 0, sum1 = 0;
double maxValue = 0;
double minValue = DBL_MAX;
double mean, variance, stdDeviation;
for (int i = 0; i < 600; i++)
{
sum += myArray[i];
if (myArray[i] > maxValue)
{
maxValue = myArray[i];
}
if (myArray[i] < minValue)
{
minValue = myArray[i]; }
}
mean = sum / dataElementsToProcess;
sum = 0;
for (int o = 0; o < 600; o++)
{
sum += myArray[o];
}
variance = sqrt(pow((sum - mean), 2) / (dataElementsToProcess));
stdDeviation = sqrt(variance);
fprintf(fp2, \"Reaction Rate Statistics:\ \ \");
fprintf(fp2, \"Number of experiments = %d\ \", dataElementsToProcess);
fprintf(fp2, \"Minimum rate\\t\\t= %f\ \", minValue);
fprintf(fp2, \"Maximum rate\\t\\t= %f\ \", maxValue);
fprintf(fp2, \"Arithmetic mean\\t\\t= %f\ \", mean);
fprintf(fp2, \"Variance\\t\\t= %f\ \", variance);
fprintf(fp2, \"Standard deviation\\t= %f\ \ \ \", stdDeviation);
}
return true;
}
Solution
Here in the function you have
average - mean
Noofelements - dataElementsTo Process
Variance formula is
V = summation over i=1 to n ((xi - mean )*(xi - mean))
So perform like this using a for loop over my_array
double V=0.0
for(int i=0;i<600;i++)
{
V=V+pow((my_Array[i]-mean),2)
}
Now calculate standard deviation using sqrt of variance
sd=0.0
sd=sqrt(V).
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I have this main.c file:
#include
#include
#include \"stats.h\"
int readGrades(char fileName[], float grades[]);
//Include this function prototype
void WelcomeToTheProgram();
int main()
{
///Declare Variables
char fileName[30];
float grades[1000]; //to store grades
int i, ArrayCount[10] = {0}; //to store histogram counts
/* Rewrite this line */
int num_grades; //for # of grades in file
///Call Intro Comment
WelcomeToTheProgram();
///Enter Filename
printf(\"\ \ \ Enter the data filename: \");
scanf(\"%s\", &fileName);
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num_grades = readGrades(fileName, grades); //get # of grades in file
///Print Results
printf(\"There are %d grades read\ \", num_grades);
printf(\"Mean = %f\ \", mean(grades, num_grades));
printf(\"Variance = %f \ \", variance(grades, num_grades));
printf(\"Maximum = %f \ \", maximum(grades, num_grades));
printf(\"Minumum = %f \ \", minimum(grades, num_grades));
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histogram(grades, ArrayCount, num_grades);
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{
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return 0;
}
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int readGrades(char fileName[], float grades[])
{
FILE *fpin = fopen(fileName, \"r\");
int num_scores = 0;
float value;
if (fpin == NULL)
{
printf(\"Cannot open the file\ \");
exit(0);
}
while ((fscanf(fpin, \"%f\", &value)) != EOF)
{
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}
void WelcomeToTheProgram()
{
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printf(\"\ The program finds the minimum value, maximum value, mean,\");
printf(\"variance, and a histogram of the grades of data by calling\");
printf(\"implementation files min(), max(), mean(), variance(), and histogram().\ \");
}
But, the only thing that prints is
Enter the data file name: grades.txt
There are 400 grades read.
Then the program just stops, it does not print the min, max, mean, variance, or histogram. Any
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Here is stats.h:
#ifndef STATS_H
#define STATS_H
float minimum(float *grades[], int Size); // NOTE: You need to complete the prototypes
float maximum(float *grades[], int Size);
float mean(float *grades[], int Size);
float variance(float *grades[], int Size);
void histogram(float *grades[], int *ArrayCount, int Size);
#endif // STATS_H
And here is the grades file:
Solution
//stats.h
#ifndef STATS_H
#define STATS_H
float minimum(float grades[], int Size); // NOTE: You need to complete the prototypes
float maximum(float grades[], int Size);
float mean(float grades[], int Size);
float variance(float grades[], int Size);
void histogram(float grades[], int ArrayCount[], int S.
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#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] .
