A population of wolves in a county is represetned by the equation p(t)=80(0.98)^t. where t is the number of yers since 1998. predict the number of wolves in the population in the year 2008.
set t = 10 (time from 1998 to 2008)
P(10) = 80(.98)^10
= 65
well this is how i think u do it
since t at the starting is 1998, take that as basically = 0. from there subtract 1998 from 2008 to get ur value for t. one u have that solve for p
To predict the number of wolves in the population in the year 2008, we need to substitute t=10 into the given equation as 2008 is 10 years since 1998.
p(t) = 80(0.98)^t
Substituting t = 10:
p(10) = 80(0.98)^10
Calculating this expression:
p(10) ≈ 80(0.817)
p(10) ≈ 65.36
Therefore, the predicted number of wolves in the population in the year 2008 is approximately 65.36.
To predict the number of wolves in the population in the year 2008, we need to substitute the value of t in the equation p(t) = 80(0.98)^t with the corresponding value for the year 2008.
In the given equation, t represents the number of years since 1998. So, to determine the value of t for the year 2008, we need to find the difference between 2008 and 1998, which is 10 years.
Now, substitute t = 10 into the equation:
p(10) = 80(0.98)^10
To evaluate this expression, we calculate the value of (0.98)^10.
Using a calculator:
(0.98)^10 ≈ 0.8179
Now, substitute this value back into the equation:
p(10) = 80(0.8179)
To calculate this expression:
p(10) ≈ 65.55
Therefore, the predicted number of wolves in the population in the year 2008 is approximately 65.55.