#include stdio.h #include stdlib.h #include time.hdouble.pdf
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#include #include #include double total(int array[], int n) { int i, sum =0; for (i = 0; i < n; i++) { sum = sum + array[i]; } return sum; } int main() { srand(time(NULL)); int i, n, rand_number; printf(\"Enter an integer: \"); scanf(\"%d\",&n); int array[n]; for (i = 0; i < n; i++) { rand_number = rand() % 1000; array[i] = rand_number; } int max = array[0]; int min = array[0]; int min_pos; int max_pos; for (i = 0; i < n; i++) { if(array[i] > max) { max = array[i]; max_pos = i; } if(array[i] < min) { min = array[i]; min_pos = i; } } int sum = total(array,n); double average = sum/n; printf(\"min: %d pos: %d\ \",min, min_pos); printf(\"max: %d pos: %d\ \",max, max_pos); printf(\"sum: %d \ \",sum); printf(\"average: %lf \ \",average); printf(\"Pos\\t|\\tval\ \"); printf(\"-----------------------\ \"); for (i = 0; i < n; ++i) { printf(\"%d\\t|\\t%d\ \",i,array[i]); } printf(\"\ \"); return 0; } /* output: Enter an integer: 30 min: 5 pos: 28 max: 998 pos: 0 sum: 14093 average: 469.000000 Pos | val ----------------------- 0 | 998 1 | 403 2 | 523 3 | 787 4 | 37 5 | 14 6 | 815 7 | 232 8 | 594 9 | 555 10 | 531 11 | 52 12 | 550 13 | 530 14 | 265 15 | 839 16 | 247 17 | 902 18 | 99 19 | 120 20 | 648 21 | 101 22 | 6 23 | 983 24 | 759 25 | 664 26 | 371 27 | 705 28 | 5 29 | 758 */ Solution #include #include #include double total(int array[], int n) { int i, sum =0; for (i = 0; i < n; i++) { sum = sum + array[i]; } return sum; } int main() { srand(time(NULL)); int i, n, rand_number; printf(\"Enter an integer: \"); scanf(\"%d\",&n); int array[n]; for (i = 0; i < n; i++) { rand_number = rand() % 1000; array[i] = rand_number; } int max = array[0]; int min = array[0]; int min_pos; int max_pos; for (i = 0; i < n; i++) { if(array[i] > max) { max = array[i]; max_pos = i; } if(array[i] < min) { min = array[i]; min_pos = i; } } int sum = total(array,n); double average = sum/n; printf(\"min: %d pos: %d\ \",min, min_pos); printf(\"max: %d pos: %d\ \",max, max_pos); printf(\"sum: %d \ \",sum); printf(\"average: %lf \ \",average); printf(\"Pos\\t|\\tval\ \"); printf(\"-----------------------\ \"); for (i = 0; i < n; ++i) { printf(\"%d\\t|\\t%d\ \",i,array[i]); } printf(\"\ \"); return 0; } /* output: Enter an integer: 30 min: 5 pos: 28 max: 998 pos: 0 sum: 14093 average: 469.000000 Pos | val ----------------------- 0 | 998 1 | 403 2 | 523 3 | 787 4 | 37 5 | 14 6 | 815 7 | 232 8 | 594 9 | 555 10 | 531 11 | 52 12 | 550 13 | 530 14 | 265 15 | 839 16 | 247 17 | 902 18 | 99 19 | 120 20 | 648 21 | 101 22 | 6 23 | 983 24 | 759 25 | 664 26 | 371 27 | 705 28 | 5 29 | 758 */.
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![#include
#include
#include
double total(int array[], int n)
{
int i, sum =0;
for (i = 0; i < n; i++)
{
sum = sum + array[i];
}
return sum;
}
int main()
{
srand(time(NULL));
int i, n, rand_number;
printf("Enter an integer: ");
scanf("%d",&n);
int array[n];
for (i = 0; i < n; i++)
{
rand_number = rand() % 1000;
array[i] = rand_number;
}
int max = array[0];
int min = array[0];
int min_pos;
int max_pos;
for (i = 0; i < n; i++)
{
if(array[i] > max)
{](https://image.slidesharecdn.com/includestdio-230411163112-6c386bfc/85/include-stdio-h-include-stdlib-h-include-time-hdouble-pdf-1-320.jpg)
![max = array[i];
max_pos = i;
}
if(array[i] < min)
{
min = array[i];
min_pos = i;
}
}
int sum = total(array,n);
double average = sum/n;
printf("min: %d pos: %d ",min, min_pos);
printf("max: %d pos: %d ",max, max_pos);
printf("sum: %d ",sum);
printf("average: %lf ",average);
printf("Post|tval ");
printf("----------------------- ");
for (i = 0; i < n; ++i)
{
printf("%dt|t%d ",i,array[i]);
}
printf(" ");
return 0;
}
/*
output:
Enter an integer: 30
min: 5 pos: 28
max: 998 pos: 0
sum: 14093
average: 469.000000
Pos | val
-----------------------
0 | 998](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Computer programming subject notes. Quick easy notes for C Programming.Cheat ...
Computer programming subject notes. Quick easy notes for C Programming.Cheat sheet
#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] .
Merge sort in CSolution#include-stdio-h- #define MAX 50 void mergeSort.docx
Merge sort in C
Solution
#include<stdio.h>
#define MAX 50
void mergeSort(int arr[],int low,int mid,int high);
void partition(int arr[],int low,int high);
int main(){
int merge[MAX],i,n;
printf(\"Enter the total number of elements: \");
scanf(\"%d\",&n);
printf(\"Enter the elements which to be sort: \");
for(i=0;i<n;i++){
scanf(\"%d\",&merge[i]);
}
partition(merge,0,n-1);
printf(\"After merge sorting elements are: \");
for(i=0;i<n;i++){
printf(\"%d \",merge[i]);
}
return 0;
}
void partition(int arr[],int low,int high){
int mid;
if(low<high){
mid=(low+high)/2;
partition(arr,low,mid);
partition(arr,mid+1,high);
mergeSort(arr,low,mid,high);
}
}
void mergeSort(int arr[],int low,int mid,int high){
int i,m,k,l,temp[MAX];
l=low;
i=low;
m=mid+1;
while((l<=mid)&&(m<=high)){
if(arr[l]<=arr[m]){
temp[i]=arr[l];
l++;
}
else{
temp[i]=arr[m];
m++;
}
i++;
}
if(l>mid){
for(k=m;k<=high;k++){
temp[i]=arr[k];
i++;
}
}
else{
for(k=l;k<=mid;k++){
temp[i]=arr[k];
i++;
}
}
for(k=low;k<=high;k++){
arr[k]=temp[k];
}
}
Sample output:
Enter the total number of elements: 5
Enter the elements which to be sort: 2 5 0 9 1
After merge sorting elements are: 0 1 2 5 9
.
