Asked by Brita
From the graph the median price paid for a home in 1968 is about $20000 and in 1993 is about $100000. Use these data points to construct a linear and an exponential model to represent the growth in median price from 1968 to 1993.
for p(t) linear I got 20000+3200t
but for the exponential I get something wrong and weird like 20000*0.97^t/25
Help, Please & Thank You
for p(t) linear I got 20000+3200t
but for the exponential I get something wrong and weird like 20000*0.97^t/25
Help, Please & Thank You
Answers
Answered by
Reiny
looks like you considered
1968 as t = 0 , that's fine
so you would two ordered pairs (0,20000) and (25,100000)
I agree with your linear equation, but you should state it as
p(t) = 20000 + 3200t, <b>where t is the number of years after 1968</b>
for the exponential:
let p(t) = 20000 e^(kt), where t is the number of years after 1968
so for (25,100000)
100000 = 20000 e^(25k)
5 = e^ 25k
take ln of both sides
ln 5 = ln (e^ 25k)
ln 5 = 25k ln e, but ln e = 1
25k = ln 5
k = ln 5/25
p(t) = 20000 e^(ln5/25 t)
or p(t) = 2000 e^ (.0643775 t)
You really could have used any base, I have no idea why you chose .97
1968 as t = 0 , that's fine
so you would two ordered pairs (0,20000) and (25,100000)
I agree with your linear equation, but you should state it as
p(t) = 20000 + 3200t, <b>where t is the number of years after 1968</b>
for the exponential:
let p(t) = 20000 e^(kt), where t is the number of years after 1968
so for (25,100000)
100000 = 20000 e^(25k)
5 = e^ 25k
take ln of both sides
ln 5 = ln (e^ 25k)
ln 5 = 25k ln e, but ln e = 1
25k = ln 5
k = ln 5/25
p(t) = 20000 e^(ln5/25 t)
or p(t) = 2000 e^ (.0643775 t)
You really could have used any base, I have no idea why you chose .97
Answered by
Steve
the "half-life" is 25 years.
(1/2)^(1/25) = 0.97265
(1/2)^(1/25) = 0.97265
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