Answer 1:
This is a compound interest model.
Model:
Value after 28 years = $2600*(1+0.092)^{28} = $22,022.64 (rounded to the nearest cent).
After how many years will the value be $9,734?
$9,734/$2600 = (1+0.092)^t, where t is the number of years.
Solving for t, we get t = 10.14 years (rounded to the nearest hundredth).
Answer 2:
This is an exponential growth model.
Model:
Let P(t) be the population at time t in years.
Then P(t) = 45,000,000*(1+0.057)^t.
To find when the population reaches 100,000,000, we need to solve the equation:
100,000,000 = 45,000,000*(1+0.057)^t
Solving for t, we get t = 21.35 years (rounded to the nearest hundredth).
How many years will it take the population to get to 103,000,000?
103,000,000 = 45,000,000*(1+0.057)^t
Solving for t, we get t = 22.12 years (rounded to the nearest hundredth).
How many years does it take to get to 161,000,000?
161,000,000 = 45,000,000*(1+0.057)^t
Solving for t, we get t = 34.86 years (rounded to the nearest hundredth).
Answer 3:
This is an exponential decay model.
Model:
Let P(t) be the population at time t in years.
Then P(t) = 168,000*(1-0.0347)^t.
To find when the population drops below 508,000, we need to solve the equation:
508,000 = 168,000*(1-0.0347)^t
Solving for t, we get t = 40.36 years (rounded to the nearest hundredth).
Population after 34 years:
P(34) = 168,000*(1-0.0347)^{34} = 76,447.23 (rounded to the nearest hundredth).
Answer 4:
This is an exponential decay model.
Model:
Let A(t) be the amount of substance at time t in years.
Then A(t) = 80*(1-0.03)^t.
Value after 16 years:
A(16) = 80*(1-0.03)^{16} = 37.09 (rounded to the nearest hundredth).
After how many years will the value be 29.28?
29.28 = 80*(1-0.03)^t
Solving for t, we get t = 24.32 years (rounded to the nearest hundredth).