Program of sorting using shell sort #include stdio.h #de.pdf
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\"Increment=%d \ \",incr);
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
incr=incr-2; /*Decrease the increment*/
}/*End of while*/
printf(\"Sorted list is :\ \");
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
}/*End of main()*/
redix sort
/*Program of sorting using radix sort*/
# include
# include
struct node
{
int info ;
struct node *link;
}*start=NULL;
main()
{
struct node *tmp,*q;
int i,n,item;
printf(\"Enter the number of elements in the list : \");
scanf(\"%d\", &n);
for(i=0;iinfo=item;
tmp->link=NULL;
if(start==NULL) /* Inserting first element */
start=tmp;
else
{
q=start;
while(q->link!=NULL)
q=q->link;
q->link=tmp;
}
}/*End of for*/
printf(\"Unsorted list is :\ \");
display();
radix_sort();
printf(\"Sorted list is :\ \");
display ();
}/*End of main()*/
display()
{
struct node *p=start;
while( p !=NULL)
{
printf(\"%d \", p->info);
p= p->link;
}
printf(\"\ \");
}/*End of display()*/
radix_sort()
{
int i,k,dig,maxdig,mindig,least_sig,most_sig;
struct node *p, *rear[10], *front[10];
least_sig=1;
most_sig=large_dig(start);
for(k = least_sig; k <= most_sig ; k++)
{
printf(\"PASS %d : Examining %dth digit from right \",k,k);
for(i = 0 ; i <= 9 ; i++)
{
rear[i] = NULL;
front[i] = NULL ;
}
maxdig=0;
mindig=9;
p = start ;
while( p != NULL)
{
/*Find kth digit in the number*/
dig = digit(p->info, k);
if(dig>maxdig)
maxdig=dig;
if(diglink = p ;
rear[dig] = p ;
p=p->link;/*Go to next number in the list*/
}/*End while */
/* maxdig and mindig are the maximum amd minimum
digits of the kth digits of all the numbers*/
printf(\"mindig=%d maxdig=%d\ \",mindig,maxdig);
/*Join all the queues to form the new linked list*/
start=front[mindig];
for(i=mindig;ilink=front[i+1];
else
rear[i+1]=rear[i];
}
rear[maxdig]->link=NULL;
printf(\"New list : \");
display();
}/* End for */
}/*End of radix_sort*/
/* This function finds number of digits in the largest element of the list */
int large_dig()
{
struct node *p=start ;
int large = 0,ndig = 0 ;
while(p != NULL)
{
if(p ->info > large)
large = p->info;
p = p->link ;
}
printf(\"Largest Element is %d , \",large);
while(large != 0)
{
ndig++;
large = large/10 ;
}
printf(\"Number of digits in it are %d\ \",ndig);
return(ndig);
} /*End of large_dig()*/
/*This function returns kth digit of a number*/
int digit(int number, int k)
{
int digit, i ;
for(i = 1 ; i <=k ; i++)
{
digit = number % 10 ;
number = number /10 ;
}
return(digit);
}/*End of digit()*/
Solution
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\.
Examples sandhiya class'
This is a simple presentation which focuses on example programs in c on arrays concept.
The products are K2SO4(aq), H2O(l)KOH(aq) + H2SO4(aq) = K2SO4(aq.pdf
The products are: K2SO4(aq), H2O(l)
KOH(aq) + H2SO4(aq) => K2SO4(aq) + H2O(l)
The reaction type is:
D.double replacement
Solution
The products are: K2SO4(aq), H2O(l)
KOH(aq) + H2SO4(aq) => K2SO4(aq) + H2O(l)
The reaction type is:
D.double replacement.
SolutionDividend per share= total dividend paid out to common stoc.pdf
Solution
Dividend per share= total dividend paid out to common stock holders/no of common stock
No of common stock = total value given as per balance sheet/face value of a common stock
= 100000/10
= 10000 common stock
Total dividend paid to common stock holders= 10000 (as per comparative retained earnings
statement)
Dividend per share = 10000/10000
= 1 per common stock
Dividend yield= annual dividend paid per common stock/current price of common stock
Annual dividend paid per common stock= 1per common stock
Current price per share is not given in this question so face price is taken as current price.=10
Dividend yield = 1/10
= 10%.
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Data Structure using C
This document contains some programs of C using Data structures, like Stack, LinkedList, queue, Fibonacci series, addition and multiplication of two matrices,etc.
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Computer programming subject notes. Quick easy notes for C Programming.Cheat ...
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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] .
