sat suite question viewer
Kao measured the temperature of a cup of hot chocolate placed in a room with a constant temperature of 70 degrees Fahrenheit (°F). The temperature of the hot chocolate was 185°F at 6:00 p.m. when it started cooling. The temperature of the hot chocolate was 156°F at 6:05 p.m. and 135°F at 6:10 p.m. The hot chocolate’s temperature continued to decrease. Of the following functions, which best models the temperature 
, in degrees Fahrenheit, of Kao’s hot chocolate m minutes after it started cooling?
Explanation
Choice D is correct. The hot chocolate cools from 185°F over time, never going lower than the room temperature, 70°F. Since the base of the exponent in this function, 0.75, is less than 1, 
decreases as time increases. Using the function, the temperature, in °F, at 6:00 p.m. can be estimated as 
and is equal to 
. The temperature, in °F, at 6:05 p.m. can be estimated as 
and is equal to 
, which is approximately 156°F. Finally, the temperature, in °F, at 6:10 p.m. can be estimated as 
and is equal to 
, which is approximately 135°F. Since these three given values of m and their corresponding values for can be verified using the function 
, this is the best function out of the given choices to model the temperature of Kao’s hot chocolate after m minutes.
Choice A is incorrect because the base of the exponent,
, results in the value of increasing over time rather than decreasing. Choice B is incorrect because when m is large enough, becomes less than 70. Choice C is incorrect because the maximum value of at 6:00 p.m. is 115°F, not 185°F.