Statistics Solutions

[See Solution] At time t=0, a bottle of juice of 90^circ F is stood in a mountain stream whose temperature is 50^circ F. After 5 minutes, its temperature is 80^circ

Question: At time \(t=0\), a bottle of juice of \(90^{\circ} \mathrm{F}\) is stood in a mountain stream whose temperature is \(50^{\circ}\) F. After 5 minutes, its temperature is \(80^{\circ} \mathrm{F}\). Let \(\mathrm{H}(t)\) denote the

temperature of the juice at time \(t\), in minutes.

a.) Write a differential equation for \(H(t)\) using Newton’s Law of cooling.

b.) Solve the differential equation

c.) When will the temperature of the juice have dropped to \(60^{\circ} \mathrm{F}\) ?

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Solution: The downloadable solution consists of 2 pages
Deliverable: Word Document


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