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A valve on a storage tank is opened for 4 hours to release a chemical in a manufacturing process. The flow rate R (in liters per hours) at a time t (in hours) is given in the table.a and c only

A valve on a storage tank is opened for 4 hours to release a chemical in a manufacturing-example-1
User Antiguru
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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

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}

(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

The answer is 639.2116 L.

A valve on a storage tank is opened for 4 hours to release a chemical in a manufacturing-example-1
A valve on a storage tank is opened for 4 hours to release a chemical in a manufacturing-example-2
User MiguelSantirso
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