Respuesta :
Answer:
(a)[tex]A(n)=5500(1.01)^{4n}[/tex]
(b)$5955.71
(c)15.02 years
Step-by-step explanation:
For an initial principal P deposited in an account at an annual interest r compounded for a number of period k, the amount in the account after n years is given by the model:
[tex]A(n)=P(1+\dfrac{r}{k})^{nk}[/tex]
(a)Aunt Ga Ga gave you $5,500 to save for college.
P=$5,500
Annual Interest, r=4%=0.04
Since interest is compounded quarterly, Number of Periods, k=4
Therefore, an exponential function modeling this situation is:
[tex]A(n)=5500(1+\dfrac{0.04}{4})^{4n}\\A(n)=5500(1+0.01)^{4n}\\$Simplified\\A(n)=5500(1.01)^{4n}[/tex]
(b)After 2 years, i.e. when n=2
[tex]A(2)=5500(1.01)^{4*2}\\=\$5955.71[/tex]
(c)When A(n)=$10000, we have:
[tex]10000=5500(1.01)^{4n}\\$Divide both sides by 5500\\(1.01)^{4n}=\dfrac{10000}{5500} \\$To solve for n, we change to logarithm form\\Log_{1.01}\dfrac{10000}{5500}=4n\\= \dfrac{ Log \dfrac{10000}{5500}}{Log 1.01}=4n\\4n=60.08\\n=60.08 \div 4\\n=15.02\\$Therefore, in 15.02 years, the account would be worth $10,000.[/tex]
