Question

the function f(t) = 349.2(0.98)^t models the relationship between t, the time an oven spends cooling and the temperature of the oven. oven cooling time time (minutes) t oven temperature (degrees fahrenheit) f(t) 5 315 10 285 15 260 20 235 25 210 for which temperature will the model most accurately predict the time spent cooling? 0 100 300 400

Explanation:

Step1: Analyze the model and data

The function is \( f(t) = 349.2(0.98)^t \), an exponential decay model. The table gives oven temperatures at different times. We need to see which temperature is closest to the range of the table's \( f(t) \) values (210 - 315) or where the model's prediction aligns best.

Step2: Evaluate each option

• For \( t = 0 \), \( f(0)=349.2(0.98)^0 = 349.2 \), which is higher than table values (max 315).
• For \( t \) leading to \( f(t)=100 \): Let's see table values are 210 - 315, 100 is lower, but check the model's range. Wait, the table's \( f(t) \) starts at 315 (t=5) and decreases. The model at t=0 is 349.2, then decreases. The table's values are 315, 285, 260, 235, 210 (decreasing). So 300: Let's see, when is \( f(t)=300 \)? Solve \( 349.2(0.98)^t = 300 \). \( (0.98)^t = 300/349.2 \approx 0.859 \). Take log: \( t = \ln(0.859)/\ln(0.98) \approx (-0.152)/(-0.0202) \approx 7.5 \). The table has t=5 (315) and t=10 (285). 300 is between 315 (t=5) and 285 (t=10), so the model is likely accurate here. For 400: \( f(t)=400 \) would require \( 349.2(0.98)^t = 400 \), but \( 349.2 < 400 \), so impossible (exponential decay, starts at 349.2 and decreases). So 300 is within the model's initial range (starts at 349.2, decreases) and the table's range (210 - 315), so the model is most accurate for 300.

Answer:

300