MHB Need help with Interest question?

  • Thread starter Thread starter Tazook
  • Start date Start date
  • Tags Tags
    Interest
Click For Summary
Lee is looking to finance a downhill ski package priced at $1894, with a $150 down payment and monthly payments of $113 for 1.5 years. The discussion revolves around calculating the annual interest charge using both monthly and yearly compounding methods. A formula for compound interest is suggested, but participants express confusion over the term "annual interest charge." Clarification is sought on the terminology and its common usage in finance. The conversation emphasizes collaborative problem-solving rather than providing direct answers.
Tazook
Messages
3
Reaction score
0
20. Lee wishes to purchase a downhill ski package that includes the skis, poles, bindings, and boats. The selling price is \$1894. The finance plan includes a \$150 down payment and payments of \$113 each at the end of the month for

1.5 years. Find the annual interest charge, if the interest charge is
a) monthly compounded b) yearly compounded

Please explain in detail... think I am a martian
 
Mathematics news on Phys.org
Hi Tazook,

Welcome to MHB! :)

We don't fully answer problems here, rather try to help you solve them so let's see what we can do.

Usually with compound interest problems we use the following formula: $$A=P \left(1 +\frac{r}{n} \right)^{nt}$$. However this seems like a slightly different situation.

I found another formula that might apply. Does the one on this Wikipedia page look familiar?

The reason I'm a bit confused is because the term "annual interest charge" is not one I'm familiar with. If we search this exact phrase on Google there are not that many results which makes me think there might be another common name for it.
 
Thanks for the wiki link... it really cleared everything...
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
Replies
3
Views
4K