#A year is considered a leap year if it abides by the #following rules: # # - Every 4th year IS a leap year, EXCEPT... # - Every
100th year is NOT a leap year, EXCEPT... # - Every 400th year IS a leap year. # #This starts at year 0. For example: # # - 1993 is not a leap year because it is not a multiple of 4. # - 1996 is a leap year because it is a multiple of 4. # - 1900 is not a leap year because it is a multiple of 100, # even though it is a multiple of 4. # - 2000 is a leap year because it is a multiple of 400, # even though it is a multiple of 100. # #Write a function called is_leap_year. is_leap_year should #take one parameter: year, an integer. It should return the #boolean True if that year is a leap year, the boolean False #if it is not. #Write your function here!
1 answer:
Answer:
To check if the year comes under each 100th year, lets check if the remainder when dividing with 100 is 0 or not.
Similarly check for 400th year and multiple 0f 4. The following C program describes the function.
#include<stdio.h>
#include<stdbool.h>
bool is_leap_year(int year);
void main()
{
int y;
bool b;
printf("Enter the year in yyyy format: e.g. 1999 \n");
scanf("%d", &y); // taking the input year in yyyy format.
b= is_leap_year(y); //calling the function and returning the output to b
if(b==true)
{
printf("Thae given year is a leap year \n");
}
else
{
printf("The given year is not a leap year \n");
}
}
bool is_leap_year(int year)
{
if(year%100==0) //every 100th year
{
if(year%400==0) //every 400th year
{
return true;
}
else
{
return false;
}
}
if(year%4==0) //is a multiple of 4
{
return true;
}
else
{
return false;
}
}
Explanation:
Output is given as image
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Answer:
I am writing a Python program.
def concordance(filename): # function that takes a file name as parameter and returns the concordance for that file
file = open(filename,"r") #opens file in read mode
unique={} #creates and empty list
for word in file.read().split(): #loops through every word in the file
if word in unique: # if words is already present in the list
unique[word] += 1 #add 1 to the count of the existing word
else: #if word is not present in the list
unique[word] = 1 #add the word to the list and add 1 to the count of word
file.close(); # close the file
for x in sorted(unique): #sort the list and displays words with frequencies
print(x, unique[x]); #prints the unique words and their frequencies in alphabetical order
#prompts user to enter the name of the file
file_name=input('Enter the input file name: ')
concordance(file_name) #calls concordance method by passing file name to it
Explanation:
The program has a function concordance that takes a text file name as parameter and returns a list of unique words and their frequencies.
The program first uses open() method to open the file in read mode. Here "r" represents the mode. Then creates an empty list named unique. Then the for loop iterates through each word in the file. Here the method read() is used to read the contents of file and split() is used to return these contents as a list of strings. The if condition inside the loop body checks each word if it is already present in the list unique. If the word is present then it adds 1 to the count of that word and if the word is not already present then it adds that unique word to the list and assign 1 as a count to that word. close() method then closes the file. The second for loop for x in sorted(unique): is used to display the list of unique words with their frequencies. sorted() method is used to sort the list and then the loop iterates through each word in the list, through x, which acts like an index variable, and the loop displays each word along with its frequency.
Answer:
So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %
Explanation:
Previous concepts
Input/output operations per second (IOPS, pronounced eye-ops) "is an input/output performance measurement used to characterize computer storage devices like hard disk drives (HDD)"
Solution to the problem
For this case since we have 4GB, but 512 MB are destinated to the operating system, we can begin finding the available RAM like this:
Available = 4096 MB - 512 MB = 3584 MB
Now we can find the maximum simultaneous process than can use with this:
And then we can find the maximum wait I/O that can be tolerated with the following formula:
The expeonent for p = 14 since we got 14 simultaneous processes, and the rate for this case would be 99% or 0.99, if we solve for p we got:
So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %