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Albert $1000 earned 1.2% annual interest compounded monthly

$500 lost 2% over the course of the 10 years
$500 grew compounded continuously at rate of 0.8% annually

What is the balance of Albert’s $2000 after 10 years?

1 Answer

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Answer:

$2159.07

Explanation:

The compound interest formula is used to find the balance for the $1000 investment:

A = P(1 +r/n)^(nt)

A = 1000(1 +.012/12)^(12·10) = 1000·1.001^120 ≈ 1127.43

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For a 2% loss, the multiplier of the investment value is 1-.02 = 0.98. The value of the first $500 investment is ...

A = 500(1 -.02) = 490.00

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The continuous compounding formula is used for the second $500 investment.

A = Pe^(rt)

A = 500e^(.008·10) = 500e^.08 = 541.64

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The total value of Albert's investments is ...

$1127.43 +490 +541.64 = $2159.07

Albert $1000 earned 1.2% annual interest compounded monthly $500 lost 2% over the-example-1
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