Merge sort in CSolution#include-stdio-h- #define MAX 50 void mergeSort.docx
Merge sort in C
Solution
#include<stdio.h>
#define MAX 50
void mergeSort(int arr[],int low,int mid,int high);
void partition(int arr[],int low,int high);
int main(){
int merge[MAX],i,n;
printf(\"Enter the total number of elements: \");
scanf(\"%d\",&n);
printf(\"Enter the elements which to be sort: \");
for(i=0;i<n;i++){
scanf(\"%d\",&merge[i]);
}
partition(merge,0,n-1);
printf(\"After merge sorting elements are: \");
for(i=0;i<n;i++){
printf(\"%d \",merge[i]);
}
return 0;
}
void partition(int arr[],int low,int high){
int mid;
if(low<high){
mid=(low+high)/2;
partition(arr,low,mid);
partition(arr,mid+1,high);
mergeSort(arr,low,mid,high);
}
}
void mergeSort(int arr[],int low,int mid,int high){
int i,m,k,l,temp[MAX];
l=low;
i=low;
m=mid+1;
while((l<=mid)&&(m<=high)){
if(arr[l]<=arr[m]){
temp[i]=arr[l];
l++;
}
else{
temp[i]=arr[m];
m++;
}
i++;
}
if(l>mid){
for(k=m;k<=high;k++){
temp[i]=arr[k];
i++;
}
}
else{
for(k=l;k<=mid;k++){
temp[i]=arr[k];
i++;
}
}
for(k=low;k<=high;k++){
arr[k]=temp[k];
}
}
Sample output:
Enter the total number of elements: 5
Enter the elements which to be sort: 2 5 0 9 1
After merge sorting elements are: 0 1 2 5 9
.
Program of sorting using shell sort #include stdio.h #de.pdf
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\"Increment=%d \ \",incr);
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
incr=incr-2; /*Decrease the increment*/
}/*End of while*/
printf(\"Sorted list is :\ \");
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
}/*End of main()*/
redix sort
/*Program of sorting using radix sort*/
# include
# include
struct node
{
int info ;
struct node *link;
}*start=NULL;
main()
{
struct node *tmp,*q;
int i,n,item;
printf(\"Enter the number of elements in the list : \");
scanf(\"%d\", &n);
for(i=0;iinfo=item;
tmp->link=NULL;
if(start==NULL) /* Inserting first element */
start=tmp;
else
{
q=start;
while(q->link!=NULL)
q=q->link;
q->link=tmp;
}
}/*End of for*/
printf(\"Unsorted list is :\ \");
display();
radix_sort();
printf(\"Sorted list is :\ \");
display ();
}/*End of main()*/
display()
{
struct node *p=start;
while( p !=NULL)
{
printf(\"%d \", p->info);
p= p->link;
}
printf(\"\ \");
}/*End of display()*/
radix_sort()
{
int i,k,dig,maxdig,mindig,least_sig,most_sig;
struct node *p, *rear[10], *front[10];
least_sig=1;
most_sig=large_dig(start);
for(k = least_sig; k <= most_sig ; k++)
{
printf(\"PASS %d : Examining %dth digit from right \",k,k);
for(i = 0 ; i <= 9 ; i++)
{
rear[i] = NULL;
front[i] = NULL ;
}
maxdig=0;
mindig=9;
p = start ;
while( p != NULL)
{
/*Find kth digit in the number*/
dig = digit(p->info, k);
if(dig>maxdig)
maxdig=dig;
if(diglink = p ;
rear[dig] = p ;
p=p->link;/*Go to next number in the list*/
}/*End while */
/* maxdig and mindig are the maximum amd minimum
digits of the kth digits of all the numbers*/
printf(\"mindig=%d maxdig=%d\ \",mindig,maxdig);
/*Join all the queues to form the new linked list*/
start=front[mindig];
for(i=mindig;ilink=front[i+1];
else
rear[i+1]=rear[i];
}
rear[maxdig]->link=NULL;
printf(\"New list : \");
display();
}/* End for */
}/*End of radix_sort*/
/* This function finds number of digits in the largest element of the list */
int large_dig()
{
struct node *p=start ;
int large = 0,ndig = 0 ;
while(p != NULL)
{
if(p ->info > large)
large = p->info;
p = p->link ;
}
printf(\"Largest Element is %d , \",large);
while(large != 0)
{
ndig++;
large = large/10 ;
}
printf(\"Number of digits in it are %d\ \",ndig);
return(ndig);
} /*End of large_dig()*/
/*This function returns kth digit of a number*/
int digit(int number, int k)
{
int digit, i ;
for(i = 1 ; i <=k ; i++)
{
digit = number % 10 ;
number = number /10 ;
}
return(digit);
}/*End of digit()*/
Solution
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\.
Examples sandhiya class'
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Merge sort in CSolution#include-stdio-h- #define MAX 50 void mergeSort.docx
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Program of sorting using shell sort #include stdio.h #de.pdf
More from arjunchetri1
The products are K2SO4(aq), H2O(l)KOH(aq) + H2SO4(aq) = K2SO4(aq.pdf
The products are: K2SO4(aq), H2O(l)
KOH(aq) + H2SO4(aq) => K2SO4(aq) + H2O(l)
The reaction type is:
D.double replacement
Solution
The products are: K2SO4(aq), H2O(l)
KOH(aq) + H2SO4(aq) => K2SO4(aq) + H2O(l)
The reaction type is:
D.double replacement.
SolutionDividend per share= total dividend paid out to common stoc.pdf
Solution
Dividend per share= total dividend paid out to common stock holders/no of common stock
No of common stock = total value given as per balance sheet/face value of a common stock
= 100000/10
= 10000 common stock
Total dividend paid to common stock holders= 10000 (as per comparative retained earnings
statement)
Dividend per share = 10000/10000
= 1 per common stock
Dividend yield= annual dividend paid per common stock/current price of common stock
Annual dividend paid per common stock= 1per common stock
Current price per share is not given in this question so face price is taken as current price.=10
Dividend yield = 1/10
= 10%.
Pure risk is a case where the outcome is strictly defined i.e. eithe.pdf
Pure risk is a case where the outcome is strictly defined i.e. either you suffer a loss or you do not
suffer loss. For example, while driving on the road , you may have an accident or you may not at
all.
Types of personal risk:
(i) Personal Risk: Personal risks are those associated with an individual directly. Personal risk
majorly involves covering for risk of premature death, risk of old age, risk of illness and risk of
unemployment.
Covering for risk of premature death protects the financial dependents on the individual in case
he/she passes away unexpectedly. Covering for old age safeguards against income needs in old
age when an individual is unable to work and earn money. Similarly, risk of illness can be
covered by taking up an health insurance should there be any chronic sickness. If you are faced
with unemployment at any point of time in life, this insurance covers the individuals financial
needs during such period.
(ii) Property Risk: Property risk is the risk of a property being stolen, damaged totally or partially
due to manmade or natural causes. Property insurance covers against losses to a tangible
property and reimburses the property in case of actual loss or damage to the property.
Two types of pure risk for which the technique cannot be used effectively:
(i) Insuring for tangible property is certain times difficult since insurers do not specifically cover
risks due to earthquakes, tornados and floods. Insurers do not find it easy to estimate the
possibility of earthquakes and risks associated with it.
(ii) Most insurers only cover a specific list of long term diseases and not all that an individual
may be affected. Therefore health insurance may not be really useful if the policy does not cover
all heath related illness.
Solution
Pure risk is a case where the outcome is strictly defined i.e. either you suffer a loss or you do not
suffer loss. For example, while driving on the road , you may have an accident or you may not at
all.
Types of personal risk:
(i) Personal Risk: Personal risks are those associated with an individual directly. Personal risk
majorly involves covering for risk of premature death, risk of old age, risk of illness and risk of
unemployment.
Covering for risk of premature death protects the financial dependents on the individual in case
he/she passes away unexpectedly. Covering for old age safeguards against income needs in old
age when an individual is unable to work and earn money. Similarly, risk of illness can be
covered by taking up an health insurance should there be any chronic sickness. If you are faced
with unemployment at any point of time in life, this insurance covers the individuals financial
needs during such period.
(ii) Property Risk: Property risk is the risk of a property being stolen, damaged totally or partially
due to manmade or natural causes. Property insurance covers against losses to a tangible
property and reimburses the property in case of actual loss or damage to the property.
.
Pauliss rule Electrons fill orbitals having lowest energy states.pdf
Paulis\'s rule : Electrons fill orbitals having lowest energy states i.e. 1s before 2s
Hund\'s principle : when multiple orbitals of same energy are available, electrons fill the
unoccupied orbitals first and then reoccupy the occupied orbitals by electrons having opposite
spin.
1, 3, and 4 come under Hund\'s rule violation and 2,5 under paulis violation.
1) Electrons should be paired by different spin - hunds rule violation
3) electrons fill the unoccupied orbitals first and then reoccupy the occupied orbitals - hunds rule
violation
4) - hunds rule violation
2) Electrons fill orbitals having lowest energy states i.e. 2s before 2p - paulis rule violation
5) Electrons fill orbitals having lowest energy states i.e.4p before 4d so paulis rule violation
2) Electrons fill orbitals having lowest energy states i.e. 2s before 2p - paulis rule violation
5) Electrons fill orbitals having lowest energy states i.e.4p before 4d so paulis rule violation
Solution
Paulis\'s rule : Electrons fill orbitals having lowest energy states i.e. 1s before 2s
Hund\'s principle : when multiple orbitals of same energy are available, electrons fill the
unoccupied orbitals first and then reoccupy the occupied orbitals by electrons having opposite
spin.
1, 3, and 4 come under Hund\'s rule violation and 2,5 under paulis violation.
1) Electrons should be paired by different spin - hunds rule violation
3) electrons fill the unoccupied orbitals first and then reoccupy the occupied orbitals - hunds rule
violation
4) - hunds rule violation
2) Electrons fill orbitals having lowest energy states i.e. 2s before 2p - paulis rule violation
5) Electrons fill orbitals having lowest energy states i.e.4p before 4d so paulis rule violation
2) Electrons fill orbitals having lowest energy states i.e. 2s before 2p - paulis rule violation
5) Electrons fill orbitals having lowest energy states i.e.4p before 4d so paulis rule violation.
Polymearase Chain Reaction and Electrophoresis can used employed in .pdf
Polymearase Chain Reaction and Electrophoresis can used employed in taxonomic identification
of DNA samples.
The test DNA is extracted -> DNA is amplified -> PCR -> Gel Electrophoresis -> Interpretation
of results -> Species identification.
The test DNA can be compared with another DNA sample to compare similarity by DNA
hybridization technique. If both the single stranded DNA hybridize and form a double helix, it
indicates, they are closely related and if hybridization does not happen, it indicates they are
distantly related. The amount of hybridization can be measured by reading the absorbance.
Single stranded DNA absorbs more UV radiation than double stranded DNA. Hybridization is of
low sensitivity. Hence RFLP is employed to compare DNA for taxonomic identification.
The RFLP analysis is widely used for the detection of interspecies variation at the DNA
sequence level. It consists in the generation of species-specific band profiles through the
digestion of DNA with one or more restriction endonucleases. These restriction enzymes cleave
the DNA molecule at specific 4-6 base pair (bp) recognition sites, originating a set of fragments
with different lengths that could be separated according to their molecular size by conventional
gel electrophoresis. The RFLP banding pattern could be visualized by hybridizing restriction
fragments with a labelled probe in a solid support (for instance, by Southern blotting) or by
treating the electrophoretic gel with ethidium bromide or silver staining.
Solution
Polymearase Chain Reaction and Electrophoresis can used employed in taxonomic identification
of DNA samples.
The test DNA is extracted -> DNA is amplified -> PCR -> Gel Electrophoresis -> Interpretation
of results -> Species identification.
The test DNA can be compared with another DNA sample to compare similarity by DNA
hybridization technique. If both the single stranded DNA hybridize and form a double helix, it
indicates, they are closely related and if hybridization does not happen, it indicates they are
distantly related. The amount of hybridization can be measured by reading the absorbance.
Single stranded DNA absorbs more UV radiation than double stranded DNA. Hybridization is of
low sensitivity. Hence RFLP is employed to compare DNA for taxonomic identification.
The RFLP analysis is widely used for the detection of interspecies variation at the DNA
sequence level. It consists in the generation of species-specific band profiles through the
digestion of DNA with one or more restriction endonucleases. These restriction enzymes cleave
the DNA molecule at specific 4-6 base pair (bp) recognition sites, originating a set of fragments
with different lengths that could be separated according to their molecular size by conventional
gel electrophoresis. The RFLP banding pattern could be visualized by hybridizing restriction
fragments with a labelled probe in a solid support (for instance, by Southern blotting) or by
treating the electrophoretic gel with .
