Ariana 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.

Ariana 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.
$$
\widehat{\log (\widehat{\text { temp }})}=2.365-0.003(\text { time })
$$

According to this model, what is the predicted temperature of a pizza that cools for 10 minutes?
You may round your answer to the nearest whole degree.
${ }^{\circ} \mathrm{C}$

Question

Ariana 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.

Ariana 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.
$$
\widehat{\log (\widehat{\text { temp }})}=2.365-0.003(\text { time })
$$

According to this model, what is the predicted temperature of a pizza that cools for 10 minutes?
You may round your answer to the nearest whole degree.
${ }^{\circ} \mathrm{C}$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the predicted temperature after 10 minutes, we first calculate the value of the logarithm of the temperature using the regression equation. ‖ ‖ step 2 ⋮ Substitute the time value into the regression equation: $\log( ext{temp}) = 2.365 - 0.003 imes 10$. ‖ ‖ step 3 ⋮ Perform the calculation: $\log( ext{temp}) = 2.365 - 0.03$. ‖ ‖ step 4 ⋮ According to the asksia-ll calculation list, the result is $\log( ext{temp}) \approx 2.365 - 0.03 = 2.335$. ‖ ‖ step 5 ⋮ To find the temperature, we need to take the antilog (base 10) of the logarithmic result: $ ext{temp} = 10^{\log( ext{temp})} = 10^{2.335}$. ‖ ‖ step 6 ⋮ Using the asksia-ll calculator result, the antilog of 2.335 is approximately 216.272. ‖ ‖ step 7 ⋮ Round the temperature to the nearest whole degree: $ ext{temp} \approx 216$. ‖ ∻Answer∻ ⚹ The predicted temperature of a pizza that cools for 10 minutes is approximately 216 degrees Celsius. ⚹ ∻Key Concept∻ ⚹Logarithmic to Exponential Conversion⚹ ∻Explanation∻ ⚹To find the temperature from the logarithmic form, we convert the log equation back to its exponential form using the antilogarithm (base 10).⚹◊What is the predicted temperature of a pizza that cools for 20 minutes according to the given model?⍭ Generate me a similar question◊

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