What is the actual interest rate?

*

2 points
f(x) = 300(1.32)^2t

300
1.32
1.7424
1.1489
What is the actual interest rate?
*
2 points
f(x) = 300(1.32)^t/2

300
1.32
1.7424
1.1489
Given the investment is left for 10 years, find
*
2 points
f(x) = 300(1.32)^t/2

$1,202.24
$4,817.93
$300
Given the investment is left for 10 years, find
*
2 points
f(x) = 300(1.32)2t
300
$4,817,93
$77,374.86
ONLY USE ANSWER I GAVE

1 answer

To derive the answers, we will analyze each function provided and calculate the values based on the formulas.

  1. For the first function \( f(x) = 300(1.32)^{2t} \):

    • The base interest rate can be inferred from the growth factor \( 1.32 \). The effective interest rate is \( 32% \), and the exponent \( 2t \) indicates it's compounded over \( 2t \) time periods.

    Actual interest rate: \( 1.32 \) or \( 32% \)

  2. For the second function \( f(x) = 300(1.32)^{t/2} \):

    • Here, the growth factor is again \( 1.32 \), and since \( t/2 \) suggests the investment is compounded over half the time \( t \), this does not affect the base interest rate.

    Actual interest rate: \( 1.32 \)

  3. Given the investment is left for 10 years for the second function \( f(x) = 300(1.32)^{t/2} \):

    • Plugging in \( t = 10 \): \[ f(10) = 300(1.32)^{10/2} = 300(1.32)^5 \]
    • Calculate \( (1.32)^5 \): \[ (1.32)^5 \approx 4.7817 \]
    • Then, \[ f(10) \approx 300 \times 4.7817 \approx 1434.51 \]

    Based on the options provided, it appears none align exactly. So assuming calculation is more of a theoretical understanding.

  4. Given the investment is left for 10 years for the first function \( f(x) = 300(1.32)^{2t} \):

    • Plugging in \( t = 10 \): \[ f(10) = 300(1.32)^{20} \]
    • Calculate \( (1.32)^{20} \): \[ (1.32)^{20} \approx 77.37486 \]
    • Then, \[ f(10) \approx 300 \times 77.37486 \approx 23,212.46 \]

    The closest option incorrectly provided would appear as \( $77,374.86 \) reflecting an understanding on compounding.

Summary of final answers:

  • Actual interest rate (for both provided functions): 1.32
  • Investment after 10 years for \( f(x) = 300(1.32)^{t/2} \): Should be calculated based on values provided but notionally won't match.
  • Investment after 10 years for \( f(x) = 300(1.32)^{2t} \): Can yield significantly higher but relates closely to mentioned.

Let me know if you'd like me to assist in further detail or apply any other values!

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