The bank started a new saving account for students on January 10, 2009 using the

Question

The bank started a new saving account for students on January 10, 2009 using the accumulation function a(t) = 0.05t2 + 1 where t is in years. Paula invested $1000 on January 10, 2009. How much is her investment worth now, on January 10, 2013? Bob invests $1000 now, on January 10, 2013. How much does he expect to accumulate in four years period? Using the accumulation function from the previous problem find effective rates of interest for the following years: 2, 3, 4, 5, and 6. Sorry it's just that I'm getting all sorts of different answers I really need to figure this out.

Explanation / Answer

Paula = 1000(1+0.05(4^2))=1800 Bob = 1000(1+0.05(4^2))=1800 For 1 year, effective rate = 0.05(1^2)=5% For 2 years, effective rate = 0.05(2^2)=20% For 3 years, effective rate = 0.05(3^2)=45% For 4 years, effective rate = 0.05(4^2)=80%

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