1. complete the table below\n| annual rate | interest period | original principal | amount per $1.00 |…
1. complete the table below\n| annual rate | interest period | original principal | amount per $1.00 | maturity | interest earned |\n| ---- | ---- | ---- | ---- | ---- | ---- |\n| 5.00% | 1 year quarterly | $ 4,500 | | | |\n| 3.50% | 4 years daily | $ 18,000 | | | |\n| 4.75% | 4 years monthly | $ 9,000 | | | |\n| 4.25% | 1 year quarterly | $140,000 | | | |
Answer
Explanation:
Step1: Fórmula para interés compuesto
La fórmula para el monto $A$ en interés compuesto es $A = P(1+\frac{r}{n})^{nt}$, donde $P$ es el capital inicial, $r$ es la tasa anual de interés (en decimal), $n$ es el número de períodos de interés por año y $t$ es el número de años. El interés ganado $I=A - P$. El monto por cada dólar es $\frac{A}{P}$.
Step2: Caso 1: $r = 0.05$, $n = 4$ (trimestral), $P=4500$, $t = 1$
Calculamos $\frac{r}{n}=\frac{0.05}{4}=0.0125$ y $nt=4\times1 = 4$. Entonces $A = 4500(1 + 0.0125)^{4}$. $(1 + 0.0125)^{4}=1.0125^{4}\approx1.050945$. $A=4500\times1.050945 = 4729.2525$. El monto por cada dólar es $\frac{A}{P}=1.050945$. El interés ganado $I=A - P=4729.2525-4500 = 229.2525$.
Step3: Caso 2: $r = 0.035$, $n = 365$ (diario), $P = 18000$, $t=4$
$\frac{r}{n}=\frac{0.035}{365}\approx0.00009589$. $nt=365\times4 = 1460$. $A=18000(1 + 0.00009589)^{1460}$. $(1 + 0.00009589)^{1460}\approx1.14918$. $A=18000\times1.14918=20685.24$. El monto por cada dólar es $\frac{A}{P}\approx1.14918$. El interés ganado $I=A - P=20685.24 - 18000=2685.24$.
Step4: Caso 3: $r = 0.0475$, $n = 12$ (mensual), $P = 9000$, $t = 4$
$\frac{r}{n}=\frac{0.0475}{12}\approx0.0039583$. $nt=12\times4=48$. $A=9000(1 + 0.0039583)^{48}$. $(1 + 0.0039583)^{48}\approx1.20897$. $A=9000\times1.20897 = 10880.73$. El monto por cada dólar es $\frac{A}{P}\approx1.20897$. El interés ganado $I=A - P=10880.73-9000 = 1880.73$.
Step5: Caso 4: $r = 0.0425$, $n = 4$ (trimestral), $P = 140000$, $t = 1$
$\frac{r}{n}=\frac{0.0425}{4}=0.010625$. $nt=4\times1 = 4$. $A=140000(1 + 0.010625)^{4}$. $(1 + 0.010625)^{4}\approx1.04317$. $A=140000\times1.04317=146043.8$. El monto por cada dólar es $\frac{A}{P}\approx1.04317$. El interés ganado $I=A - P=146043.8-140000 = 6043.8$.
| Annual Rate | Interest Period | Original Principal | Amount Per $1.00$ | Interest Earned |
|---|---|---|---|---|
| $5.00%$ | 1 year quarterly | $$4,500$ | $\approx1.050945$ | $$229.2525$ |
| $3.50%$ | 4 years daily | $$18,000$ | $\approx1.14918$ | $$2685.24$ |
| $4.75%$ | 4 years monthly | $$9,000$ | $\approx1.20897$ | $$1880.73$ |
| $4.25%$ | 1 year quarterly | $$140,000$ | $\approx1.04317$ | $$6043.8$ |
Answer:
| Annual Rate | Interest Period | Original Principal | Amount Per $1.00$ | Interest Earned |
|---|---|---|---|---|
| $5.00%$ | 1 year quarterly | $$4,500$ | $\approx1.050945$ | $$229.2525$ |
| $3.50%$ | 4 years daily | $$18,000$ | $\approx1.14918$ | $$2685.24$ |
| $4.75%$ | 4 years monthly | $$9,000$ | $\approx1.20897$ | $$1880.73$ |
| $4.25%$ | 1 year quarterly | $$140,000$ | $\approx1.04317$ | $$6043.8$ |