Question

At 7:00 A.M., Alicia pours a cup of tea whose temperature is 200°F. The tea starts to cool to room temperature (72°F). At 7:02 A.M. the temperature of the tea is 197°F. At 7:15 A.M. the temperature of the tea is 180°F. Alicia will drink the tea when its temperature is 172°F.
Part A
Which of the following shows an exponential cooling equation that models the temperature of the tea?

Part B
When can Ella drink the tea?

Answer 1

Given :  At 7:00 A.M., Alicia pours a cup of tea whose temperature is 200°F. The tea starts to cool to room temperature (72°F). At 7:02 A.M. the temperature of the tea is 197°F. At 7:15 A.M. the temperature of the tea is 180°F. Alicia will drink the tea when its temperature is 172°F.

To find : Which of the  exponential cooling equation that models the temperature of the tea.  When can Ella drink the tea?

Solution:

Given options : ( Missing Data)

A. y = 72(0.989)ˣ + 128

B. y = 128(0.989)ˣ + 72

C. y = 200(0.989)ˣ + 72

D. y = 128(0.989)ˣ + 200

at x = 0  Temperature = 200

C.  y = 200(0.989)ˣ + 72   using x = 0  y =  272  does not satisfy

D.   y = 128(0.989)ˣ + 200 using x = 0  y =  328  does not satisfy

A. y = 72(0.989)ˣ + 128  using x = 0  y =  200

B. y = 128(0.989)ˣ + 72  using x = 0  y =  200

now at x = 2  temperature is 197

A. y = 72(0.989)ˣ + 128  using x = 2   y = 198.4

B. y = 128(0.989)ˣ +72  using x = 2   y = 197.2  ( closer to 180)

at x = 15  temperature = 180

A. y = 72(0.989)ˣ + 128  using x = 15   y = 189

B. y = 128(0.989)ˣ +72  using x = 15   y = 180.4  (closer to 180)

Hence y = 128(0.989)ˣ +72 shows an exponential cooling equation that models the temperature of the tea

y = 172  

=> 172  =  128(0.989)ˣ +72

=> (0.989)ˣ = 100/128

using log

 x  ≈ 22

Ella can drink the tea 7 :22 AM

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