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More from irshadoptical
In practice, a RAM is used to simulate the operation of a stack by m.pdf
In nonvascular plants the _ is usually no photosynthetic and is_ dependent on the garnet
zoophyte nutritionally, sporophyte sporophyte reproducing. gametophyte gametophyte
structurally, sporophyte gametophyte reproductively. sporophyte sporophyte. nutritional.
gametophyte
Solution
sporophyte,nutrionally,gametophyte
In non vascular plants the sporophyte is usually non-photosynthetic and is nutrionally dependant
on the gametophyte
Explination
In bryophytes like mosses gametophyte n is the dominant generation and sporophyte 2n lives on
the gametophyte .....
i need help to edit the two methods below so that i can have them wo.pdf
I have to make a program for my C class and im stuck on the part where i have to calculate the
variance with elements from my array. This is exactly what im suppose to do:
Reset the sum variable to 0
Using another for loop (which needs its own loop variable), run through all valid elements of the
array, starting from index 0. Inside the loop:
Use the sum variable to accumulate the numerator of the variance formula shown on page 1.
After the end of this loop, compute the variance by dividing the new sum by the
dataElementsToProcess parameter.
this is the entire function
bool GenerateStatistics(FILE *fp2, double myArray[], int dataElementsToProcess)
{
if (fp2 == NULL && myArray == NULL && dataElementsToProcess < 1)
{
return false;
}
else
{
int i = 0,o = 0;
int sum = 0, sum1 = 0;
double maxValue = 0;
double minValue = DBL_MAX;
double mean, variance, stdDeviation;
for (int i = 0; i < 600; i++)
{
sum += myArray[i];
if (myArray[i] > maxValue)
{
maxValue = myArray[i];
}
if (myArray[i] < minValue)
{
minValue = myArray[i]; }
}
mean = sum / dataElementsToProcess;
sum = 0;
for (int o = 0; o < 600; o++)
{
sum += myArray[o];
}
variance = sqrt(pow((sum - mean), 2) / (dataElementsToProcess));
stdDeviation = sqrt(variance);
fprintf(fp2, \"Reaction Rate Statistics:\ \ \");
fprintf(fp2, \"Number of experiments = %d\ \", dataElementsToProcess);
fprintf(fp2, \"Minimum rate\\t\\t= %f\ \", minValue);
fprintf(fp2, \"Maximum rate\\t\\t= %f\ \", maxValue);
fprintf(fp2, \"Arithmetic mean\\t\\t= %f\ \", mean);
fprintf(fp2, \"Variance\\t\\t= %f\ \", variance);
fprintf(fp2, \"Standard deviation\\t= %f\ \ \ \", stdDeviation);
}
return true;
}
Solution
Here in the function you have
average - mean
Noofelements - dataElementsTo Process
Variance formula is
V = summation over i=1 to n ((xi - mean )*(xi - mean))
So perform like this using a for loop over my_array
double V=0.0
for(int i=0;i<600;i++)
{
V=V+pow((my_Array[i]-mean),2)
}
Now calculate standard deviation using sqrt of variance
sd=0.0
sd=sqrt(V).
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entries and a trial balance. 1 May Invests £8,000 in his business bank account. 2 May Purchases
£2,000 goods for cash from E. Skimo. £4,000 of goods on credit from S. Eal. £1,000 of goods on
credit from P. Olar. 4 May Sells £1,500 goods for cash to A.R.C. Tic. £5,000 goods on credit to
H. Unter. £3,000 goods on credit to M. Dick. 7 May M. Dick returns £2,000 goods unpaid saying
that they were rotting. 8 May R. Poon sends M. Dick\'s £2,000 goods (that he had originally
purchased from S. Eal for £1,800) back unpaid because he can\'t stand the smelt 10 May
Purchases another £1,200 of goods on credit from S. Eal. 11 May H. Unter buys on credit £1,600
of goods from R. Poon.
