The population of a pack of wolves is 88. the population is expected to grow at a rate of 2.5% each year. what function equation
represents the population of the pack of wolves after t years?
a.f(t)=88(0.025)^t
b.f(t)=88(1.25^)t
c.f(t)=88(2.5)^t
d.f(t)=88(1.025)^t
2 answers:
D
First year: 88 + 88 * 2.5% = 88 + 88 * 0.025 = 88 * 1.025
Second year: 88 * 1.025 + 88 * 1.025 * 2.5% = 88 * 1.025 + 88 * 1.025 * 0.025 = 88 * 1.025 * 1.025 = 88 * 1.025^2
Answer:
The population model of wolves after t years is given by
D is the correct option.
Step-by-step explanation:
The exponential function can be represented as
a = initial amount
r = rate
t = time
Now, we have been given that
r = 2.5% = 0.025
a = 88
On substituting these values in the above exponential model
The population model of wolves after t years is given by
You might be interested in
Answer:
Step-by-step explanation:
Yeshgggghhgfgggggggy. B be cc
6 + 9x.....a common factor in both terms is 3
3(2 + 3x) <==
4. 3(3) + 2
(3 x 3) + 2 = 11
5. -(-6) -7
(+)6 -7 = -1
Divide 100.5 by 1.5 to get 67.67