Question 1071612
Decrease of 2% each year means keeping 98% each year.



{{{y=p*e^(kt)}}}
y population
p initial population
t time in years


In one year,
{{{98=100e^(k*1)}}}, assuming if starting with population 100;
{{{ln(98)=ln(100*e^k)}}}

{{{ln(98)=ln(100)+k*ln(e)}}}

{{{k*1=ln(98)-ln(100)}}}

{{{k=ln(98)-ln(100)}}}

{{{k=-0.0202}}}

MODEL:  {{{highlight_green(y=27200*e^(-0.0202t))}}}



What is the population in 10 years?
t=10; find y.
{{{y=27200*e^(-0.0202*10)}}}

{{{y=27200*e^-0.2020}}}

To the nearest WHOLE number, y=22225