<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>
1 Whole is bigger than a fraction
1 part of the 6 equal groups is shaded
1/3
Answer:
Step-by-step explanation:
s = n(a + 1)
n(a + 1) = s
a + 1 = s/n
a = s/n - 1
W=4 that the answer !!!!!
