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.
To bleep it out! hopefully this helped
Answer:
(b) (int)(Math.random() * 101)
Explanation:
Given
--- minimum
--- maximum
Required
Java expression to generate random integer between the above interval
The syntax to do this is:
(int)(Math.random((max - min) + 1) + min)
Substitute the values of max and min
(int)(Math.random((100 - 0) + 1) + 0)
Simplify the expression
(int)(Math.random(100 + 1) + 0)
(int)(Math.random(101) + 0)
(int)(Math.random(101))
Hence, the right option is:
(b) (int)(Math.random() * 101)
Answer:
Answer explained
Explanation:
From the previous question we know that while searching for n^(1/r) we don't have to look for guesses less than 0 and greater than n. Because for less than 0 it will be an imaginary number and for rth root of a non negative number can never be greater than itself. Hence lowEnough = 0 and tooHigh = n.
we need to find 5th root of 47226. The computation of root is costlier than computing power of a number. Therefore, we will look for a number whose 5th power is 47226. lowEnough = 0 and tooHigh = 47226 + 1. Question that should be asked on each step would be "Is 5th power of number < 47227?" we will stop when we find a number whose 5th power is 47226.
Answer:
Yes
Explanation:
Depending on the ethernet standard used, the IEEE 802.3 is faster than the WIFI (IEEE 802.11ac).
The ethernet protocol on cabled networks are of different speed based on ethernet standard which ranges from 10 Mega-bits to 100 Giga-bits per second. This protocol is found in the physical layer of the OSI model.
The wifi 802.11ac also known as wifi 5 is a wireless connection medium in the physical layer of the OSI model. It has a range of aggregate speed capacity of 433 mega-bits per second to 6.77 giga-bits per second.