P& loss ac drTo provision for social securituy taxes (amount of g.pdf
P& loss a/c dr
To provision for social securituy taxes (amount of gross pay paid is not mentioned)
Provision for social security taxes are made.
2- profit & loss A/c debit 110.25
To provision for medicare taxes 110.25
Provision for medicare expenses are made
3- profit & loss a/c. debit 245
to provision for FUTA taxes 42
to provision for suta taxes 203
Provision for suta and futa taxes made
(Provisions for SUTA and FUTA taxes made)
4-provision for social security taxes debit (amount)
Provision for medical care taxes debit 110.24
Provision for suta taxes debit 203
Provision for futa taxes debit 42
to various taxes paid 345.24
(Taxes paid to tax authorities)
Total tax liability is 345.24 except social security taxes as amount is not specified
Solution
P& loss a/c dr
To provision for social securituy taxes (amount of gross pay paid is not mentioned)
Provision for social security taxes are made.
2- profit & loss A/c debit 110.25
To provision for medicare taxes 110.25
Provision for medicare expenses are made
3- profit & loss a/c. debit 245
to provision for FUTA taxes 42
to provision for suta taxes 203
Provision for suta and futa taxes made
(Provisions for SUTA and FUTA taxes made)
4-provision for social security taxes debit (amount)
Provision for medical care taxes debit 110.24
Provision for suta taxes debit 203
Provision for futa taxes debit 42
to various taxes paid 345.24
(Taxes paid to tax authorities)
Total tax liability is 345.24 except social security taxes as amount is not specified.
One month liquidity index=10(.981)+90(.85.92)=0.93Liquidity .pdf
One month liquidity index=10%*(.98/1)+90%*(.85/.92)=0.93
Liquidity index value lies between 0 to 1. Smaller the liquidity index signifies higher liquidity
risk.
Solution
One month liquidity index=10%*(.98/1)+90%*(.85/.92)=0.93
Liquidity index value lies between 0 to 1. Smaller the liquidity index signifies higher liquidity
risk..
Intranet Portals1) Intranet portals now a days are at the epicent.pdf
Intranet Portals:
1) Intranet portals now a days are at the epicenter of the enterprose universe.
2) They provide the both utility as well as the usability necessities for empployees success.
3) popular enterprise portal are used via mobile phones and home computers.
4) Most of the organisations use in the digital workplace, intranet portals are beginning to serve
as the hub of the corporate wheel.
5) Not all organisations have achiebed the dream of total integration of all the enterprose
applications due to resource constraints and security concerns.
6) By updating the Intranet portals, most of the organisations want to offer thier employees the
best possible tools and experience with their enterprise portals.
7) Portals included in the Report are:
7.1) The Carle Foundation
7.2) City of Olathe, Kansas
7.3) Coca-Cola Enterprises Ltd.
7.4) Consumer Financial Protection Bureau (CFPB)
7.5) FDC
Solution
s, Inc.
7.6) Fraunhofer-Gesellschaft zur Förderung der angewandten Forschung e.V.
7.7) Fraunhofer Heinrich Hertz Institute
7.8) Municipal Design and Survey Unitary Enterprise “Minskinzhproekt”
7.9) National Archives and Records Administration (NARA)
7.10) Northern Arizona University (NAU)
7.11) Palm Beach County Board of County Commissioners
7.12) Persistent Systems Limited
7.13) Resource Data, Inc.
7.14) Think Mutual Bank
7.15) Department of Transport (Canada)
7.16) Yara International ASA
WDA\'s:
1) WDAS-FM is an Urban Adult Contemporary radio station that features R&B and Classic
Soul.
2) It is licensed to the city of Philadelphia.
3) It is under ownership of iHeartMedia.
4) The station is widely regarded as one of the originators of the Urban AC format which mixes
R&B oldies with non-rap contemporary R&B.
5) It is found in many major markets across the United States.
6) WDAS broadcasts at an effective radiated power of 16,500 watts (16.5 kilowatts (kW)) from a
class B signal from a tower located in the Roxborough section of Philadelphia.
7) Its studios are located in Bala Cynwyd.
8) The main competitor of WDAS in the Philadelphia market is WRNB and its format
complements WUSL and WDAS-AM..
import java.awt.; Using AWT container and component classes im.pdf
import java.awt.*; // Using AWT container and component classes
import java.awt.event.*; // Using AWT event classes and listener interfaces
// An AWT program inherits from the top-level container java.awt.Frame
public class GradeCalculator extends Frame implements ActionListener {
private Label e1,e2,e3,e4,a1,a2,a3,a4,grade; // Declare a Label component
private TextField es1,es2,es3,es4,as1,as2,as3,as4,gradeText; // Declare a TextField component
private int maxScore = 550;
private Button btnCount; // Declare a Button component
// Constructor to setup GUI components and event handlers
public GradeCalculator() {
setLayout(new FlowLayout());
e1 = new Label(\"Exam1 Score(0-50):\"); // construct the Label component
add(e1); // \"super\" Frame adds Label
es1 = new TextField(\"0\", 10); // construct the TextField component
es1.setEditable(true);
add(es1); // \"super\" Frame adds TextField
e2 = new Label(\"Exam2 Score(0-50):\"); // construct the Label component
add(e2); // \"super\" Frame adds Label
es2 = new TextField(\"0\", 10); // construct the TextField component
es2.setEditable(true);
add(es2); // \"super\" Frame adds TextField
e3 = new Label(\"Exam3 Score(0-50)\"); // construct the Label component
add(e3); // \"super\" Frame adds Label
es3 = new TextField(\"0\", 10); // construct the TextField component
es3.setEditable(true);
add(es3); // \"super\" Frame adds TextField
e4 = new Label(\"Exam4 Score(0-50)\"); // construct the Label component
add(e4); // \"super\" Frame adds Label
es4 = new TextField(\"0\", 10); // construct the TextField component
es4.setEditable(true);
add(es4); // \"super\" Frame adds TextField
a1 = new Label(\"Assignment1 Score(0-75):\"); // construct the Label component
add(a1); // \"super\" Frame adds Label
as1 = new TextField(\"0\", 10); // construct the TextField component
as1.setEditable(true);
add(as1); // \"super\" Frame adds TextField
a2 = new Label(\"Assignment2 Score(0-75):\"); // construct the Label component
add(a2); // \"super\" Frame adds Label
as2 = new TextField(\"0\", 10); // construct the TextField component
as2.setEditable(true);
add(as2); // \"super\" Frame adds TextField
a3 = new Label(\"Assignment3 Score(0-75):\"); // construct the Label component
add(a3); // \"super\" Frame adds Label
as3 = new TextField(\"0\", 10); // construct the TextField component
as3.setEditable(true);
add(as3); // \"super\" Frame adds TextField
a4 = new Label(\"Assignment4 Score(0-75):\"); // construct the Label component
add(a4); // \"super\" Frame adds Label
as4 = new TextField(\"0\", 15); // construct the TextField component
as4.setEditable(true);
add(as4); // \"super\" Frame adds TextField
grade = new Label(\"Grade\");
add(grade);
gradeText = new TextField(\"Grade\",15);
add(gradeText);
btnCount = new Button(\"Calculate\");
add(btnCount);
btnCount.addActionListener(this);
setTitle(\"Calculate Grade\"); // Frame sets its title
setSize(175, 700); // Frame sets its initial window size
setVisible(true); // Frame shows
}
// The ent.
