Abottle of white wine at room temperature (68°f) is placed in a refrigerator at 4 p. m. its temperature after t hr is changing at the rate of −18e−0.6t °f/hour. by how many degrees will the temperature of the wine have dropped by 5 p. m.? (round your answer to one decimal place.) 17.0 changed: your submitted answer was incorrect. your current answer has not been submitted. your answer has the wrong sign.°f what will the temperature of the wine be at 5 p. m.? (round your answer to one decimal place.) °f

The temperature of the wine have dropped by 0.9°F.

The temperature of the wine will be 58.1 °F at 5 PM.

Step-by-step explanation:

We have been given that a bottle of white wine at room temperature (68°f) is placed in a refrigerator at 4 p.m. its temperature after t hr is changing at the rate of °f/hour.

We will use newton'e law of cooling to solve our given problem.

, where,

= Initial temperature.

Upon substituting and , we will get:

To find the temperature at 5 p.m. we will substitute in our function as:

Now, we will subtract 67.10 from 68 to find the temperature dropped by 5 PM.

Therefore, the temperature dropped by 0.9°F.

To find the temperature of wine at 5 PM, we will substitute as 5 PM is 1 hour later 4 PM.

Therefore, the temperature of the wine will be 58.1 °F at 5 PM.