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luda_lava [24]
3 years ago
9

It took her 9 more months but Marina has managed to save the full $725 plus more to cover fees to pay off the pay-day loan compa

ny. However, she forgot to account for the interest that had been compounding over time. Consider it is now 275 days later, the remaining loan was $725 and the APR is 47% compounded daily.
What is the total amount that Marina must now pay in order to pay off her the loan, accounting for interest?

What is the total amount of interest paid (not including fees)?
Computers and Technology
1 answer:
Marianna [84]3 years ago
6 0

Answer:

she loaned 750 dollars.

it took her 9 months to repay it.

it was 275 days since she took out the loan.

the interest rate was 47% per year compounded daily.

the daily interest rate would be 47% / 365 / 100 = .0012876712 per day.

this assumes 365 days in a year, which is the standard assumption that i know of.

the future value of 750 for 275 days would be based on the formula of f = p * (1 + r) ^ n

f is the future value

p is the present value

r is the interest rate per time period (days in this case)

 

n is the number of time periods (days in this case).

the formula becomes:

f = 750 * (1 + .0012876712) ^ 275

solve for f to get:

f = 1068.440089.

that's the future value of the loan.

it's what she owes.

the interest rate on the loan is that value minus 750 = 318.440089.

that's how much extra she needs to pay in addition to whatever fees she was charged.

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Answer:

The Following are the code to this question:

def cost(miles, milesPerGallon, dollarsPerGallon):##defining a method cost that accepts three parameter

   return miles / milesPerGallon * dollarsPerGallon# calculate cost and return it value  

milesPerGallon = float(input('Enter miles/Gallon value: '))#defining variable milesPerGallon for user input  

dollarsPerGallon = float(input('Enter dollars/ Gallon value: '))#defining variable dollarsPerGallon for user input

print('{:.5f}'.format(cost(10, milesPerGallon, dollarsPerGallon)))#call method and print its value

print('{:.5f}'.format(cost(50, milesPerGallon, dollarsPerGallon)))#call method and print its value

print('{:.3f}'.format(cost(400, milesPerGallon, dollarsPerGallon)))#call method and print its value

Output:

Enter miles/Gallon value: 20.0

Enter dollars/ Gallon value: 3.1599

1.57995

7.89975

63.198

Explanation:

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In the next step, two variable "milesPerGallon, and dollarsPerGallon" is declared, that use the input method with float keyword for input floating-point value.

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