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- ?rmt the given patterfi 1n C l‘anguageCan you annotate this code. cars = ["Buick","Ford","Honda"]colors = ["Red","Blue","Black","White","Green"]aType = ["Sedan","SUV","Sports","Coupe","Truck"]arr = [[0 for i in range(len(cars))] for j in range(len(cars)*len(colors))]for i in range(0, len(colors)*len(cars)):for j in range(0, len(cars)):if j == 0:arr[i][j] = cars[int(i/5)]if j == 1:arr[i][j] = colors[int(i % 5)]if j == 2:arr[i][j] = aType[int(i % 5)]for i in range(0, len(arr)):print(arr[i])Modify this code in to Mofify the code to display distribution of a continous variable like range of counts and percentage import matplotlib.pyplot as plt # Sample data referral_sources = ['Website', 'Word of Mouth', 'Advertisement', 'Social Media'] counts = [250, 400, 300, 350] plt.bar(referral_sources, counts) plt.title('Referral Sources Count') plt.xlabel('Referral Source') plt.ylabel('Count')
- Fix the following code:Dim dblRegularPay As DoubleDim dblOvertimePay As DoubleintRegularPay = 783.87intOvertimePay = 105.92lbl.TotalPay = (dblRegularPay + dblOvertimePay).ToString ('C')please code in pythonis_phone_num: This Boolean function takes a string and returns True if it is in the formatddd-ddd-dddd, where d is a digit, False otherwise. Example calls:is_phone_num('123-456-7890') --> Trueis_phone_num('123-4556-7890') --> Falseis_phone_num('(123)456-7890') --> FalsePlease written by computer source Comsider the following Inear equstion. y=(1)/(4)x+4
- Accept from the user today’s date and your Date of birth in the format dd-mm-yyyy. Assuming yousleep 6 hours a day, read 10 hours and spend 1.5 hours a day for eating. How many minutes have youspent as of this date for (i) sleeping (ii) eating. The number of years you have spent for reading as ofthis date (amount of time spent for reading in years).Can you fix the error in the python code. import numpy as npimport matplotlib.pyplot as plt N = 20 # number of points to discretizeL = 0.15X = np.linspace(0, L, N) # position along the rodh = L / (N - 1) # discretization spacing C0t = 0.200 # concentration at x = 0Cth = 0.00000409D = 0.0000025t_avg = 0n = 0 tfinal = 300Ntsteps = 1000dt = tfinal / (Ntsteps - 1)t = np.linspace(0, tfinal, Ntsteps) alpha =( D * dt / h**2) C_xt = [] # container for all the time steps # initial condition at t = 0C = np.zeros(X.shape)C[0] = C0t C_xt += [C] for j in range(1, Ntsteps): N = np.zeros(C.shape) N[0] = C0t N[1:-1] = alpha*C[2:] + (1 - 2 * alpha) * C[1:-1] + alpha * C[0:-2] N[-1] = N[-2] # derivative boundary condition flux = 0 C[:] = N C_xt += [N] if ((Cth-0.000001) < N[-1]<(Cth+0.000001)): print ('Time=',t[j],'conc=',[N[-1]],'Cthr=',[Cth]) t_avg = t_avg+t[j] n = n + 1 # plot selective solutions if j in…Convert the following strings into appropriate numeric variables. Hint: you have to call ato* functions. “0” “3.14159” “21 PGECHS” “PGECHS 21” “PGECHS21” “3..14159” “.987” "124z3yu87" "-3.4" – convert to integer “e24.5”
- Print a line of num characters that are made up of balanced curly braces or square brackets with either "*" or "++" in the center. num is a positive integer the desired pattern is displayed on cout no iteration is allowed on this function, but you can print multiple characters together (at one time). Do not add a linefeed. Examples: num pattern // 1 * // 2 ++ // 3 [*] // 4 {++} // 5 [[*]] // 6 {{++}} // 7 [[[*]]] void printPattern2(size_t num){}Given the following declaration : char msg[100] = "Department of Computer Science"; What is printed by: strcpy_s(msg, "University"); int len = strlen(msg); for (int i = len-3; i >0; i--) { msg[i] = 'x'; } cout << msg;String: abcdefghijklmnopqrstuvwxyzyxwvutsr qponmlkjihgfedcba Make a program that: reads a letter from standard input print the above string up to this letter For example, if the letter was c, the output would be: abcba Must use a function and Python