\(\mathrm{C(h) = 150(0.92)^{(h/4)}}\)The function C gives the concentration, in milligrams per liter, of a medication in the bloodstream h hours...

1. TRANSLATE the function components

• Given: \(\mathrm{C(h) = 150(0.92)^{(h/4)}}\)
• This is an exponential decay function where:
  • 150 = initial concentration
  • 0.92 = decay factor (applied every 4 hours due to h/4)
  • h = time in hours

2. INFER what the decay factor means

• The decay factor \(\mathrm{0.92}\) tells us what fraction remains after each 4-hour period
• If \(\mathrm{0.92}\) (or \(92\%\)) remains, then the rest has decreased
• Percentage decrease = \(100\% - 92\% = 8\%\)

Answer: C (8)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students misinterpret what \(\mathrm{0.92}\) represents

Many students see \(\mathrm{0.92}\) and think "this must be the percentage that decreases," leading them to select Choice B (4) by somehow connecting \(\mathrm{0.92}\) to a 4% decrease, or they get confused about the relationship entirely.

Second Most Common Error:

Poor TRANSLATE reasoning: Students focus on the wrong part of the function

Some students see the "/4" in the exponent and think this directly represents the percentage decrease, leading them to select Choice B (4). They miss that the "/4" just indicates the time period over which the decay factor applies.

The Bottom Line:

This problem tests whether students understand that in exponential decay, the base represents what remains after each time period, not what decreases. The key insight is that \(\mathrm{0.92}\) means 92% stays, so 8% goes away.