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posledela
3 years ago
13

For this portion of the lab you will design the solution so that you perform some conditional tests. For this lab: 1. You will v

alidate input to ensure that the user enters inputs within a certain range or larger than a certain minimum value. You will validate the inputs as follows: (LO 1, 2, 3) a. The user cannot enter a negative number for: i. Miles to kilometers ii. Gallons to liters iii. Pounds to kilograms iv. Inches to centimeters b. The user cannot enter a value above 1000 degrees for Fahrenheit to Celsius (LO1)c. You MUST design a logical program exit. You may NOT use exit, break, quit, or system exit, or ANY OTHER forced exit. Do not use a menu. Use LOGIC to exit the program. 2. If the user enters an invalid value, then the program will issue an error message and terminate immediately. (Do NOT accept further data).3. Save the program as firstname_lastname_Lab3a.py where you will replace firstname and lastname with your actual first and last name.4. Test all conditions prior to submitting.
Computers and Technology
1 answer:
PIT_PIT [208]3 years ago
4 0

Answer:

#Prompt the user to enter the miles.

Miles = float(input('Enter the miles to convert into kilometer: '))

#Check the miles.

if Miles >= 0 :

   #Convert the miles to kilometers.

   Miles_To_km = Miles*1.6

   #Display the result.

   print (Miles, "miles equivalent to", Miles_To_km, "kilometer.")

   #Prompt the user to enter the gallons.

   Gallon = float(input('Enter the gallons to convert into liter: '))

   #Check the validity of the Gallons entered.

   if Gallon >= 0:

       #Convert gallons into liters.

       Gal_To_Lit = Gallon*3.9

       #Display the result.

       print (Gallon, "gallons equivalent to", Gal_To_Lit, "liters.")

       #Prompt the user to enter the pounds.

       Pound = float(input('Enter the pounds to convert into kilograms: '))

       #Check the validity of the Pounds entered.

       if Pound >= 0:

           #Convert pounds into kilograms.

           Pounds_To_Kg = Pound*0.45

           #Display the result.

           print (Pound, "pounds equivalent to", Pounds_To_Kg, "kilograms.")

           #Prompt the user to enter the temperature in Fahrenheit.

           f = float(input('Enter the temperature in Fahrenheit: '))

           #Check the value to be not greater than 1000.

           if f < 1000:

               #Convert Fahrenheit into celsius.

               F_To_C = (f -32)*5/9

               #Display the result.

               print (f, "Fahrenheit equivalent to", F_To_C, "celsius.")

           #Otherwise.

           else:

               

               #Display the error message.

               print ("Invalid temperature (greater than 1000) !!!")

       else:

           #Display the error message.

           print ("Pounds cannot be negative !!!")

   else:

           #Display the error message.

           print ("Gallons cannot be negative !!!")

else:

   #Display the error message.

   print ("Miles cannot be negative !!!")

Explanation:

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You are asked to simulate a binary search algorithm on an array of random values.An array is the list of similar type of element
Alex Ar [27]

Answer:

Explanation:

Problem statement:

to simulate a binary search algorithm on an array of random values.

Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.

Input/output description

Input:

Size of array: 4

Enter array:10  20 30 40

Enter element to be searched:40

The Output will look like this:

Element is present at index 3

Algorithm and Flowchart:

We basically ignore half of the elements just after one comparison.

Compare x with the middle element.

If x matches with middle element, we return the mid index.

Else If x is greater than the mid element, then x can only lie in right half subarray after the mid element. So we recur for right half.

Else (x is smaller) recur for the left half.

The Flowchart can be seen in the first attached image below:

Program listing:

// C++ program to implement recursive Binary Search

#include <bits/stdc++.h>

using namespace std;

// A recursive binary search function. It returns

// location of x in given array arr[l..r] is present,

// otherwise -1

int binarySearch(int arr[], int l, int r, int x)

{

   if (r >= l) {

       int mid = l + (r - l) / 2;

       // If the element is present at the middle

       // itself

       if (arr[mid] == x)

           return mid;

       // If element is smaller than mid, then

       // it can only be present in left subarray

       if (arr[mid] > x)

           return binarySearch(arr, l, mid - 1, x);

       // Else the element can only be present

       // in right subarray

       return binarySearch(arr, mid + 1, r, x);

   }

   // We reach here when element is not

   // present in array

   return -1;

}

int main(void)

{ int n,x;

cout<<"Size of array:\n";

cin >> n;

int arr[n];

cout<<"Enter array:\n";

for (int i = 0; i < n; ++i)

{ cin >> arr[i]; }

cout<<"Enter element to be searched:\n";

cin>>x;

int result = binarySearch(arr, 0, n - 1, x);

   (result == -1) ? cout << "Element is not present in array"

                  : cout << "Element is present at index " << result;

   return 0;

}

The Sample test run of the program can be seen in the second attached image below.

Time(sec) :

0

Memory(MB) :

3.3752604177856

The Output:

Size of array:4

Enter array:10  20 30 40

Enter element to be searched:40

Element is present at index 3

Conclusions:

Time Complexity:

The time complexity of Binary Search can be written as

T(n) = T(n/2) + c  

The above recurrence can be solved either using Recurrence T ree method or Master method. It falls in case II of Master Method and solution of the recurrence is Theta(Logn).

Auxiliary Space: O(1) in case of iterative implementation. In case of recursive implementation, O(Logn) recursion call stack space.

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Active Directory and __________ are critical in centralizing control of authentication and security policies in a domain environ
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Answer:

DNS(Domain Name Services) is the correct answer.

Explanation:

DNS is the type of directory system which is distributed that finds the solution of a hostname which is readable by the human being. It is also critical in centralizing control of the confirmation and policies of the security in the environment of the domain with Active Directory. That's why the following answer is correct.

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Free wifi is typically offered on _ network
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Free Wi-Fi is typically offered on public network

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What kind of server is another computer that screens all your incoming and outgoing messages?
Rom4ik [11]
A proxy server
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3 years ago
Modern computers are fixed-program computers (as opposed to stored program computers). True or False?
hjlf

Answer:

False.

Explanation:

The fixed program computer was used earlier which used wired processor and it was very laborious to make changes in processors. In the fixed program, the computer was following the build-in instructions and were processing only those data which were stored in memory.

Stored program computers are used in modern computers that are those computers that have single process structure and which will have instructions on how to perform the function on the computer and how to keep track of the processed data in computation.

Stored program computers keep program instruction in electronic memory.

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3 years ago
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