Answer:
Increasing [tex]=2522000[/tex]
[tex]3[/tex] years ago population [tex]=16000000[/tex]
Step-by-step explanation:
Let [tex]3[/tex] years ago population [tex]=x[/tex]
If initial amount [tex]=P,[/tex] rate of increase [tex]=r[/tex] and time [tex]=t[/tex]
after [tex]t[/tex] years amount [tex](A)=P(1+\frac{r}{100})^t[/tex]
Hence [tex]A=18522000[/tex]
[tex]P=x\\r=5\%\\t=3\ years[/tex]
[tex]18522000=x(1=\frac{5}{100})^3\\\\18522000=x(1.05)^3\\\x=\frac{18522000}{(1.05)^3}\\\\x=\frac{18522000}{1.1576625}\\\\x=16000000[/tex]
[tex]3\ years\ ago\ population\ =16000000\\\\Increase=1852000-16000000\\\\=2522000[/tex]
