Q2: Answer the fallowing Suppose you have the Degree, and then find the Result As fallow: * C 1 Degree Result 77 ? 89 4 45 ? 66 6. 95 ? 7 55 8. 9 max degree ? 10 1- Find the result depending on Degree as fallow: Degree in [0:49] → "F" Degree in [50:59] → "E" Degree in [60..69] →"D" Degree in [70..79] →"C" Degree in [80..89] →"B" Degree in [100..90 →"A" 2-Find the max degree value B.
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- c) Using the string T = “EVERY GOOD BOY DESERVES FAVOUR”, find the following:SUBSTRING(T,24,6)//” “// SUBSTRING(T,7,8)// SUBSTRING(T,18,1)Computer Science In Erlang, how do i create this: % in1to10(n, outside_mode)% Given a number n, return true if n is in the range 1..10, inclusive. Unless "outsideMode" is true, in which case return true if the number is less or equal to 1, or greater or equal to 10. in1to10(_,_) -> true.dont reject
- ?rmt the given patterfi 1n C l‘anguageThis Code print the first occurrences of x, Modify the code to print the last occurrences of x and print how many times it exists? ( with explanation for each line if possible) many thanks.int const MULTIPLIER = 5; is a valid way to declare a constant integer variable. Group of answer choices True False ------- Given the following code segment, what is output to the screen? int num1 = 6; int num2 = 4 * num1++; cout << "num1=" << num1 << " num2=" << num2; Group of answer choices num1=7 num2=24 num1=6 num2=24 num1=6 num2=28 Nothing, because the code does not compile.
- 57. What is the postfix of the following infix expression? 3 / A + (B + 12) - C. Group of answer choices 3 A + B + 12 C - 3+A B 12 + C - 3 A + B 12 + C - 3 A + B 12 + - C 3 A + + B 12 C -How can i cout a double if it has .0000 as decimal places as a int or whole number? how can i check if the double is a perfect whole number to then turn into an int? double Student::getAverage() { double avg = ((quiz1 + quiz2 + quiz3) / 3.0);if( avg = )//has all zeros as decimals ){ //turn double into int}return avg; // if 65.0000 = cout << 65} cout << "The student's quiz average is: " << fixed << setprecision(4) << getAverage() <<endl;I got an error message for the answer Bartleby gave me last night as follows Input Answer alphabet = "abcdefghijklmnopqrstuvwxyz"test_dups = ["zzz","dog","bookkeeper","subdermatoglyphic","subdermatoglyphics"]def histogram(s):d=dict()for c in s:if c not in d:d[c]=1 else:d[c]+=1return d;def has_duplicate(string):# return false if each letter in s is not distincth=histogram(string)for k,v in h.items():if v>1:return truereturn falsefor string in test_dups:if has_duplicate(string):printf(string,"has duplicates")else:print(string,"has no duplicates")def main():test_dups_loop()if_name=='_main_'main() THE ERROR MESSAGE said ***TRUE NEEDS TO BE DEFINED. this is all I need from the answer code I received.
- Reorder the following to melts faster to slowest to melt 1. Ice with Sand 2. Ice with nothing 3. Ice with Sugar 4. Ice with SaltCorrect answer will be upvoted else downvoted. you can choose any substring of a containing exactly k characters 1 (and arbitrary number of characters 0) and reverse it. Formally, if a=a1a2…an, you can choose any integers l and r (1≤l≤r≤n) such that there are exactly k ones among characters al,al+1,…,ar, and set a to a1a2…al−1arar−1…alar+1ar+2…an. Find a way to make a equal to b using at most 4n reversals of the above kind, or determine that such a way doesn't exist. The number of reversals doesn't have to be minimized. Input Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤2000). Description of the test cases follows. Each test case consists of three lines. The first line of each test case contains two integers n and k (1≤n≤2000; 0≤k≤n). The second line contains string a of length n. The third line contains string b of the same length. Both strings consist of characters 0 and 1. It is guaranteed that the sum of n over all…Blackout Math is a math puzzle in which you are given an incorrect arithmetic equation. The goal of the puzzle is to remove two of the digits and/or operators in the equation so that the resulting equation is correct. For example, given the equation 6 - 5 = 15 ^ 4/2we can remove the digit 5 and the / operator from the right-hand side in order to obtain the correct equality 6 - 5 = 1 ^ 42. Both sides of the equation now equal to 1. Observe how removing an operator between two numbers (4 and 2) causes the digits of the numbers to be concatenated (42). Here is a more complicated example: 288 / 24 x 6 = 18 x 13 x 8 We can remove digits and operators from either side of the equals sign (either both from one side, or one on each side). In this case, we can remove the 2 from the number 24 on the left-hand side and the 1 from the number 13 on the right-hand side to obtain the correct equality 288 / 4 x 6 = 18 x 3 x 8 Both sides of the equation now equal to 432. Here is another puzzle for you…