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- Please use java Using two nested loops, initialize the board so that zeros and ones’ alternate, as on a checkerboard:0 1 0 1 0 1 0 11 0 1 0 1 0 1 00 1 0 1 0 1 0 11 0 1 0 1 0 1 00 1 0 1 0 1 0 1…1 0 1 0 1 0 1 0Hint: check whether i + j is evenYou will select one of the problems below and create a program that uses nested selections to solve the problem.Do not use Boolean operators or Java!Write a program that reads three edges for a triangle and computes the perimeter if the input is valid. Otherwise, display that the input is invalid. The input is valid if the sum of every pair of two edges is greater than the remaining edge.The following code examines two variables -- x and y -- and prints the largest odd number among them. If none of them is odd, it prints a message to that effect. def largestOdd(x,y): """ x and y are numbers Returns the maximum odd number of x and y. If they all are even, return None """ foundOdd = False largest = 0 if x % 2 == 1 : largest = x foundOdd = True if y % 2 == 1 : if not foundOdd : largest = y foundOdd = True elif y > largest : largest = y if foundOdd : return largest else : return None Assume that largestOdd is called with numbers as arguments.Which of the following test suites would make a path-complete glass box test suite for largestOdd? Question 6 options: Test Suite A: foundOdd(2,4), foundOdd(2,5), foundOdd(11,12), foundOdd(11,13), foundOdd(11,7) Test Suite B: foundOdd(2,4), foundOdd(2,6), foundOdd(11,12),…
- Hi, Im trying complete my lab for my programming class and im getting and error i cannot disifer. The purpose of this code is to simulate a coin flip and you can determine how many times you want the coin to be filped. import randomdef heads_or_tails(n): for n in range(n): if random.randint(0,1) == 1: print("tails") else: print ("heads") if __name__=="__main__": random.seed(1) number_of_flips = int(input()) heads_or_tails(number_of_flips) i get this error when submitting TypeError: heads_or_tails() missing 1 required positional argument: 'n'Write this using java Write a complete version of the Bresenham Midpoint algorithm to handle ALL slope values. m = 0 m > 0 m < 0 m = 1 m > 1 swap the rolls of x and y m < 1 m = infinity ( needs special test case) Include appropriate tests for ‘special case’ conditions. Insteadof “WritePixel” write to the screen the coordinates of the pixel that would be drawn. Please provide photos that it works and so I can understand the code betterWhen analyzing data sets, such as data for human heights or for human weights, a common step is to adjust the data. This can be done by normalizing to values between 0 and 1, or throwing away outliers. For this program, adjust the values by subtracting the smallest value from all the values. The input begins with an integer indicating the number of integers that follow. Ex: If the input is: 5 30 50 10 70 65 the output is: 20 40 0 60 55 The 5 indicates that there are five values in the list, namely 30, 50, 10, 70, and 65. 10 is the smallest value in the list, so is subtracted from each value in the list. For coding simplicity, follow every output value by a space, including the last one. 272772.1571588
- This assignment does not exactly match the one in the textbook on which it is based. I don't want any counter-controlled loops in your solution. Your solution should be able to detect when the accuracy isn't improving and then display the results of the work (as outlined below): Look at Programming Exercise **5.25 in the textbook on approximating the value of π. It is described as: π = 4 ( 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ... + (-1)^(i+1)/(2i-1) ) Notice that Java has the best possible value of π that can be stored as a type double available as Math.PI (part of the java.lang package). See how close you can get to the best value of π using a loop with the above technique. Notice that the above implies you are testing the value of your own estimate as it approaches the best value available in Java (as a type double). This would not be a pre-determined, counter-controlled loop. You are testing the results (sentinel-controlled) until you reach a value with the minimum discrepancy…A square is divided into four smaller regions as shown in (a). If you throw a dart into the square one million times, what is the probability for the dart to fall into an odd-numbered region? Write a program to simulate the process and display the result.In the following code stream, identify if there are any hazards. Crisply describe if it is possible to get around the hazards, and what the resulting steps will be. lw r2, 0(r1) add r3, r2, r3 sub r1, r1, r4
- Hi, please help with the following java for loop. Rewrite the following for loop so that no variables are used.for (int roll = 1; roll <= ROLLS; rol += 1) { int num1 = die1.roll(); int num2 = die2.roll(); if (num1 == 1 && num2 == 1) { // check for snake eyes count += 1; }}You will select one of the problems below and create a program that uses nested selections to solve the problem.Do not use Boolean operators Do not use java.util.Scanner; Write a program that reads three edges for a triangle and computes the perimeter if the input is valid. Otherwise, display that the input is invalid. The input is valid if the sum of every pair of two edges is greater than the remaining edge.Here is the sample run: Enter three edges: 1,1,1 The perimeter is 3 Enter three edges 1,3,1 The input is invalidThe concentration (C) of insulin in the body is described by the following equation C = C0e^-40t / M Where C0 = 85 pmol/L is the initial insulin concentration, t is time in minutes, M is the person mass in kg. a) Write code that will calculate the insulin concentration of two people with weighs 60 Kg and 100 Kg. Simulate the insulin concentration for 4 minutes in steps of 0.1 minute. b) Plot the insulin concentration of the two people on the same graph. For the first curve plot a magenta solid line, square marker, cyan for MarkerFaceColor, and MarkerSize of 6. For the second curve plot a black dashed line with LineWidth of 2. c) Your plot should have a title, a legend, and appropriate labels for the x and y axis.
