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.
Answer:
0 1 2 However many you like
✓ I think its 2x they will start to define the program
When resizing an image or an object in a presentation, a user should not utilize the sizing handles in the middle of the sides or the bottom of the image. There's nothing wrong with resizing this way.
There's nothing wrong with resizing this way
<u>Explanation:</u>
When the sizing handles in the middle of the sides are used, the image stretches out reflecting an increase in the width maintaining There's nothing wrong with resizing this way.
On the other hand, while using the sizing handles at the bottom, the image increases in height whilst maintaining the same width again disturbing the aspect ratio.
Answer:
/*
I don't know what language you're using, so I'll write it in javascript which is usually legible enough.
*/
console.log(buildSequence(30));
function buildSequence(maxVal){
maxVal = Math.abs(maxVal);
var n, list = [];
for(n = 1; n < maxVal; n++){
/*
to check for odd numbers, we only need to know if the last bit
is a 1 or 0:
*/
if(n & 1){ // <-- note the binary &, as opposed to the logical &&
list[list.length] = n;
}else{
list[list.length] = -n;
}
}
return list.implode(',');
}
Answer:
Create web and mobile applacations
Explanation:
