Answer:
A. The growth rate of the CO² concentration in the atmosphere, from 1866 to 2006 was 0.26% annually
B. The CO² concentration in the atmosphere will double its 1866 level after 266 years and 5 months approximately.
Step-by-step explanation:
1. Let's review the information provided to us to answer the question correctly:
Parts per million of CO² concentration in 1866 = 262
Parts per million of CO² concentration in 2006 = 377
Duration of the model = 140 years (2006 - 1866)
2. Let's find the growth rate (r) of this model after 140 years, using the following exponential function:
FV = PV * (1 + r) ⁿ
PV = Parts per million of CO² concentration in 1866 = 262
FV = Parts per million of CO² concentration in 2006 = 377
number of periods (n) = 140 (140 years compounded annually)
Growth rate (r) = r
Replacing with the real values, we have:
377 = 262 * (1 + r) ¹⁴⁰
377/262 = (1 + r) ¹⁴⁰
¹⁴⁰√ 377/262 = 1 + r
r = ¹⁴⁰√ 377/262 - 1
r = 1.002602672 - 1
r = 0.002602672 = 0.26% (Rounding to two decimal places)
The growth rate from 1866 to 2006 was 0.26% annually
3. Use this exponential model to predict when the CO² concentration will double its 1866 level
FV = PV * (1 + r) ⁿ
PV = Parts per million of CO² concentration in 1866 = 262
FV = Parts per million of CO² concentration in ? = 262 * 2 = 524
number of periods (n) = n
Growth rate (r) = 0.26% = 0.0026
Replacing with the real values, we have:
524 = 262 * (1 + 0.0026) ⁿ
524/262 = (1 + 0.0026) ⁿ
2 = 1.0026ⁿ
n = ㏒ 2/ ㏒ 1.0026
n = 0.30103/0.00113
n = 266.4 (Rounding to the next tenth)
0.4 of a year = 0.4 * 12 = 5 months (Rounding to the next whole)
The CO² concentration in the atmosphere will double its 1866 level after 266 years and 5 months approximately.