If possible let T+e1+e2+e2 be non planarThen either one of e1, e2 .pdf
If possible let T+e1+e2+e2 be non planar
Then either one of e1, e2 or e3 or T must be in another plane
T is a tree of order atleast 4.
i.e maximum number of children per root is more than or equal to 4.
Hence vertices containing e1, e2, e3 have to be planar.
T being a tree with root is also in the same plane
Hence proved
Solution
If possible let T+e1+e2+e2 be non planar
Then either one of e1, e2 or e3 or T must be in another plane
T is a tree of order atleast 4.
i.e maximum number of children per root is more than or equal to 4.
Hence vertices containing e1, e2, e3 have to be planar.
T being a tree with root is also in the same plane
Hence proved.
How should the judicial system treat juvenile criminal offendersJ.pdf
How should the judicial system treat juvenile criminal offenders?
Juveniles must be prevented from having frequent contact with the criminal justice system and
intervening to support juveniles desist from crime. They should be assisted to grow out of crime
are the ways in which the judicial system should treat juvenile criminal offenders.
How does adolescent brain science influence juvenile justice cases?
Juveniles usually offense against property than offense against the person. The juvenile offenses
are mostly attention seeking and need driven. Hence the adolescent brain science treats the
juvenile justice cases as a crime where they are mature and do not think about the immature
brain.
What are some of the consequences of relying on brain science to defend adolescents?
Some defendants deserve special consideration because they have brains that are immature or
impaired, due to which a specific age group cannot be brought into consideration, some
individuals brain will be completely matured by the age of 16 hence can be considered as adult.
It will also provide way for many individuals to escape from the justice for their offense. For
example, if a prosecutor used an MRI scan to show that a 16-year-old who committed a capital
crime had a very mature brain, \"Would we then insist that we execute that juvenile?\"
Solution
How should the judicial system treat juvenile criminal offenders?
Juveniles must be prevented from having frequent contact with the criminal justice system and
intervening to support juveniles desist from crime. They should be assisted to grow out of crime
are the ways in which the judicial system should treat juvenile criminal offenders.
How does adolescent brain science influence juvenile justice cases?
Juveniles usually offense against property than offense against the person. The juvenile offenses
are mostly attention seeking and need driven. Hence the adolescent brain science treats the
juvenile justice cases as a crime where they are mature and do not think about the immature
brain.
What are some of the consequences of relying on brain science to defend adolescents?
Some defendants deserve special consideration because they have brains that are immature or
impaired, due to which a specific age group cannot be brought into consideration, some
individuals brain will be completely matured by the age of 16 hence can be considered as adult.
It will also provide way for many individuals to escape from the justice for their offense. For
example, if a prosecutor used an MRI scan to show that a 16-year-old who committed a capital
crime had a very mature brain, \"Would we then insist that we execute that juvenile?\".
coordinated movements of gastrulation begin. This dramatic process t.pdf
coordinated movements of gastrulation begin. This dramatic process transforms the simple
hollow ball of cells into a multilayered structure with a central gut tube and bilateral
symmetry.many of the cells on the outside of the embryo are moved inside it. Subsequent
development depends on the interactions of the inner, outer, and middle layers of cells thus
formed: the endoderm on the inside, consisting of the cells that have moved into the interior to
form the primitive gut; the ectoderm on the outside, consisting of cells that have remained
external; and the mesoderm between them, consisting of cells that detach from the epithelium to
form a more loosely organized embryonic connective tissue.
eventual pattern of muscles-in the limbs, for example-is determined by the routes that the
migrant cells follow and the selection of sites that they colonize. The embryonic connective
tissues form the framework through which the myoblasts travel and provide signals that guide
their distribution. No matter which somite they come from, myoblasts that migrate into a
forelimb bud will form the pattern of muscles appropriate to a forelimb, and those that migrate
into a hindlimb bud will form the pattern appropriate to a hindlimb.
Limb formation results from series of epithelial-mesenchymal inductions between the
mesenchymal cells of the lateral plate mesoderm and the overlying ectodermal cells. Cells from
the lateral plate mesoderm and the myotome migrate to the limb field and proliferate to create the
limb bud. The lateral plate cells produce the cartilaginous and skeletal portions of the limb while
the myotome cells produce the muscle components. The lateral plate mesodermal cells secrete a
fibroblast growth factor (FGF7 andFGF10, presumably) to induce the overlying ectoderm to
form an important organizing structure called the apical ectodermal ridge(AER).The AER
reciprocatively secretes FGF8 and FGF4 which maintains the FGF10 signal and induces
proliferation in the mesoderm.[citation needed] The position of FGF10 expression is regulated
by Wnt8c in the hindlimb and Wnt2b in the forelimb. The forelimb and the hindlimb are
specified by their position along the anterior/posterior axis and possibly by two T-box containing
transcription factors: Tbx5 and Tbx4, respectively.
