SOLUTION: First-class postage (for the first ounce) was 20cent in 1981 and 37cent in 2002. Assume the cost increases according to an exponential growth function.
a. Let t =0 correspond t
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Question 173281This question is from textbook introduction to college mathematics
: First-class postage (for the first ounce) was 20cent in 1981 and 37cent in 2002. Assume the cost increases according to an exponential growth function.
a. Let t =0 correspond to 1981 abd t= 21 correspond to 2002. Then t is the number of years since 1981. Use the data points (0,20) and (21,37) to find the exponential growth rate and fit and exponential growth function
P(t)= P0e ^kt to the data where P(t) is the cost of first-class postage, in cents, t years after 1981.
b. Use the function found in part (a) to predict the cost of first-class postage in 2008.
c. When will the cost of first-class postage be $1.00 or 100cent
This question is from textbook introduction to college mathematics
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
First-class postage (for the first ounce) was 20cent in 1981 and 37cent in 2002. Assume the cost increases according to an exponential growth function.
a. Let t =0 correspond to 1981 abd t= 21 correspond to 2002. Then t is the number of years since 1981. Use the data points (0,20) and (21,37) to find the exponential growth rate and fit and exponential growth function
P(t)= Poe ^kt to the data where P(t) is the cost of first-class postage, in cents, t years after 1981.
Using the point (0,20)
20 = Po*e^(k*0)
20 = Po
--------
Now P(t) = 20*e^(kt)
Using (21,37
37 = 20e^(k*21)
1.85 = e^(21k)
21k = ln(1.85)
k = (1/21)*0.6152
k = 0.0293
=====================
Now P(t) = 20*e^(0.0293t)
===================================
b. Use the function found in part (a) to predict the cost of first-class postage in 2008.
1981 to 2008 = 27
P(27) = 20*e^(0.0293*27) = 44.12
----------------------------------------
c. When will the cost of first-class postage be $1.00 or 100cent
100 = 20e^(0.0293t)
e^(0.0293t) = 5
0.0293t = ln5
t = 54.93
--------------------
1981 + 54.93 = 2035.93
Ans: $1.00 in the year 2036
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Cheers,
Stan H.
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