Implement the simulation of a biased 6-sided die which takes the values 1,2,3,4,5,6 with probabilities 1/8,1/12,1/8,1/12,1/12,1/
hjlf
Answer:
see explaination
Explanation:
import numpy as np
import matplotlib.pyplot as plt
a = [1, 2, 3, 4, 5, 6]
prob = [1.0/8.0, 1.0/12.0, 1.0/8.0, 1.0/12.0, 1.0/12.0, 1.0/2.0]
smls = 1000000
rolls = list(np.random.choice(a, smls, p=prob))
counts = [rolls.count(i) for i in a]
prob_exper = [float(counts[i])/1000000.0 for i in range(6)]
print("\nProbabilities from experiment : \n\n", prob_exper, end = "\n\n")
plt.hist(rolls)
plt.title("Histogram with counts")
plt.show()
check attachment output and histogram
Answer:I really don’t know
Explanation:
Um You can look it up though
Answer:
Explanation:
In Javascript, you can accept an input value by using the prompt() function and saving the input into a variable. In the following lines of code, I have declared the three variables at the beginning and then prompted the user to enter a value for each and saved the values in the correct variables. In Javascript length is a keyword so I used len instead.
let base, height, len;
base = prompt("Enter Base value: ");
height = prompt("Enter Height value: ");
len = prompt("Enter Length value: ");
Answer:
See explanation below
Explanation:
Previos concepts
First Come First Serve (FCFS) "is an operating system scheduling algorithm that automatically executes queued requests and processes in order of their arrival".
Shortest job next (SJN), or the shortest job first (SJF) or shortest "is a scheduling policy that selects for execution the waiting process with the smallest execution time".
Shortest remaining time (SRF) "is a scheduling method that is a preemptive version of shortest job next scheduling'".
Round robin (RR) is an algorithm where the time parts "are assigned to each process in equal portions and in circular order, handling all processes without priority"
Solution for the problem
Assuming the dataset given on the plot attached.
Part a
For this algorithm the result would be:
Job A 0-6
Job B 6-(6+3) = 6-9
Job C 9-(9+1) = 9-10
Job D 10-(10+4) = 10-14
Part b
For this algorithm the result would be:
Job A 0-6
Job C 6-(6+1) = 6-7
Job B 7-(7+3) = 7-10
Job D 10-(10+4) = 10-14
Part c
For this algorithm the result would be:
Job A 0-1 until 14
Job B 2-(2+3) = 2-5
Job C 3-(3+2) = 3-5
Job D 9-(9+5) = 9-14
Part d
For this algorithm the result would be:
Job A 0-2 , 7-9, 12-14
Job B 2-4, 9-10
Job C 4-(4+1) = 4-5
Job D 5-7, 10-12
The answer is (c) infinite loop