Solution
PURCHASE ACCOUNT S.EAL A/C (CREDITOR) P.OLAR A/C
(*CREDITOR) DATE PARTICULAR AMOUNT (DR) AMOUNT (CR) DATE
PARTICULAR AMOUNT (DR) AMOUNT (CR) DATE PARTICULAR AMOUNT (DR)
AMOUNT (CR) 2-May TO CASH 2000 2-May BY PURCH A/C 4000 2-May BY PURCH
A/C 1000 2-May TO S .EAL 4000 10-May BY PURCH A/C 1200 20-May TO BANK 750
2-May TO P.OLAR 1000 8-May TO PUR. RETURN 1800 31-May BY CL BALANCE
250 10-May TO SEAL 1200 20-May TO BANK 3400 31-May P&L A/C 8200 31-
May CL BAL - - SALES ACCOUNT H.UNTERA/C (DEBTOR) M.DICKI
A/C DEBTOR DATE PARTICULAR AMOUNT (DR) AMOUNT (CR) DATE
PARTICULAR AMOUNT (DR) AMOUNT (CR) DATE PARTICULAR AMOUNT (DR)
AMOUNT (CR) 4-May BY CASH 1500 4-May SALES A/C 5000 4-May SALES A/C
3000 4-May BY H.UNTER 5000 11-May SALES A/C 1600 7-May SALES RETURN
2000 4-May BY M.DICKI 3000 21-May BY BANK 2000 21-May BY BANK 800 11-
May TO H.UNTER 1600 31-May CL. BAL 4600 (5000+1600-2000) 31-May TO CL. BAL
200 31-May P&L A/C 11100 SALES RETURN ACCOUNT
DATE PARTICULAR AMOUNT (DR) AMOUNT (CR) 7-May TO DICKI 2000
P&L A/C 2000 PURCHASE RETURN ACCOUNT DATE
PARTICULAR AMOUNT (DR) AMOUNT (CR) 8-May BY S.EAL 1800 31-
May P&L A/C 1800 WAGES A/C DATE PARTICULAR
AMOUNT (DR) AMOUNT (CR) 31-May BANK A/C 800 31-May P&L A/C
800 RENT A/C DATE PARTICULAR AMOUNT (DR)
AMOUNT (CR) 31-May BANK A/C 500 31-May P&L A/C 500
ELECTRICITY EXP A/C DATE PARTICULAR AMOUNT (DR) AMOUNT
(CR) 31-May BANK A/C 75 31-May P&L A/C 75
STATIONERY A/C BANK INTEREST A/C DATE PARTICULAR AMOUNT
(DR) AMOUNT (CR) 15-May BY BANK - 250 31-May BANK A/C 25 31-May P&L
A/C 250 31-May P&L A/C 25 BANK A/C CAPITAL A/C
DATE PARTICULAR AMOUNT (DR) AMOUNT (CR) 1-May BY BANK 8000 1-May
CAPITAL A/C 8000 3-May CLOSING BALANCE 8000 15-May BANK INT . A/C 250
20-May S.EAL 3400 20-May P.OLA 750 CASH A/C 21-May h.unter
2000 DATE PARTICULAR AMOUNT (DR) AMOUNT (CR) 21-May M.DICKI 800 2-
May PURCHASE 2000 31-May WAGES 800 4-May SALES 1500 31-May RENT
500 CL. BALNANCE 500 31-May ELEC .
How does feasibility differ in an agile environment in comparison to.pdf
Traditional development assumes systems are fully specifiable through extensive planning, while agile development develops high quality adaptive software through small teams that continuously improve design and testing using fast feedback. Management shifts from command and control to leadership and collaboration, while knowledge moves from explicit to tacit. Other differences include communication style, life cycle model, organizational structure, quality control approach, and client involvement. Both approaches have merits and demerits, and neither is inherently simpler - what matters is properly understanding the methodology for the given situation. In-house software provides knowledgeable support from dedicated developers but can lack scalability, while outsourcing decreases coding burden but may increase costs.
describe a real-life example of selection against a homozygous reces.pdf
describe a real-life example of selection against a homozygous recessive genotype
Solution
When both the alleles of a particular chromosome are recessive, it’s called homozygous
recessive for that chromosome pair. There are few real life examples of homozygous recessive
such as albinism, cystic fibrosis etc.
Albinism is congenital disorder in humans that is a characterized by absence of pigment in skin,
eyes, and hairs. Absence of pigment may be partial or complete. Lack of skin pigmentation in
albinism makes individual more sensitive to sunlight and susceptible skin sunburn and skin
cancers. Albinism is also associated with visions defects such as amblyopia, photophobia and
nystagmus. This disorder is due to inheritance of recessive alleles which has defect/ absence of
tyrosinase involved in the production of melanin..
Define a motor unit. What roles do motor units play in graded contrac.pdf
Define a motor unit. What roles do motor units play in graded contractions in skeletal muscles?
Solution
Motor unit is made of Motor Neuron, and its branches extended to the myofibrils of muscle
fibres. Signal is passed through the neuron to the muscle fibres. Motor neuron reaches to the
muscle fibre at neuromuscular junctions. Motor neurons work in group to assist the contraction
of single muscle in body. Force of Muscle contraction is controlled by the number of activated
motor neurons..
