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Stels [109]
3 years ago
5

We will pass in a value N. Ouput every even value between 0 and N. Tip: you may remember the modulo operator. 10 % 3 returns the

remainder of 10 divided by 3. You can use this in a decision to determine whether a number is odd or even. Just think about what makes an even number even and all should become clear!
Computers and Technology
1 answer:
Taya2010 [7]3 years ago
6 0

Answer:

def odd_even(N):

   for number in range(0, N+1):

       if number % 2 == 0:

           print(str(number) + " is even")

       elif number % 2 == 1:

           print(str(number) +" is odd")

odd_even(5)

Explanation:

- Create a method called odd_even that takes one parameter, N

- Initialize a for loop that iterates through 0 to N

Inside the loop, check if module of the number with respect to 2 is equal to 0. If yes, print it as even number. If module of the number with respect to 2 is equal to 1, print it as odd number

Call the method

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Write a program called DeliveryCharges for the package delivery service. The program should use an array that holds the 10 zip c
Vlad [161]

Answer:

import java.util.Scanner;

public class DeliveryCharges

{

public static void main(String[] args) {

 

       String[] zips = {"01234", "11234", "21234", "31234", "41234", "51234", "61234", "71234", "81234", "91234"};

    double[] prices = {2.2, 1.0, 3.6, 6, 9, 7.1, 0.8, 4.7, 3.3, 5.2};

    int index = -1;

    Scanner ob = new Scanner(System.in);

    System.out.print("Enter the zip code for your delivery: ");

    String zip = ob.next();

   

    for (int i=0; i<10; i++) {

        if (zip.equals(zips[i])) {

            index = i;

        }

    }

       

    if (index!= -1)

       System.out.println("Delivery charge to " + zips[index] + " is: " + prices[index]);

    else

       System.out.println("No delivery to " + zip);

}

}

Explanation:

Initialize the zips and prices

Initialize index that represents the index of the zip code. If the entered zip code is in the array

Ask the user to enter the zip code

Create a for loop that iterates through the zips. If the entered zip is in the zips, set its index to index

When the loop is done, check if the index. If the index is not -1, that means zip code was found in the zips array, print the corresponding price from the price array. Otherwise, print no delivery for the entered zip code

8 0
4 years ago
3. with a dui charge on a driver’s record ______________. a. the price of his or her insurance greatly increases b. he or she ma
djyliett [7]
<span>The answer to your question is D. both A and C</span>
7 0
4 years ago
Which type of financial institution typically has membership requirements?
Anon25 [30]
Credit Union typically has membership requirements
8 0
4 years ago
Pick the 3 correct Python functions that parse and output only the date and time part of each log in log.txt as follows.
Mila [183]

Answer:

1, 4, 5

Explanation:

parse2:

. In this case it's passing "r" as an argument, which really does absolutely nothing, because whenever you call open("file.txt") it defaults to reading mode, so all you're doing is explicitly passing the "r". So let's look at the first line. Whenever you call str.split() without any arguments, by default it splits it by empty text, and filters any empty text. So str.split() is not the same as str.split(" ") although it has similar behavior. "     ".split(" ") will output ['', '', '', '', '', ''], while "     ".split() will output []. So in this case the line.split() will split the string '10.1.2.1 - car [01/Mar/2022:13:05:05  +0900] "GET /python HTTP/1.0" 200 2222' into the list<em> ['10.1.2.1', '-', 'car', '[01/Mar/2022:13:05:05', '+0900]', '"GET', '/python', 'HTTP/1.0"', '200', '2222'].</em> As you can see the the data is split into two pieces of text, AND they include the brackets in both strings. So when it gets the 3 index and strips it of the "[]" it will have the incomplete date

parse3:

 In this instance the "r" does nothing as mentioned before the "r" is already defaulted whenever you call open("file.txt") so open("file.txt") is the same as open("file.txt", "r"). So in this case we won't be working left to right, we're going inside the brackets first, kind of like in math you don't don't work left to right in equation 3 + 3(2+3). You work in the brackets first (inside brackets you do left to right). So the first piece of code to run is the line.split("[" or "]"). I actually kind of misspoke here. Technically the "[" or "]" runs first because this doesn't do what you may think it does. The or will only return one value. this is not splitting the line by both "[" and "]". The, or will evaluate which is true from left to right, and if it is true, it returns that. Since strings are evaluated on their length to determine if they're true. the "[" will evaluate to true, because any string that is not empty is true, if a string is empty it's false. So the "[" will evaluate to true this the "[" or "]" will evaluate to "[". So after that the code will run line.split("[") which makes the list: <em>['10.1.2.1 - car ', '01/Mar/2022:13:05:05  +0900] "GET /python HTTP/1.0" 200 2222']</em>. Now the [3:5] will splice the list so that it returns a list with the elements at index 3 (including 3) to 5 (excluding 5). This returns the list: [], because the previous list only has 2 elements. There are no elements at index 3 to 5 (excluding 5). So when you join the list by " ", you'll get an empty string

parse4:

  So I'm actually a bit confused here, I thought the "r+" would open the file in read-writing mode, but maybe this is a different version of python I have no idea, so I'm going to assume it is reading/writing mode, which just means you can read and write to the file. Anyways when you split the line by doing line.split(), as mentioned before it will split by empty spaces and filter any empty spaces. This line will return: <em>['10.1.2.1', '-', 'car', '[01/Mar/2022:13:05:05', '+0900]', '"GET', '/python', 'HTTP/1.0"', '200', '2222']</em>. and then you splice the list from indexes 3 to 5 (excluding 5). This will return the list: <em>['[01/Mar/2022:13:05:05', '+0900]']</em> which has the two pieces of information you need for the date. Now it joins them by a space which will output: '[01/Mar/2022:13:05:05 +0900]'. Now when you strip the "[]" you get the string: '01/Mar/2022:13:05:05 +0900' which is the correct output!

parse 5:

 So in this example it's using re.split. And the re.split is splitting by "[" or "]" which is what re.split can be used for, to split by multiple strings, which may be confused by string.split("[" or "]") which is not the same thing as explained above what the latter does. Also the reason there is a backslash in front of the [ and ] is to escape it, because normally those two characters would be used to define a set, but by using a \ in front of it, you're essentially telling regex to interpret it literally. So in splitting the string by "[" and "]" you'll get the list: <em>['10.1.2.1 - car ', '01/Mar/2022:13:05:05  +0900', ' "GET /python HTTP/1.0" 200 2222']</em> which has 3 elements, since it was split by the [ and the ]. The second element has the date, so all you need to do is index the list using the index 1, which is exactly what the code does

8 0
2 years ago
Imagine you have borrowed your friend's computer to work on a class project. Checking
castortr0y [4]

Answer:

Yes.It would be considered as a computer Crime.

Explanation:

This is because you are an unauthorised user

6 0
4 years ago
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