Answer:
Let's use the variable y to represent the number of years passed since the first year.
The population on the first year (y = 0) was 800
The population in the second year (y = 1) was 600.
The ratio in which the population decreased can be calculated as:
R = 600/800 = 0.75
This means that 600 is the 75% of 800, this also means that between the first year and the second year, the population decreased by the 25%
Now let's look at the ratio between the second and third year (y = 2), the population the third year was 450
Now the ratio is:
R = 450/600 = 0.75
Same as before, then we already can see that the population will decrease by 25% each year.
The generic exponential decay equation is:
f(y) = A*(1 - r)^y
where:
A = initial population, in this case, is 800
y = our variable, in this case, represents the number of years
r = the amount that decreases per each unit of our variable, this must be written in decimal form. In this case, we know that the population decreases by 25% each year, and 25% written in decimal form is 0.25
Also, (1 - r) = R, where R is the ratio we found earlier.
To do this, we just divide 25% by 100% to get (25%/100% = 0.25)
Then our equation will be:
f(y) = 800*(1 - 0.25)^y
f(y) = 800*( 0.75)^y
1) We want to predict when the population will be 200.
to do this, we set:
f(y) = 200 = 800*( 0.75)^y
and solve it for y.
(200/800) = 0.75^y
(1/4) = 0.75^y
Now we can use the relationship:
Ln(a^x) = x*ln(a)
Then let's apply Ln( ) in both sides to get
ln(1/4) = y*ln(0.75)
ln(1/4)/ln(0.75) = y = 4.8 years.
This means that 4.8 years after the first year, the population will be around 200.
2) The population in the 25 th year (this is 24 years after the first one, so we take y = 24) is:
f(24) = 800*(0.75)^(24) = 0.80
Around this point, we will have no more sunfish in the lake.
is the slope of the line, and
is the y-intercept (the y-value of the point where the line crosses the y-axis.)