DKK1 activation in osteoblasts is the underlying cause of glucocorticoid- and estrogen
deficiency–mediated osteoporosis, and at least partially underlies the teratogenic effects of
thalidomide on limb development.
both thalidomide-induced PCD and limb deformities could be partially inhibited by blocking
Dkk1 or activating Wnt signaling downstream of the ligand-receptor interaction
Solution
coordinated movements of gastrulation begin. This dramatic process transforms the simple
hollow ball of cells into a multilayered structure with a central gut tube and bilateral
symmetry.many of the cells on the outside of the embryo are moved inside it. Subsequent
development depends on the interactions .
Cu(II)Oxide has solid with mono clinic structure .m.p is 12010 C and.pdf
Cu(II)Oxide has solid with mono clinic structure .m.p is 12010 C and b.p is 2000 0C.
It is insoluble in water.it appears as block powder.
Cu (II) nitrate has solid with orthorhobic structure . it appears in blue color.
m.p 2560C an d b.p is 1700 C.it is very soluble in water.
Cu (II) hydroxide is a blue green solid and m.p is 800 C. it is soluble in NH4OH.
Cu(II) sulfate is blue color solid with m.p is 1100 C . It has triclinic crystal structure.
it is soluble in water and it has octahedral geometry.
Solution
Cu(II)Oxide has solid with mono clinic structure .m.p is 12010 C and b.p is 2000 0C.
It is insoluble in water.it appears as block powder.
Cu (II) nitrate has solid with orthorhobic structure . it appears in blue color.
m.p 2560C an d b.p is 1700 C.it is very soluble in water.
Cu (II) hydroxide is a blue green solid and m.p is 800 C. it is soluble in NH4OH.
Cu(II) sulfate is blue color solid with m.p is 1100 C . It has triclinic crystal structure.
it is soluble in water and it has octahedral geometry..
Assign a major brain structure to each of the cehalon belowdiencep.pdf
Assign a major brain structure to each of the cehalon below
diencephalon
The diencephalon is a piece of the prosencephalon (forebrain), which creates from the first
essential cerebral vesicle. The prosencephalon separates into a caudal diencephalon and rostral
telencephalon. The cerebral sides of the equator create from the sides of the telencephalon, each
containing a parallel ventricle. The diencephalon comprises of structures that are sidelong to the
third ventricle, and incorporates the thalamus, the hypothalamus, the epithalamus and the
subthalamus.The diencephalon comprises of the accompanying structures:
Myelencephalon
The myelencephalon or afterbrain is the most back area of the embryonic hindbrain, from which
the medulla oblongata creates
it comprises of the accompanying structures: Tectum, Cerebral peduncle, Pretectum, Cerebral
water system.
metacephalon
The metencephalon is the embryonic part of the hindbrain that separates into the pons and the
cerebellum. It contains
diencephalon
The diencephalon is a piece of the prosencephalon (forebrain), which creates from the first
essential cerebral vesicle. The prosencephalon separates into a caudal diencephalon and rostral
telencephalon. The cerebral sides of the equator create from the sides of the telencephalon, each
containing a parallel ventricle. The diencephalon comprises of structures that are sidelong to the
third ventricle, and incorporates the thalamus, the hypothalamus, the epithalamus and the
subthalamus.The diencephalon comprises of the accompanying
structures:ThalamusHypothalamus including the NeurohypophysisEpithalamus which comprises
ofForemost and Posterior Paraventricular coresAverage and parallel Habenular coresStria
medullaris thalamiBack commissurePineal bodySubthalamus
Myelencephalon
The myelencephalon or afterbrain is the most back area of the embryonic hindbrain, from which
the medulla oblongata creates
it comprises of the accompanying structures: Tectum, Cerebral peduncle, Pretectum, Cerebral
water system.
metacephalon
The metencephalon is the embryonic part of the hindbrain that separates into the pons and the
cerebellum. It containsa segment of the fourth ventricle andthe trigeminal nerve (CN
V),abducens nerve (CN VI),facial nerve (CN VII), anda bit of the vestibulocochlear nerve (CN
VIII).
Solution
Assign a major brain structure to each of the cehalon below
diencephalon
The diencephalon is a piece of the prosencephalon (forebrain), which creates from the first
essential cerebral vesicle. The prosencephalon separates into a caudal diencephalon and rostral
telencephalon. The cerebral sides of the equator create from the sides of the telencephalon, each
containing a parallel ventricle. The diencephalon comprises of structures that are sidelong to the
third ventricle, and incorporates the thalamus, the hypothalamus, the epithalamus and the
subthalamus.The diencephalon comprises of the accompanying structures:
Myelencephalon
The myelencephalon or afterbrain is the m.
B. Assets are listed in descending order of liquidity.The most liq.pdf
B. Assets are listed in descending order of liquidity.
The most liquid assets are presented first, followed by less liquid assets.
Current assets = net working capital - current liabilities.(so A is false)
Share holders equity = total assets - total liabilities (so C D and E are incorrect)
Solution
B. Assets are listed in descending order of liquidity.
The most liquid assets are presented first, followed by less liquid assets.
Current assets = net working capital - current liabilities.(so A is false)
Share holders equity = total assets - total liabilities (so C D and E are incorrect).
ans) 2 is the true oneone parentheses one thied close parentheses .pdf
ans) 2 is the true one
one parentheses one thied close parentheses to the power of n equals theta left parenthesis one
right parentheses
Solution
ans) 2 is the true one
one parentheses one thied close parentheses to the power of n equals theta left parenthesis one
right parentheses.
Answer B. FiberExplanationFiber optic cable is used by the Int.pdf
Answer: B. Fiber
Explanation:
Fiber optic cable is used by the Internet backbone and transmits digital data at a rate of billions
of bits per second using laser technology
Solution
Answer: B. Fiber
Explanation:
Fiber optic cable is used by the Internet backbone and transmits digital data at a rate of billions
of bits per second using laser technology.
