A car originally worth 25,000 depreciates 20% per year. Which equation models this situation? y=25,000(.8)^t y=25,000(80)^t C y=25,000(.2)^t y=25,000(1.2)^t

To model the depreciation of the car, we need to use an exponential decay equation. The general form of an exponential decay equation is:<br /><br />\[ y = a(b)^t \]<br /><br />where:<br />- \( y \) is the value of the car after \( t \) years,<br />- \( a \) is the initial value of the car,<br />- \( b \) is the decay factor,<br />- \( t \) is the time in years.<br /><br />Given:<br />- The initial value of the car, \( a \), is \$25,000.<br />- The car depreciates by 20% per year, which means it retains 80% of its value each year.<br /><br />The decay factor \( b \) is calculated as:<br />\[ b = 1 - \text{depreciation rate} = 1 - 0.20 = 0.80 \]<br /><br />So, the equation that models this situation is:<br />\[ y = 25,000(0.8)^t \]<br /><br />Therefore, the correct answer is:<br />\[ y = 25,000(0.8)^t \]