Question

A pot of soup, currently 16°C above room temperature, is left out to cool. If that temperature

difference decreases by 6% per minute, then what will the difference be in 16 minutes?
If necessary, round your answer to the nearest tenth.

°C

Answer 1

5.9 C

Explanation:

Plug in 16°C for the initial amount, 0.06 for the rate of decrease, and 16 minutes for the time elapsed.

y= a(1–r)t

y= 16(1–0.06)16 Plug in values

y= 16(0.94)16 Subtract

y= 16(0.37157…) Simplify

y= 5.94518… Multiply

Round to the nearest tenth.

5.94518… → 5.9

To the nearest tenth, the difference in temperature will be 5.9°C.