Question 1119682
 Between 1858 and 2011​, carbon dioxide ​(CO2​) concentration in the atmosphere rose from roughly 259 parts per million to 379 parts per million. Assume that this growth can be modeled with an exponential function Upper Q equals Upper Q 0 times left parenthesis 1 plus r right parenthesis Superscript t. 
By experimenting with various values of the growth rate​ r, find an exponential function that fits the data for 1858 and 2011. 
r almost equals______________________
​(Round to five decimal places as​ needed.) 
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t = 2011-1850 = 
Qo = 259
Q = 379
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Q = Qo(1+r)^t
379 = 259(1+r)^161
(1+r)^161 = 1.4633
161 = log(1.4633)/log(1+r)
log(1+r) = log(1.4633)/161 = 0.0010
1+r = 10^0.0010.. = 1.002367..
r = 0.002367..
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Cheers,
Stan H.
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