Shawna works for a company that makes frozen pizzas. She cooked a sample of pizzas in
different ovens and let them cool in different rooms. She noticed an exponential relationship
between cooling times and pizza temperatures after cooking.
Shawna took the base 10 logarithm for the temperatures only, and she noticed a linear
relationship between the cooling times and the transformed temperatures.
Here's the least-squares regression equation for the transformed data, where "time"
represents minutes spent cooling, and "temp" is the pizza's temperature in degrees Celsius.
log(temp) = 2.390 -0.004(time)
According to this model, what is the predicted temperature of a pizza that cools for 20
minutes?
You may round your answer to the nearest whole degree.

Respuesta :

Answer:

  204 °C

Step-by-step explanation:

You want the predicted temperature of a pizza that has cooled for 20 minutes, as predicted by the formula ...

  log(temp) = 2.39 -0.004·time

where temp is in degrees Celsius, and time is in minutes.

Evaluating the formula

Using time = 20 in the formula, we find ...

  log(temp) = 2.39 - 0.004·20 = 2.39 -0.08 = 2.31

Taking the antilog, we find the temperature to be ...

  temp = 10^2.31 ≈ 204.2 . . . . . degrees C

The predicted temperature of the pizza is 204 °C.