Question

Felipe works for a company that makes frozen pizzas. He cooked a sample

of pizzas in different ovens and let them cool in different rooms. He
noticed an exponential relationship between cooling times and pizza
temperatures after cooking.
Felipe took the base 10 logarithm for the temperatures only, and he
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) = 2.402 - 0.005 (time)
According to this model, what is the predicted temperature of a pizza that
cools for 15 minutes?
You may round your answer to the nearest whole degree.

Answer 1

To predict the temperature of a pizza that cools for 15 minutes using the given model, we can substitute the value of time into the equation and solve for the transformed temperature.

The equation provided is:

log(temp) = 2.402 - 0.005(time)

Substituting time = 15 minutes:

log(temp) = 2.402 - 0.005(15)

log(temp) = 2.402 - 0.075

log(temp) = 2.327

To find the actual temperature, we need to take the inverse logarithm (base 10) of both sides of the equation:

temp = 10^(2.327)

temp ≈ 199.4 degrees Celsius

Therefore, according to the given model, the predicted temperature of a pizza that cools for 15 minutes is approximately 199.4 degrees Celsius.

Explanation: