Kao measured the temperature of a cup of hot chocolate placed in a room with a constant temperature of 70

Question

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 $T of m$ , in degrees Fahrenheit, of Kao’s hot chocolate m minutes after it started cooling?

A.

$T of m equals, 185 times, 1 point 2 5 to the m power$

B.

$T of m equals, 185 times, 0 point 8 5 to the m power$

C.

$T of m equals, open parenthesis, 185 minus 70, close parenthesis, times, 0 point 7 5 raised to the m over 5 power$

D.

$T of m equals, 70 plus, 115 times, 0 point 7 5 raised to the m over 5 power$

Accepted Answer

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, $T of m$ decreases as time increases. Using the function, the temperature, in °F, at 6:00 p.m. can be estimated as $T of 0$ and is equal to $70 plus 115, times, open parenthesis, 0 point 7 5, close parenthesis, raised to the power zero fifths, equals 185$ . The temperature, in °F, at 6:05 p.m. can be estimated as $T of 5$ and is equal to $70 plus 115, times, open parenthesis, 0 point 7 5, close parenthesis raised to the power five fifths$ , which is approximately 156°F. Finally, the temperature, in °F, at 6:10 p.m. can be estimated as $T of 10$ and is equal to $70 plus 115, times, open parenthesis, 0 point 7 5, close parenthesis, raised to the power ten fifths$ , which is approximately 135°F. Since these three given values of m and their corresponding values for $T of m$ can be verified using the function $T of m equals, 70 plus 115, times, open parenthesis, 0 point 7 5, close parenthesis, raised to the power the fraction m over 5$ , 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, $1 point 2 5$ , results in the value of $T of m$ increasing over time rather than decreasing. Choice B is incorrect because when m is large enough, $T of m$ becomes less than 70. Choice C is incorrect because the maximum value of $T of m$ at 6:00 p.m. is 115°F, not 185°F.

Question kao-me

Kao measured the temperature of a cup of hot chocolate placed in a room with a

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