A table is given. It is required to use the linear regression capabilities to find a linear model for (t, ln(R)) with an equation of the form ln(R)=at+b, and then write in exponential form.
It is also required to use a graphing utility to draw a graph.
(a) Rewrite the table values for t and ln(R) as follows:
Input the values into a graphing utility to find the linear model:
![\ln(R)=-0.6259t+6.0770](https://img.qammunity.org/2023/formulas/mathematics/college/cguhwuqflb3wx453m4rtj521e6j9tyarcx.png)
Take the exponent of both sides of the equation to write in exponential form as required:
![\begin{gathered} \Rightarrow e^(\ln(R))=e^(-0.6259t+6.0770) \\ \text{ Using the powers of sum property:} \\ \Rightarrow R=e^(6.0770)\cdot e^(-0.6259t) \\ \Rightarrow R=435.7201(e^(-0.6259t)) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/61as9adzwu9r92iwz5y9hk1120fhp45ulf.png)
(b) Use the graphing utility to graph the exponential function:
Notice that the graph at the top-right best fits the graph.
(c) Evaluate the definite integral using the graphing utility:
![\int_0^4435.7201e^(-0.6259t)dt=639.2116](https://img.qammunity.org/2023/formulas/mathematics/college/ejyj6gqgdgi4kmp4susp6pnm5ne65eet7g.png)
The answer is 639.2116 L.