If $4000 is borrowed at a rate of 5.75% interest per year, compounded quarterly, find the amount due at the end of the given number of years. (Round your answers to the nearest cent.)
5 Years
7 Years
9 Years
+6251 If $4000 is borrowed at a rate of 5.75% interest per year, compounded quarterly, find the amount due at the end of the given number of years. (Round your answers to the nearest cent.)
Let N be the number of times per year the interest is compounded and let T be in years.
$$FV=PV\left(1+\dfrac {rate}{N}\right)^{T \cdot N}$$
$$FV=4000\left(1+\dfrac{0.0575}{4}\right)^{4T}$$
$${\mathtt{4\,000}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.057\: \!5}}}{{\mathtt{4}}}}\right)}^{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{5}}\right)} = {\mathtt{5\,321.459\: \!175\: \!499\: \!329\: \!081\: \!7}}$$
$${\mathtt{4\,000}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.057\: \!5}}}{{\mathtt{4}}}}\right)}^{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{7}}\right)} = {\mathtt{5\,965.117\: \!902\: \!958\: \!065\: \!897\: \!6}}$$
$${\mathtt{4\,000}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.057\: \!5}}}{{\mathtt{4}}}}\right)}^{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{9}}\right)} = {\mathtt{6\,686.630\: \!569\: \!302\: \!827\: \!466\: \!9}}$$
so
5yrs -> $5321.46
7yrs -> $5965.12
9yrs -> $6686.63
+6251 If $4000 is borrowed at a rate of 5.75% interest per year, compounded quarterly, find the amount due at the end of the given number of years. (Round your answers to the nearest cent.)
Let N be the number of times per year the interest is compounded and let T be in years.
$$FV=PV\left(1+\dfrac {rate}{N}\right)^{T \cdot N}$$
$$FV=4000\left(1+\dfrac{0.0575}{4}\right)^{4T}$$
$${\mathtt{4\,000}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.057\: \!5}}}{{\mathtt{4}}}}\right)}^{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{5}}\right)} = {\mathtt{5\,321.459\: \!175\: \!499\: \!329\: \!081\: \!7}}$$
$${\mathtt{4\,000}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.057\: \!5}}}{{\mathtt{4}}}}\right)}^{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{7}}\right)} = {\mathtt{5\,965.117\: \!902\: \!958\: \!065\: \!897\: \!6}}$$
$${\mathtt{4\,000}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.057\: \!5}}}{{\mathtt{4}}}}\right)}^{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{9}}\right)} = {\mathtt{6\,686.630\: \!569\: \!302\: \!827\: \!466\: \!9}}$$
so
5yrs -> $5321.46
7yrs -> $5965.12
9yrs -> $6686.63
