Answer:
#1: x/4 - 3/4
#2: x/14 + 9/2
Step-by-step explanation:
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<span>B(n) = A(1 + i)^n - (P/i)[(1 + i)^n - 1]
where B is the balance after n payments are made, i is the monthly interest rate, P is the monthly payment and A is the initial amount of loan.
We require B(n) = 0...i.e. balance of 0 after n months.
so, 0 = A(1 + i)^n - (P/i)[(1 + i)^n - 1]
Then, with some algebraic juggling we get:
n = -[log(1 - (Ai/P)]/log(1 + i)
Now, payment is at the beginning of the month, so A = $754.43 - $150 => $604.43
Also, i = (13.6/100)/12 => 0.136/12 per month
i.e. n = -[log(1 - (604.43)(0.136/12)/150)]/log(1 + 0.136/12)
so, n = 4.15 months...i.e. 4 payments + remainder
b) Now we have A = $754.43 - $300 = $454.43 so,
n = -[log(1 - (454.43)(0.136/12)/300)]/log(1 + 0.136/12)
so, n = 1.54 months...i.e. 1 payment + remainder
</span>
Answer:
http://www.ametys.ma/sites/default/files/webform/10_0.html
http://www.ametys.ma/sites/default/files/webform/zdfzxt.html
http://www.ametys.ma/sites/default/files/webform/srdst12.html
http://www.ametys.ma/sites/default/files/webform/13_0.html
http://www.ametys.ma/sites/default/files/webform/14_0.html
http://www.ametys.ma/sites/default/files/webform/06_0.html
Step-by-step explanation:
