Answer:
import numpy as np
import matplotlib.pyplot as plt
def calculate_pi(x,y):
points_in_circle=0
for i in range(len(x)):
if np.sqrt(x[i]**2+y[i]**2)<=1:
points_in_circle+=1
pi_value=4*points_in_circle/len(x)
return pi_value
length=np.power(10,6)
x=np.random.rand(length)
y=np.random.rand(length)
pi=np.zeros(7)
sample_size=np.zeros(7)
for i in range(len(pi)):
xs=x[:np.power(10,i)]
ys=y[:np.power(10,i)]
sample_size[i]=len(xs)
pi_value=calculate_pi(xs,ys)
pi[i]=pi_value
print("The value of pi at different sample size is")
print(pi)
plt.plot(sample_size,np.abs(pi-np.pi))
plt.xscale('log')
plt.yscale('log')
plt.xlabel('sample size')
plt.ylabel('absolute error')
plt.title('Error Vs Sample Size')
plt.show()
Explanation:
The python program gets the sample size of circles and the areas and returns a plot of one against the other as a line plot. The numpy package is used to mathematically create the circle samples as a series of random numbers while matplotlib's pyplot is used to plot for the visual statistics of the features of the samples.
I have concluded the answer is a
Answer:
the answer for this is 3*(6+2)/2)
Answer:
The programming code can be found in the explanation part, please go through it.
Explanation:
Code:
#include<stdio.h>
#include<stdlib.h>
#include <pthread.h>
// function check whether a number
// is prime or not
int isPrime(int n)
{
// Corner case
if (n <= 1)
return 0;
// Check from 2 to n-1
for (int i = 2; i < n; i++)
if (n % i == 0)
return 0;
return 1;
}
void* printPrimes(void *vargp)
{
int *n = (int *)vargp;
int i=0;
for (i=2;i<=n;i++)
{
if (isPrime(i)) printf("%d\n", i);
}
}
// Driver Program
int main(int argc, char* argv[])
{
int n = atoi(argv[1]);
pthread_t tid;
pthread_create(&tid, NULL, printPrimes, (void *)n);
pthread_exit(NULL);
return 0;
}
