Investing $1000 at an annual interest rate of r %, compounded continuously, for

Question

Investing $1000 at an annual interest rate of r %, compounded continuously, for 10 years gives you a balance of $B, where B = g(r). Give a financial interpretation of the following statements. The balance in the account, at 9% interest, would increase by about $1460 if the interest rate were increased by 1%. The balance in the account, at 9% interest, would increase by about $2460 if the interest rate were increased by 1%. The balance in the account, at 9% interest, would increase by about $1000 if the interest rate were increased by 1%. Investing $1000 at 9% would yield about $2460 after 10 years. The balance in the account, at 9% interest, would increase by about $1246 if the interest rate were increased by 1%. Investing $1000 at 9% would yield about $246 after 10 years. Investing $10 at 9% would yield about $246 after 10 years. The balance in the account, at 9% interest, would increase by about $246 if the interest rate were increased by 1%. The balance in the account, at 9% interest, would increase by about $1000 if the interest rate were increased by 246%. What are the units of g'(9)? dollars percent interest dollars per percent interest G percent interest per dollar If f(t) is the quantity in kilograms of a chemical produced after t minutes and g(t) is the quantity in kilograms produced after t seconds then f'(t) = 60g'(t).

Explanation / Answer

a) investing $1000 today will yeild $ 2460 in 10 years @ 9 %

b) the balance in the account would increase by 246 if the interest rate increased by 1 percent

c) dollars per percent interest

d) false

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