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/ \(A = 1000 \left(1.025\right)^{12}\)
\(A = 1000 \left(1.025\right)^{12}\)
February 22, 2026
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\(A = P \left(1 + \frac{r}{n}\right)^{nt}\)
Where \(P = 1000\), \(r = 0.10\), \(n = 4\) (quarterly), and \(t = 3\).
\(A = 1000 \left(1 + \frac{0.10}{4}\right)^{4 \times 3}\)
\(A = 1000 \times 1.34489 \approx 1344.89\)
The balance after 3 years will be approximately $1344.89.
#### 1344.89
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