numberThe amount of cameras that can be assembled by the employee after training is given by;
![Q=60-30e^(-0.5t)](https://img.qammunity.org/2023/formulas/mathematics/college/2o3cnra5cmzgtfwouvn2dd52awjju8p1tb.png)
1. To find the number of cameras that can be assembled by an employee who has worked for two months, we insert t = 2 in the equation, we get;
![\begin{gathered} Q=60-30e^(-0.5*2) \\ Q=60-30e^(-1) \\ Q=48.96\approx49\text{ cameras} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5ut1yl4a6ua72xddhj7kwgwjhbs3pbxn6s.png)
Theref
2. The maximum number of cameras can be expected from an employee that has worked a long time ( close to "infinity" time), so as t approaches infinity, we get;
![\begin{gathered} Q(\infty)=60-30e^(-0.5*\infty) \\ -0.5*\infty\text{ is simply }\infty,\text{ so} \\ Q=60-30e^(-\infty) \\ e^(-\infty)=0,\text{ so} \\ Q=60-30(0) \\ Q=60\text{ cameras} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/dated2hfpth7xqpcwgugpkwcz8pwpz7bu5.png)
Thus, the maximum number of cameras that can be expected is 60