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The function below describes the population of caribou in a tundra, where f(t) represents the number of caribou, in hundreds, and t represents the time, in years.

f(t)=1.8(1.2)^t

Initially, the tundra has (180,120,1.8,1.2.1)
caribou, and every (1.2years, 1.8years, 1year, 120years, 180years)
, the number of caribou increases by a factor of (1.8, 1.2, 180, 120, 1)
.

The initial population is at time t = 0. Put 0 in for t to get [tex]f(0)=1.8(2.1)^0=1.8(1)=1.8[/tex] (measured in hundreds) so this population is 1.8(100) = 180

The time t is in years, so increasing t by 1 increases it by 1 year.

As t increases by 1, the factor that changes is 1.2 . For example,

At t = 2, the population is [tex]1.8(1.2)^2[/tex]

At t = 3, the population is [tex]1.8(2.1)^3[/tex]

There is just one more factor of 2.1, so the population in year 3 is 2.1 times the population in year 2.

aydengoad6730 aydengoad6730 · Answer 2 · 2022-02-25T18:58:35+01:00

Answer:

Answer:

Initial pop. = 1.8 (hundreds) = 180, 1 year, 1.2

Step-by-step explanation:

The initial population is at time t = 0. Put 0 in for t to get (measured in hundreds) so this population is 1.8(100) = 180

The time t is in years, so increasing t by 1 increases it by 1 year.

As t increases by 1, the factor that changes is 1.2 . For example,

At t = 2, the population is

At t = 3, the population is

There is just one more factor of 2.1, so the population in year 3 is 2.1 times the population in year 2.