(a)
25 represents the room temperature (°C)

In both models, \( T(t) = at^b + 25 \) and \( T(t) = kc^t + 25 \)

As \( t \to \infty \), \( at^b \to 0 \) (since \( b < 0 \)) and \( kc^t \to 0 \) (since \( c < 1 \))

\( T(t) \to 25 \), the equilibrium temperature

Result: Room temperature in °C [2]

(b)
\( a = 244 \), \( b = -1.03 \)

Model: \( T(t) = at^b + 25 \)

Adjust temperatures: \( (85-25, 55-25, 43-25, 36-25, 31-25, 28-25) = (60, 30, 18, 11, 6, 3) \)

Data: \( t = (0, 5, 10, 15, 20, 25) \), \( y = (60, 30, 18, 11, 6, 3) \)

Fit \( y = at^b \) using power regression

\( a \approx 243.920 \approx 244 \)

\( b \approx -1.02965 \approx -1.03 \)

Result: \( a = 244 \), \( b = -1.03 \) [3]

(c)
\( k = 61.1 \), \( c = 0.923 \)

Model: \( T(t) = kc^t + 25 \)

Use adjusted data: \( t = (0, 5, 10, 15, 20, 25) \), \( y = (60, 30, 18, 11, 6, 3) \)

Fit \( y = kc^t \) using exponential regression

\( k \approx 61.0848 \approx 61.1 \)

\( c \approx 0.923029 \approx 0.923 \)

Result: \( k = 61.1 \), \( c = 0.923 \) [3]

(d)
Soo Min’s model has a higher \( R^2 \) value

Method 1
Akira’s model: \( T(t) = 244 t^{-1.03} + 25 \), \( R^2 \approx 0.7657 \)

Soo Min’s model: \( T(t) = 61.1 \times 0.923^t + 25 \), \( R^2 \approx 0.9233 \)

Higher \( R^2 \) (0.9233 vs. 0.7657) indicates better fit

Method 2
Exponential decay in Soo Min’s model aligns with Newton’s law of cooling

Power model does not match physical cooling process

Result: Soo Min’s model is better [2]