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vovangra [49]
2 years ago
12

Driving is expensive. Write a program with a car's miles/gallon and gas dollars/gallon (both floats) as input, and output the ga

s cost for 10 miles, 50 miles, and 400 miles. Output each floating-point value with two digits after the decimal point, which can be achieved as follows: print('{:.2f}'.format(your_value)) Ex: If the input is: 20.0 3.1599 the output is: 1.58 7.90 63.20 Your program must define and call the following driving_cost() function. Given input parameters driven_miles, miles_per_gallon, and dollars_per_gallon, the function returns the dollar cost to drive those miles. Ex: If the function is called with: 50 20.0 3.1599 the function returns: 7.89975 def driving_cost(driven_miles, miles_per_gallon, dollars_per_gallon) Your program should call the function three times to determine the gas cost for 10 miles, 50 miles, and 400 miles. Note: This is a lab from a previous chapter that now requires the use of a function.
Computers and Technology
2 answers:
Aneli [31]2 years ago
6 0

Answer:

def driving_cost(driven_miles, miles_per_gallon, dollars_per_gallon):

  gallon_used = driven_miles / miles_per_gallon

  cost = gallon_used * dollars_per_gallon  

  return cost  

miles_per_gallon = float(input(""))

dollars_per_gallon = float(input(""))

cost1 = driving_cost(10, miles_per_gallon, dollars_per_gallon)

cost2 = driving_cost(50, miles_per_gallon, dollars_per_gallon)

cost3 = driving_cost(400, miles_per_gallon, dollars_per_gallon)

print("%.2f" % cost1)

print("%.2f" % cost2)

print("%.2f" % cost3)

Explanation:

Alexus [3.1K]2 years ago
3 0

Answer:

  1. def driving_cost(driven_miles, miles_per_gallon, dollars_per_gallon):
  2.    gallon_used = driven_miles / miles_per_gallon
  3.    cost = gallon_used * dollars_per_gallon  
  4.    return cost  
  5. miles_per_gallon = float(input("Input miles per gallon: "))
  6. dollars_per_gallon = float(input("Input dollar per gallon: "))
  7. cost1 = driving_cost(10, miles_per_gallon, dollars_per_gallon)
  8. cost2 = driving_cost(50, miles_per_gallon, dollars_per_gallon)
  9. cost3 = driving_cost(400, miles_per_gallon, dollars_per_gallon)
  10. print("$ %.2f" % cost1)
  11. print("$ %.2f" % cost2)
  12. print("$ %.2f" % cost3)

Explanation:

The solution is written in Python 3.

Firstly, create a function driving_cost that takes three parameters, driven_miles, miles_per_gallon and dollars_per_gallon (Line 1). In the function, calculate the gallon consumption by applying formula driven_miles / miles_per_gallon and then use it to calculate the cost (Line 2 - 3). Return the cost as output (Line 4).

In the main program, prompt user to input miles per gallon and dollars per gallon and then use these input values as arguments to call the function driving_cost function for three times with each time with different driven_miles value (Line 6 - 11).

At last, use formatted print to display the output to two decimal points (Line 13 - 15).

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Which decimal value (base 10) is equal to the binary number 1012?
GaryK [48]

Answer:

The decimal value of 101₂² base 2 = 25 base 10

Explanation:

The number 101₂² in decimal value is found as follows;

101₂ × 101₂ = 101₂ + 0₂ + 10100₂

We note that in 101 + 10100 the resultant 2 in the hundred position will have to be converted to a zero while carrying over 1 to the thousand position to give;

101₂ + 0₂ + 10100₂ = 11001₂

Therefore;

101₂² =  11001₂

We now convert the result of the square of the base 2 number, 101², which is 11001₂ to base 10 as follows;

Therefore converting 11001₂ to base 10 gives;

11001₂= 1 × 2⁴ + 1 × 2³ + 0 × 2² + 0 × 2 ¹ + 1 × 2⁰

Which gives;

16 + 8 + 0 + 0 + 1 = 25₁₀.

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The groups_per_user function receives a dictionary, which contains group names with the list of users. Users can belong to multi
Akimi4 [234]

The groups_per_user function receives a dictionary, which contains group names with the list of users.

Explanation:

The blanks to return a dictionary with the users as keys and a list of their groups as values is shown below :

def groups_per_user(group_dictionary):

   user_groups = {}

   # Go through group_dictionary

   for group,users in group_dictionary.items():

       # Now go through the users in the group

       for user in users:

       # Now add the group to the the list of

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         # in the dictionary if necessary

         user_groups[user] = user_groups.get(user,[]) + [group]

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3 0
2 years ago
Complete the function to replace any period by an exclamation point. Ex: "Hello. I'm Miley. Nice to meet you." becomes:
vodomira [7]

Answer:

Here is the complete function:

void MakeSentenceExcited(char* sentenceText) {  // function that takes the text as parameter and replaces any period by an exclamation point in that text

int size = strlen(sentenceText);  //returns the length of sentenceText string and assigns it to size variable

char * ptr;  // character type pointer ptr

ptr = sentenceText;  // ptr points to the sentenceText string

for (int i=0; i<size; i++){  //iterates through the sentenceText string using i as an index

    if (sentenceText[i]=='.'){  // if the character at i-th index of sentenceText is a period

        sentenceText[i]='!'; } } } //places exclamation mark when it finds a period at i-th index of sentenceText

Explanation:

The program works as follows:

Suppose we have the string:

sentenceText = "Hello. I'm Miley. Nice to meet you."

The MakeSentenceExcited method takes this sentenceText as parameter

int size = strlen(sentenceText) this returns the length of sentenceText

The size of sentenceText is 35 as this string contains 35 characters

size =  35

Then a pointer ptr is declared which is set to point to sentenceText

for (int i=0; i<size; i++) loop works as follows:    

1st iteration:

i=0

i<size is true because i=0 and size = 35 so 0<35

So the body of loop executes:

 if (sentenceText[i]=='.') statement checks :

if (sentenceText[0]=='.')

The first element of sentenceText is H

H is not a period sign so the statement inside if statement does not execute and value of i increments to 1. Now i = 1

2nd iteration:

i=1

i<size is true because i=1 and size = 35 so 1<35

So the body of loop executes:

 if (sentenceText[i]=='.') statement checks :

if (sentenceText[1]=='.')

This is the second element of sentenceText i.e. e

e is not a period sign so the statement inside if statement does not execute and value of i increments to 1. Now i = 2

So at each iteration the if condition checks if the character at i-th index of string sentenceText is a period.

Now lets see a case where the element at i-th index is a period:

6th iteration:

i=5

i<size is true because i=5 and size = 35 so 5<35

So the body of loop executes:

 if (sentenceText[i]=='.') statement checks :

if (sentenceText[5]=='.')

This is the character at 5th index of sentenceText i.e. "."

So the if condition evaluates to true and the statement inside if part executes:

sentenceText[i]='!'; statement becomes:

sentenceText[5]='!'; this means that the character at 5th position of sentenceText string is assigned an exclamation mark.

So from above 6 iterations the result is:

Hello!

This loop continues to execute until all the characters of sentenceText are checked and when the value of i gets greater than or equal to the length of sentenceText then the loop breaks.

The screenshot of the program along with its output is attached.

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2 years ago
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