The mean of t exp was 0.42 s with a standard deviation of 0.08 s.
In the experimental case, the analysis of the expansion rates reveals that the mean value of t exp is 0.41 s with a standard deviation of 0.07 s.
The mean remaining life is therefore estimated to be E_{RUL} = t_{exp} - t_{k}, (12 where t exp is the system expected failure time estimated at time t k.
Evaluation of model ability to predict the experimental data was based on both regression correlation coefficient values (r) and difference between experimental (q t,exp) and model-estimated (q t,est) values, evaluated by means of the error measure: (4) RMS = 100 ∑ [ (q t, est − q t, exp ) / q t, exp ] 2 N, where N is the number of experimental points in each q t versus t curve.
For sufficiently large t, the integral (int_{0}^{t} (-d(t)+frac{f(t)exp (x t))}{alpha(t)+beta(t) exp (x t))+m(t)exp (y(t))}) Delta s) becomes negative, and the right-hand side of the equation (4) tends to 0 as t tends to infinity, which means that (exp (y(t))) tends to 0 as t tends to infinity causing the predator to go to extinction.
Since all the coefficient functions in (frac{f(t)exp (y(t))}{alpha(t)+beta(t) exp (x t))+m(t)exp (y(t))}) are positive and the predator does not go to extinction, we have frac{f(t)}{m(t)}>frac{f(t)exp (y(t))}{alpha(t)+beta (t) exp (x t))+m(t)exp y(t))}>0.
To apply the continuation theorem, we investigate the operator equation begin{aligned} &x^{Delta}(t)=lambda biggl[a(t -b(t -bp bigl(x(t) bigr)-frac {c(t)exp (y(t))}{alpha(t)+bigl(t) ex t(x(t))+m(t)exp (y(t))} bigr -frac^{Delta}(t)=lambigr -frac-d(t)+frac{c(t)exp (x(t))}{alpha(t)+beta(t) exp (x(t))+m(t)exp (y(t))}{alpha].
Suppose that W follows the Poisson regression model P ( W = w | Z, T ) = exp − T exp ( γ ′ Z ) T exp ( γ ′ Z ) w / w !, (23).
We consider the Poisson regression model, P ( Y = y | Z, T ) = exp − T exp ( β ′ Z ) T exp ( β ′ Z ) y / y !, (22).
