0.191=1/(1+19e^-0.25t)
1+19e^-0.25x=1/0.191
1+19e^-.25x=5.24
19e^-0.25x=4.24
e^-0.25x=0.223
ln0.223=-0.25x
-1.5=0.25x
x=6
6years to rech 2.82
Yes
No
I cant do it so i dunno
Why am i here?
so far, g'(t) = 95e^-0.25t * [1+19e^-0.25t]^-2
where g'(t) is the growth rate formula
the thing i have to do is find the mean height of the tree's and how many years that i expect the process would occur.
mean = 2.821
Integrate g'(t) = 95e^-0.25t * [1+19e^-0.25t]^-1
95e^-0.25t = -380e^-0.25t+C
[1+19e^-0.25t]^-1 = -4.75e^-0.25t(1+19e^-0.25t)^-1 +c
-4.75e(sigma ) [1+19e^-0.25t]^-1.dt = 20(1+19e^-0.25t)^-1+c
at this point you are to find c
let g=0 & t=0
0(0) = 20(1+19e^-0.25t)^-1+c
0(0) = 20(20)^-1+c
0(0) = 1+c
therefore c=-1
at this time you are to find the value of t when g=2.821
2.821(t [dont know if this has to be here]) = 20(1+19e^-0.25t)^(-2or-1) -1 (dunno if its supposed to be -2or-1)
3.821(t) = 20(1+19e^-0.25t)^(-2or-1)
0.19105(t)=(1+19e^-0.25t)^(-2or-1)
0.19105(t)= 1 / (1+19e^-0.25t)^(2or1)
if ^2 = 1/(1+19e^-0.25t)x(1+19e^-0.25t)
if ^1 = 1/1+19e^-0.25t
Either way im lost, please help!