Answer The outer earPinnaEar canalThe Middle EarTympanic .pdf
Answer :
The outer ear
Pinna
Ear canal
The Middle Ear
Tympanic Membrane
The tympanic membrane or eardrum serves as a divider between the outer ear and the middle
ear structures. It is gray-pink in color when healthy and consists of three very thin layers of
living tissue.
The eardrum is very sensitive to sound waves and vibrates back and forth as the sound waves
strike it.
Middle Ear Cavity
The middle ear cavity is located in the mastoid process of the temporal bone. The middle ear
cavity is actually an extension of the nasopharynx via the eustachian tube.
Eustachian Tube
The eustachian tube acts as an air pressure equalizer and ventilates the middle ear. When the air
pressure between the outer and middle ear is unequal, the eardrum is forced outward or inward
causing discomfort and the ability of the eardrum to transmit sound is reduced.
Ossicular Chain
The middle ear is connected and transmits sound to the inner ear via the ossicular chain. The
ossicular chain amplifies a signal approximately 25 decibels as it transfers signals from the
tympanic membrane to the inner ear.
THE INNER EAR
The inner ear is composed of the sensory organ for hearing—the cochlea, as well as for
balance—the vestibular system. The systems are separate, yet both are encased in the same bony
capsule and share the same fluid systems.
Vestibular or Balance System
The balance part of the ear is referred to as the vestibular apparatus. It is composed, in part, of
three semicircular canals located within the inner ear. The vestibular system helps to maintain
balance, regardless of head position or gravity, in conjunction with eye movement and
somatosensory input. The semicircular canals are innervated by the VIIIth cranial nerve.
Cochlea
The hearing part of the inner ear is the cochlea. The cochlea is spiral-shaped, similar to the
shape of a snail.
The cochlear duct contains the Basilar membrane upon which lies the Organ of Corti. The Organ
of Corti is a sensory organ essential to hearing. It consists of approximately 30,000 finger-like
projections of cilia that are arranged in rows. These cilia are referred to as hair cells. Each hair
cell is connected to a nerve fiber that relays various impulses to the cochlear branch of the VIIIth
cranial nerve or auditory nerve.
The apical portion of the basilar membrane (the most curled area of the cochlea) transfers lower
frequency impulses. The basal end relays higher frequency impulses.
The VIII cranial nerve (VIII C.N.) or auditory C.N. carries the impulses generated from the
Organ of Corti to the brainstem. From the brainstem, nerve pathways extend through numerous
nuclei to the cerebral cortex in the temporal lobes of the brain.
Solution
Answer :
The outer ear
Pinna
Ear canal
The Middle Ear
Tympanic Membrane
The tympanic membrane or eardrum serves as a divider between the outer ear and the middle
ear structures. It is gray-pink in color when healthy and consists of three very thin layers of
living tissue.
The eardrum is very se.
the carbon 2 and carbon 3 are having one double b.pdf
the carbon 2 and carbon 3 are having one double bond each so the correct
hybrization is :- b. sp2 - sp2
Solution
the carbon 2 and carbon 3 are having one double bond each so the correct
hybrization is :- b. sp2 - sp2.
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#include stdio.h #include stdlib.h #include time.hdouble.pdf
- 1. #include #include #include double total(int array[], int n) { int i, sum =0; for (i = 0; i < n; i++) { sum = sum + array[i]; } return sum; } int main() { srand(time(NULL)); int i, n, rand_number; printf("Enter an integer: "); scanf("%d",&n); int array[n]; for (i = 0; i < n; i++) { rand_number = rand() % 1000; array[i] = rand_number; } int max = array[0]; int min = array[0]; int min_pos; int max_pos; for (i = 0; i < n; i++) { if(array[i] > max) {
- 2. max = array[i]; max_pos = i; } if(array[i] < min) { min = array[i]; min_pos = i; } } int sum = total(array,n); double average = sum/n; printf("min: %d pos: %d ",min, min_pos); printf("max: %d pos: %d ",max, max_pos); printf("sum: %d ",sum); printf("average: %lf ",average); printf("Post|tval "); printf("----------------------- "); for (i = 0; i < n; ++i) { printf("%dt|t%d ",i,array[i]); } printf(" "); return 0; } /* output: Enter an integer: 30 min: 5 pos: 28 max: 998 pos: 0 sum: 14093 average: 469.000000 Pos | val ----------------------- 0 | 998
- 3. 1 | 403 2 | 523 3 | 787 4 | 37 5 | 14 6 | 815 7 | 232 8 | 594 9 | 555 10 | 531 11 | 52 12 | 550 13 | 530 14 | 265 15 | 839 16 | 247 17 | 902 18 | 99 19 | 120 20 | 648 21 | 101 22 | 6 23 | 983 24 | 759 25 | 664 26 | 371 27 | 705 28 | 5 29 | 758 */ Solution #include #include #include
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- 5. { min = array[i]; min_pos = i; } } int sum = total(array,n); double average = sum/n; printf("min: %d pos: %d ",min, min_pos); printf("max: %d pos: %d ",max, max_pos); printf("sum: %d ",sum); printf("average: %lf ",average); printf("Post|tval "); printf("----------------------- "); for (i = 0; i < n; ++i) { printf("%dt|t%d ",i,array[i]); } printf(" "); return 0; } /* output: Enter an integer: 30 min: 5 pos: 28 max: 998 pos: 0 sum: 14093 average: 469.000000 Pos | val ----------------------- 0 | 998 1 | 403 2 | 523 3 | 787 4 | 37
- 6. 5 | 14 6 | 815 7 | 232 8 | 594 9 | 555 10 | 531 11 | 52 12 | 550 13 | 530 14 | 265 15 | 839 16 | 247 17 | 902 18 | 99 19 | 120 20 | 648 21 | 101 22 | 6 23 | 983 24 | 759 25 | 664 26 | 371 27 | 705 28 | 5 29 | 758 */