An antibody is a protein that has which level of protein structure .pdf
An antibody is a protein that has which level of protein structure? Primary Quaternary Tertiary
Secondary Which cell type does the HIV infect? Cytotoxic T lymphocytes B lymphocytes
Helper T lymphocytes Suppressor T lymphocytes
Solution
antibody is a protein which is in quaternary structure because it consist of more than one
polypeptide chains (two heavy chain and two light chain ) joint by disulphide bond.
HIV infect Helper T lymphocyte(CD4+T cells).It causes the lower level of CD4+ cells by
pyroptosis.When CD4 cell number is declined below a normal level than cell mediated immunity
is lost and body become suceptible to various infection..
You are in charge of the investigation regarding possible exposur.pdf
You are conducting a study to see if students do better when they study all at once or in intervals.
One group of 12 participants took a test after studying for one hour continuously. The other
group of 12 participants took a test after studying for three twenty minute sessions. The first
group had a mean score of75 and a variance of 120. The second group had a mean score of 86
and a variance of 100.
a. What is the calculated t value? Are the mean test scores of these two groups significantly
different at the .05 level?
b. What would the t value be if there were only 6 participants in each group? Would the scores
be significant at the .05 level?
Solution
A)
Formulating the null and alternative hypotheses,
Ho: u1 - u2 = 0
Ha: u1 - u2 =/ 0
At level of significance = 0.05
As we can see, this is a two tailed test.
Calculating the means of each group,
X1 = 75
X2 = 86
Calculating the standard deviations of each group,
s1 = 10.95445115
s2 = 10
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 12
n2 = sample size of group 2 = 12
Thus, df = n1 + n2 - 2 = 22
Also, sD = 4.281744193
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = -2.569046516 [ANSWER, T VALUE]
********************************
b)
Calculating the means of each group,
X1 = 75
X2 = 86
Calculating the standard deviations of each group,
s1 = 10.95445115
s2 = 10
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 6
n2 = sample size of group 2 = 6
Also, sD = 6.055300708
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = -1.816590212 [ANSWER]
For part A, the P value is
p = 0.01750305
For part B, the P value is
p = 0.099328388
Hence, Part A would be significant, but part B wouldn\'t. [ANSWER]
**********************************
*****************************************************
Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please
resubmit this question together with the formula/method you use in determining the degrees of
freedom. That way we can continue helping you! Thanks!
********************************************.
Which of the following is not a product of the light reactions of ph.pdf
Which of the following developments have not helped to increase the importance of packaging
cover time Which stage is the most costly one? In which stage are criteria such as adequacy of
the technical - financial -manpower resources of the company and suitability to company
objectives and mission are used to evaluate new product Which stage involves a feasibility study
Which stage is most appropriate for \"informative\" advertising? At which stage should
management make a choice between skimming and penetration In which stage is
\"rejuvenation\" most likely to occur (if h does)? At which stage is heaviest price competition
most likely to occur? Which stage is the right time for starting to plan a new product
introduction? At which stage should product differentiation be at its lowest level? At which
stage do profits reach their highest level? A small confectionery (tath, Sekerleme, recel)
manufacturer markets its jams and jel an agreed upon price every year provided that the label is
changed into \"\"Sultan Ha following would be true
Solution
(8) (a)
Need for better packaging has nothing to do with economic crises - the crises can impact the
purchase decision with or without better & more attractive packaging.
(9) (C)
Concept development & testing is costliest because if the prototype developed fails, entire cost
incurred goes waste.
(10) (B)
These are tested at Screening stage for suitability of ideas generated.
(11) (D)
It involves technical, economic & business feasibility analysis.
(12) (A)
(13) (B)
(14) (D)
(15) (C)
(16) (C)
(17) (A)
(18) (B)
(19) Question is partially visible & can\'t be answered properly..
Write a line of code to create a Print Writer to write to the file .pdf
Write a class named Deque20, placed in the file Deque20.java. It should behave exactly as
described in the Java interface Deque (Google: \"Java 8 Deque\") except that it should only
implement the following methods: a no-argument constructor, addFirst, addLast, removeFirst,
removeLast, peekFirst, peekLast, and size. Your class should not extend or implement any class
or interface. It must use an array as its storage, allow generic types (E), and be as efficient as
possible. The Deque javadoc mentions several exceptions, but you should only return
NullPointerException (if someone tries to add a null reference to the Deque) and
NoSuchElementException (if someone tries to remove an item from an empty Deque).
Your deque should store objects in a circular array.
Probably the simplest implementation has objects with fields
The meaning being that the elements currently in the deque are at positions list[first_index],
list[(first_index+1)%list.length], list[(first_index+2)%list.length],...,
list[(first_index+num_elems-1)%list.length]. The modulus is necessary because first_index
might be near the end of the array and adding to it might result in an index that\'s out of bounds.
When an add operation would cause the capacity of the array to be exceeded, you should double
the size of the array by creating a new array of twice the size and copying the old array contents
into the new. Do not worry about ever reducing the size of an array. Also, whenever a reference
is removed from your deque, the old array location should be set to null so that the garbage
collector knows that the old item is no longer referred to by the deque.
Solution
public class Deque20 { private int first_index; private int num_elems; private Object[]
list; Deque20(){ first_index = 0; num_elems = 0; list = new Object[20]; }
private void doubleSize(){ Object[] arr = new Object[2 * list.length]; for(int i = 0; i <
list.length; ++i){ arr[i] = list[(i + first_index) % list.length]; } list = arr; }
public void addFirst(E val){ if(list.length == num_elems){ doubleSize(); }
int index = (first_index + list.length - 1) % list.length; list[index] = val; ++num_elems;
} public void addLast(E val){ if(list.length == num_elems){ doubleSize();
} int index = (first_index + num_elems) % list.length; list[index] = val;
++num_elems; } public E removeFirst(){ int index = first_index; E ret = (E)
list[index]; first_index = (first_index + 1) % list.length; return ret; } public E
removeLast(){ int index = ((first_index + num_elems) % list.length) - 1; E ret = (E)
list[index]; --num_elems; return ret; } public E peekFirst(){ return (E)
list[first_index]; } public E peekLast(){ int index = ((first_index + num_elems) %
list.length) - 1; return (E) list[index]; } public int size(){ return num_elems; }
}.
what is the difference bettween cotranslational import and posttrans.pdf
What is meant by ecological fallacy?
Solution
Answer:
The term \"fallacy\" means the invalid. ecological fallacy tells about the individuals in the
ecosystem which are invalid, it tell statistically about the group of individuals that are not part of
the group which they should belong to.
Eg. The earnings of people in particular district is high, the assumption that every person in that
distict is rich would be fallacy. There are few people with low earnings..
What is the meaning of critical periods in neonatal development Wha.pdf
what is the difference bettween cotranslational import and posttranslational import of proteins
into organelles? of the organelles listed below, which make use of which type of import? 1. ER
2. Mitochondria 3. nucleus 4. peroxisome
Solution
Cotranslational transport involves translocation of a growing peptide chain within the cell
simultaneously like to the ER.
Where as, post translational transport involves initially protein synthesis in the cytosol followed
by transport to ER.
Cotranslational - ER,
Post translational- Mitochondria, nucleus and peroxisome.
What are the major differences between foreign bonds and Eurobonds .pdf
what are some necessary organizational steps to successfully offer acupressure as part of nursing
care in the labor and delivery setting?
Solution
Principles of acupuncture forms the basis of acupressure.
During labour it can be used to relieve pain and increase the efficiency of the labor. From 37th
week onwards all points are safe for acupressure. Lesser pressure as compared to the nirmal
times has to be applied during labor. It should be started during early stages of labor and shall
intermittently be given during the entire process.
There are few specific acupressure points during the delivery and labor. These are lower back
points, large intestine, urinary bladder, kidney and spleen..
What is meant by ecological fallacySolutionAnswerThe term .pdf
What is Incorporation by Reference? and How is it used in a contract?
Solution
Incorporation by reference is an act of including second document within the first document by
giving only reference in first document. Ie if in first document only a single line is kept giving
reference of second document. than the secound document will be part of first document.
Incorporation by reference is often done in creating law as well in contract law nad trust and
estate law
Incorporation by reference is used in contract by way of terms and condition. For example there
are general set of terms and condition to contarct now instead of including all the tems and
condtion to contract a company can just give reference to it, making it binding to the companies
entering the contract.
Thus it is very important to read document if it is included by reference.
What are the two main branches of Non-Euclidean Geometry What is the.pdf
What are the major differences between foreign bonds and Eurobonds? Why do some US
companies prefer to issue Eurobonds to raise US dollars from foreign markets?
Solution
A Eurobond is long term bond issued and sold by the compnies outside the country in which it is
dominated. If US company issued a bond in dollar and selling it to Japanese investors in Japan
then it is know as Eurobond.
Foreign is also sold in the other courtry but it is issued in that country currency as well that\'s a
difference from Eurobond. US company selling and issuing a bond in Yen for Japanese
investors.
US companies issue Eurobonds because in that way US companies can hedge its currency risk..
By what transport method does oxygen enter the blood from the alveol.pdf
By what transport method does oxygen enter the blood from the alveoli? By what method is
oxygen transported to the body tissues from the lungs? By what method does oxygen enter the
body tissues from the blood?
a. Diffusion, bulk flow, diffusion
b. Bulk flow, bulk flow, diffusion
c. Bulk flow, diffusion, carrier proteins
d. Diffusion, diffusion, carrier proteins
e. Bulk flow, diffusion, bulk flow
Solution
b. Bulk flow, bulk flow, diffusion
bulk flow carries air or water, relatively high in O2 and low in CO2, past a respiratory surface.
The flow is usually propelled by muscular movements. Bulk flow of blood transports gases
between the respiratory system and the tissues. Blood is pumped through the body by the heart.
Diffusion transfers O2 out of the capillaries and into nearby tissues, and transfers CO2 from the
tissues into capillaries.
All of these are examples of evidence used to support the hypothesis.pdf
All of these are examples of evidence used to support the hypothesis of endosymbiosis,
EXCEPT:a. Mitochondria and plastids contain circular DNA molecules similar to the circular
chromosomes of bacteriab. Mitochondria and plastids, like prokaryotes, only contain genetic
material in the form of RNAc. Mitochondria and plastids can replicate themselves in a process
similar to prokaryotesd. Mitochondria and plastids have their own ribosomes and enzymes that
can be used to transcribe and translate DNA into proteinse. Mitochondria and plastids have a
double membrane (inner and outer)
Solution
Out of the above options , the correct will be
B. Mitochondria and plastids a have their genetic material in the form of DNA not RNA.
They donot have any such machinary for their expression , so this option is incorrect regarding
endosymbiosis.
Otherwise rest all the sentenses are true.
They have circular genome like Prokaryotes, and can replicate and also have double membrane
layer..
A 28 year-old biologist makes a trip to study life forms in a distan.pdf
A 28 year-old biologist makes a trip to study life forms in a distant planet 17 light-years away.
The round trip including a stay of 1 year in the distant planet takes 38 years according to the
clock on earth. The biologist\'s son is 2.0 year-old when she left. Assume that the planet is not
moving with respect to earth. How fast is the spaceship traveling? Give your answer in terms of
the speed of light. You do not need to enter \"c\" as unit.
How old is she when she returns according to her clock?How old is her son according to the
clock on earth?
What is the total distance traveled according to the biologist?
Give your answer in light-years. You do not need to enter the units.
Solution
As it takes 38 - 1 = 37 years for the biologist to make the trip
Speed = 17 / 38 = 0.45 c
Time dialation that is occurred t\' = t * sqrt(1 - (v2/c2))
t\' = 37 * sqrt(1 - 0.452) = 33.04 years
Biologists age = 28 + 33.04 + 1 = 62.04 years
Her son\'s age = 2.0 + 38 = 40 years.
4. Name the following compounds CO2 CH CH3 CH20H CH3COOH 5. Water ha.pdf
4. Name the following compounds: CO2 CH CH3 CH20H CH3COOH 5. Water has a molecular
weight of 18. A milliliter of liquid water has a mass of one gram. Calculate the molar
concentration of liquid water: 6. What kind of bonds hold together the following compounds
(match): hydrogen bonds water ionic bonds table salt covalent bonds nucleotides in RNA james
bonds strands in dsDNA 7. Fill in the blanks: is transcribed by to make is translated by to make
Solution
Answer:
4)
CO2 : carbondioxide
O2: Oxygen
N2: NItrogen
CH4 : Methane
CH3CH2OH : Ethyl alcohol
CH3COOH: Acetic acid.
4. When examining whether group differences occur on more than one d.pdf
4. When examining whether group differences occur on more than one dependent variable, what
should be used? a. factor analysis b. MANOVA c. t-test for independent means d. t-test for
dependent means
Solution
For examining whether group differences occur on more than one dependent variable MANOVA
or the multivariate analysis of variance is used.
Hope this helps..
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C++ PROGRAMThis program builds on the code below#include iostr.pdf
- 1. C++ PROGRAM This program builds on the code below: #include #include #include "shirtType.h" using namespace std; void getInput(shirtType s[],ifstream& infile, int& numShirts); int main() { int numShirts; shirtType shirts[500]; ifstream infile; infile.open("h:shirtlab.txt"); getInput(shirts,infile,numShirts); for(int i =0; i>id; infile.ignore(100,' '); getline(infile,c); infile>>size>>p>>q; while(infile) { s[index].setAll(id, c, size, p, q); index++; infile>>id; infile.ignore(100,' '); getline(infile,c); infile>>size>>p>>q; } numShirts=index; } See also the description of the class shirtType below: #include #include using namespace std; class shirtType
- 2. { private: int shirtID; string color; string size; double price; int quantity; public: void setAll(int i,string c,string s, double p, int q); void print() const; void newQuan(); shirtType(); shirtType(int,string,string,double,int); }; shirtType::shirtType(int i,string c,string s, double p, int q) { color = c; size = s; if(i>0) shirtID=i; else shirtID=0; if(p>0.0) price = p; else price=0.0; if (q>=0) quantity=q; else quantity = 0; } shirtType::shirtType() {
- 3. shirtID=0; color=""; size=""; price=0.0; quantity=0; } void shirtType::setAll(int i,string c,string s, double p, int q) { color = c; size = s; if(i>0) shirtID=i; else shirtID=0; if(p>0.0) price = p; else price=0.0; if (q>=0) quantity=q; else quantity = 0; } void shirtType::newQuan() { cout<<"Enter the new quantity of shirts: "<>quantity; } void shirtType::print() const { cout<<"Shirt ID: "< Solution #include #include
- 4. struct node { int data; struct node *next; }*head; void append(int nkkm) { struct node *temp,*right; temp= (struct node *)malloc(sizeof(struct node)); temp->data=nkkm; right=(struct node *)head; while(right->next != NULL) right=right->next; right->next =temp; right=temp; right->next=NULL; } void add( int nkkm ) { struct node *temp; temp=(struct node *)malloc(sizeof(struct node)); temp->data=nkkm; if (head== NULL) { head=temp; head->next=NULL; } else {
- 5. temp->next=head; head=temp; } } void addafter(int nkkm, int loc) { int i; struct node *temp,*left,*right; right=head; for(i=1;inext; } temp=(struct node *)malloc(sizeof(struct node)); temp->data=nkkm; left->next=temp; left=temp; left->next=right; return; } void addaShirt(int nkkm) { int c=0; struct node *temp; temp=head; if(temp==NULL) { add(nkkm); } else { while(temp!=NULL) { if(temp->datanext; }
- 6. if(c==0) add(nkkm); else if(cdata==nkkm) { if(temp==head) { head=temp->next; free(temp); return 1; } else { prev->next=temp->next; free(temp); return 1; } } else { prev=temp; temp= temp->next; } } return 0; } void display(struct node *r) { r=head; if(r==NULL) { return; } while(r!=NULL) {
- 7. printf("%d ",r->data); r=r->next; } printf(" "); } int count() { struct node *n; int c=0; n=head; while(n!=NULL) { n=n->next; c++; } return c; } int main() { int i,nkkm; struct node *n; head=NULL; while(1) { printf("1.Add a shirt "); printf("2.Sell a shirt "); printf("3.Restock a shirt "); printf("4.Print a list of all shirts "); printf("5.Quit "); printf("Enter your choice : "); if(scanf("%d",&i)<=0){
- 8. printf("Enter only an Integer "); exit(0); } else { switch(i) { case 1: printf("Enter the number to addaShirt : "); scanf("%d",&nkkm); addaShirt(nkkm); break; case 2: if(head==NULL) { printf("List is Empty "); } else { printf("Element(s) in the list are : "); } display(n); break; case 3: printf("Size of the list is %d ",count()); break; case 4: if(head==NULL) printf("List is Empty "); else{ printf("Enter the number to delete : "); scanf("%d",&nkkm); if(delete(nkkm)) printf("%d deleted successfully ",nkkm); else printf("%d not found in the list ",nkkm); } break; case 5: return 0; default: printf("Invalid option "); } }