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Karolina [17]
3 years ago
6

Anyone free to inbox me plz....​

Computers and Technology
2 answers:
MArishka [77]3 years ago
8 0

Answer:

To message you? Here is my alt number (240-918-2355) you can text me if you want ¯\_(ツ)_/¯

fenix001 [56]3 years ago
6 0
I’m only doing this to get more points
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Differences between analog computer and hybrid computer​
aleksandrvk [35]

Answer:

Analog computers only work with continuous numerical data in analog quantities, digital computers can process both non-numerical and numerical data and a hybrid computer is a combination of both analog and digital. A hybrid computer has the accuracy of a digital computer paired with speed of an analog one.

5 0
3 years ago
Read 2 more answers
To decrease the size of text so it will fit into a text box without overflowing, select the
Trava [24]

Answer:

It is is A

Explanation:

Becuase if you shrink a text thats just the answer

5 0
4 years ago
Read 2 more answers
Write a basic program that performs simple file and mathematical operations.
Ostrovityanka [42]

Answer:

Explanation:

I have written the Python program based on your requirements.

Just give the path of the input file and the path for the output file correctly where you want to place the output file.

In, my code - I have given my computer's path to the input and output file.

You just change the path correctly

My code works perfectly and I have tested the code with your inputs.

It gives the exact output that you need.

I have attached the Output that I got by running the below program.

Code:

month_list={ "january":"1","february":"2", "march":"3","april":"4", "may":"5", "june":"6","july":"7", "august":"8", "september":"9","october":"10", "november":"11", "december":"12"} input_file=open('C:\\Users\\Desktop\\inputDates.txt', 'r') output_file=open('C:\\Users\\Desktop\\parsedDates.txt','w') for each in input_file: if each!="-1": lis=each.split() if len (lis) >=3: month=lis [0] day=lis[1] year=lis [2] if month.lower() in month_list: comma=day[-1] if comma==',': day=day[:len (day)-1] month_number=month_list[month.lower()] ans-month_number+"/"+day+"/"+year output_file.write (ans) output_file.write("\n") output_file.close() input_file.close()

- O X parsedDates - Notepad File Edit Format View Help 3/1/1990 12/13/2003

- X inputDates - Notepad File Edit Format View Help March 1, 1990 April 2 1995 7/15/20 December 13, 2003 -1

cheers i hope this helped !!

7 0
4 years ago
Assume the following variable definition appears in a program:
AnnZ [28]

Answer:

cout << setprecision(2)<< fixed << number;

Explanation:

The above statement returns 12.35 as output

Though, the statement can be split to multiple statements; but the question requires the use of a cout statement.

The statement starts by setting precision to 2 using setprecision(2)

This is immediately followed by the fixed manipulator;

The essence of the fixed manipulator is to ensure that the number returns 2 digits after the decimal point;

Using only setprecision(2) in the cout statement will on return the 2 digits (12) before the decimal point.

The fixed manipulator is then followed by the variable to be printed.

See code snippet below

<em>#include <iostream>  </em>

<em>#include <iomanip> </em>

<em>using namespace std;  </em>

<em>int main()  </em>

<em>{  </em>

<em> // Initializing the double value</em>

<em> double number = 12.3456;  </em>

<em> //Print  result</em>

<em> cout << setprecision(2)<< fixed << number;  </em>

<em> return 0;  </em>

<em>}  </em>

<em />

4 0
3 years ago
Janice has decided that the best approach to programming a new social media app is to start with a small portion of it by writin
Maru [420]

Answer:

blacc

bhhj

Explanation:

4 0
3 years ago
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