Answered

At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

WILL GIVE 5 STAR IF CORRECT: Amelia uses her credit card to buy some clothes for $678.34. She can pay off up to $235 per month. The card has an annual rate of 21.7% compounded monthly. How much total interest will she pay? (4 points)


$11.93


$12.27


$24.65


$29.72

Sagot :

980092279

Answer:

chetta es polla airemeris di quana premeria

Step-by-step explanation:

lexyjordan

Answer:

Month 1:

Calculate Balance (by adding interest to existing balance): 678.34*(1 + 0.217/12) = 690.606648333333 = 690.61

Make payment: 690.61 - 235 = 455.61

Old Balance: 690.61

New Balance: 455.61

--------------------------------------------------------------

Month 2:

Calculate Balance (by adding interest to existing balance): 455.61*(1 + 0.217/12) = 463.8489475 = 463.85

Make payment: 463.85 - 235 = 228.85

Old Balance: 463.85

New Balance: 228.85

--------------------------------------------------------------

Month 3:

Calculate Balance (by adding interest to existing balance): 228.85*(1 + 0.217/12) = 232.988370833333 = 232.99

Make payment: 232.99 - 232.99 = 0

Old Balance: 232.99

New Balance: 0

--------------------------------------------------------------

Payments Made: 235, 235, 232.99

Number of Payments Made: 3

Total Cost: 235 + 235 + 232.99 = 702.99

Total Cost: $702.99

Total Interest: 702.99 - 678.34 = 24.65

Total Interest: $24.65

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.