Question

Please I could really use some help on this (50 points, 5 stars and Brainliest)

Answer 1

Answer:

(a)$A(n)=5500(1.01)^{4n}$

(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:

$A(n)=P(1+\dfrac{r}{k})^{nk}$

(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:

$A(n)=5500(1+\dfrac{0.04}{4})^{4n}\\A(n)=5500(1+0.01)^{4n}\\ Simplified\\A(n)=5500(1.01)^{4n}$

(b)After 2 years, i.e. when n=2

$A(2)=5500(1.01)^{4*2}\\=\ 5955.71$

(c)When A(n)=$10000, we have:

$10000=5500(1.01)^{4n}\\ Divide both sides by 5500\\(1.01)^{4n}=\dfrac{10000}{5500} \\ To solve for n, we change to logarithm form\\Log_{1.01}\dfrac{10000}{5500}=4n\\= \dfrac{ Log \dfrac{10000}{5500}}{Log 1.01}=4n\\4n=60.08\\n=60.08 \div 4\\n=15.02\\ Therefore, in 15.02 years, the account would be worth 10,000.$