Answer:
(5). [tex]Finial\ amount(A)=\$3619.80[/tex]
(6). [tex]a=5,\ b=0.5[/tex]
(7). [tex]A=33236[/tex]
Step-by-step explanation:
(5).
[tex]Monthly\ compound\ interst\ is\ given\ by\\\\A=P\left(1+\frac{r}{n}\right)^{nt}\\\\Where\ A= Final\ amount\\\\P= Initial\ amount\\\\r=annual\ interest\ rate\\\\n=number\ of\ times\ interest\ is\ compounded\ per\ unit\\\\t=time\ in\ year\\\\Given,\\\\Initial\ amount\ (P)=\$ 3500\\\\annual\ interest\ rate\ (r)=6.75\%\\\\t=6\ month=\frac{1}{2}\ year\\\\n=12\\\\A=3500\left(1+\frac{6.75}{12\times 100}\right)^{\frac{1}{2}\times 12}\\\\A=3500\left(1+0.005625\right)^6\\\\A=3500(1.005625)^6\\\\[/tex]
[tex]A=3500\times 1.03422818\\\\A=\$3619.80[/tex]
(6).
[tex]Exponential\ decay\ formula\ is\ given\ by\\\\y=a(1-b)^x\\\\Where\ (a)\ is\ initial\ amount\ and\ (b)\ is\ decay\ factor\\\\Given,\\\\y=5\times(0.5)^x\\\\y=5\times(1-0.5)^x\\\\compare\ with\ exponential\ decay\ formula\\\\a=5,\ b=0.5[/tex]
(7).
[tex]Given,\\\\Initial\ population=45000\\\\annual\ decrease=2\%\\\\n=15\ years\\\\ Decrease\ population\ is\ given\ by\\\\A=P\left(1-\frac{r}{100}\right)^n\\\\Where\ P\ is\ initial\ population\ r\ is\ decrease\ rate\ and\ n\ is\ number\ of\ years\\\\A=45000\left(1-\frac{2}{100}\right)^{15}\\\\A=45000(0.98)^{15}\\\\A=45000\times 0.738569\\\\A=33235.60\\\\A\approx 33236[/tex]
