exercises\nround to the nearest cent wherever necessary.\n1. mr. mady opens a savings account with principal…

exercises\nround to the nearest cent wherever necessary.\n1. mr. mady opens a savings account with principal p dollars that pays 2.11% interest compounded quarterly. express his ending balance after one year algebraically.\n2. jeff deposits $2,300 at 1.13% interest compounded weekly. what will be his ending balance after one year?\n3. nancy has $4,111 in an account that pays 1.07% interest compounded monthly. what is her ending balance after two years?\n4. mr. weinstein has a savings account with a balance of $19,211.34. it pays 1.1% interest compounded daily. what is his ending balance after three years, if no other deposits or withdrawals are made? how much interest does he earn over the three years?\n5. if you invested $10,000 at 3.8% compounded hourly for five years, what would be your ending balance?\n6.

exercises\nround to the nearest cent wherever necessary.\n1. mr. mady opens a savings account with principal p dollars that pays 2.11% interest compounded quarterly. express his ending balance after one year algebraically.\n2. jeff deposits $2,300 at 1.13% interest compounded weekly. what will be his ending balance after one year?\n3. nancy has $4,111 in an account that pays 1.07% interest compounded monthly. what is her ending balance after two years?\n4. mr. weinstein has a savings account with a balance of $19,211.34. it pays 1.1% interest compounded daily. what is his ending balance after three years, if no other deposits or withdrawals are made? how much interest does he earn over the three years?\n5. if you invested $10,000 at 3.8% compounded hourly for five years, what would be your ending balance?\n6.

Answer

Explanation:

Step1: Usar fórmula de interés compuesto

La fórmula de interés compuesto es $A = P(1+\frac{r}{n})^{nt}$, donde $A$ es el saldo final, $P$ es el principal, $r$ es la tasa de interés anual (en decimal), $n$ es el número de veces que se compone el interés por año y $t$ es el número de años.

Ejercicio 1

$P$ es el principal, $r = 0.0211$, $n = 4$ (compuesto trimestralmente), $t = 1$. $A = P(1+\frac{0.0211}{4})^{4\times1}=P(1 + 0.005275)^{4}=P(1.005275)^{4}$

Ejercicio 2

$P=2300$, $r = 0.0113$, $n = 52$ (compuesto semanalmente), $t = 1$ $A=2300(1+\frac{0.0113}{52})^{52\times1}=2300(1+\frac{0.0113}{52})^{52}$ $1+\frac{0.0113}{52}=1 + 0.0002173$ $A = 2300(1.0002173)^{52}$ Calculando $(1.0002173)^{52}\approx1.01136$ $A=2300\times1.01136 = 2326.128\approx2326.13$

Ejercicio 3

$P = 4111$, $r=0.0107$, $n = 12$ (compuesto mensualmente), $t = 2$ $A=4111(1+\frac{0.0107}{12})^{12\times2}=4111(1+\frac{0.0107}{12})^{24}$ $1+\frac{0.0107}{12}\approx1.0008917$ $A = 4111\times(1.0008917)^{24}$ Calculando $(1.0008917)^{24}\approx1.02157$ $A=4111\times1.02157=4199.77427\approx4199.77$

Ejercicio 4

$P = 19211.34$, $r = 0.011$, $n = 365$ (compuesto diariamente), $t = 3$ $A=19211.34(1+\frac{0.011}{365})^{365\times3}=19211.34(1+\frac{0.011}{365})^{1095}$ $1+\frac{0.011}{365}\approx1.00003014$ $A=19211.34\times(1.00003014)^{1095}$ Calculando $(1.00003014)^{1095}\approx1.03377$ $A=19211.34\times1.03377\approx19850.17$ Interés ganado $I=A - P=19850.17- 19211.34 = 638.83$

Ejercicio 5

$P = 10000$, $r = 0.038$, $n = 24$ (compuesto por hora, 24 horas al día), $t = 5$ $A=10000(1+\frac{0.038}{24})^{24\times5}=10000(1+\frac{0.038}{24})^{120}$ $1+\frac{0.038}{24}\approx1.0015833$ $A=10000\times(1.0015833)^{120}$ Calculando $(1.0015833)^{120}\approx1.21097$ $A = 10000\times1.21097=12109.70$

Answer:

  1. $A = P(1.005275)^{4}$
  2. $2326.13$
  3. $4199.77$
  4. Saldo final: $19850.17$, Interés: $638.83$
  5. $12109.70$