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 179°F. Alicia will drink the tea

when its temperature is 170°F. Use exponential regression to find a model for the

temperature of the tea. When can Ella drink the tea?

Step-by-step explanation:

Here we have the temperature variation with time given in an exponential equation as follows;

Let the temperature at time x (in minutes) = y

Therefore, y = a × mˣ + b

Where:

b = Shift of the curve or the limit value of the decreasing exponential function as x → ∞

When y = 200, x = 0

Therefore. 200 = a × m⁰ + c = a + c

We note that c is the shift of the graph, the value upon which temperature increases = final temperature = 72°F

Hence a = 200 - 72 = 128°F

When y = 197, x = 2 minutes

Therefore, 197 = 128·m² + 72 =

m² = (197 - 72)/128 = 125/128

m = √(125/128) = 0.98821

Hence the exponential equation of cooling is presented in the following equation;

y = 128 × (0.98821)ˣ + 72

Therefore, y = 128(0.989)x + 72

When the temperature is 172°F we have;

170 = 128 × (0.989)ˣ + 72

∴ (0.989)ˣ = (170 - 72)/128

= 98/128

= 0.7656

log(0.989)ˣ = log(0.7656)

x·log(0.989) = log(0.7656)

x = log(0.7656)/log(0.989)

= 24.15 minutes