This calculation involves exponential growth over multiple time periods. Starting with 160 people and multiplying by 1.25 each day, after 3 days we have $160(1.25)^3$. Calculate $(1.25)^3$ = 1.953125, then 160 × 1.953125 = 312.5 people. This represents compound growth where each day's increase is 25% of the previous day's total, not the original amount. Students sometimes multiply 160 × 1.25 × 3 = 600, incorrectly treating it as simple interest. For exponential growth, always raise the growth factor to the power of time periods, then multiply by the initial value.

This problem requires finding a linear equation from two points and then extrapolating. Given points (1,6) and (5,14), first find the slope: m = (14-6)/(5-1) = 8/4 = 2. Using point-slope form with (1,6): y - 6 = 2(x - 1), which simplifies to y = 2x + 4. For a 9-mile ride, y = 2(9) + 4 = 18 + 4 = 22 dollars. The key steps are calculating slope from two points, finding the equation, then evaluating at the desired x-value. A common error is using the slope incorrectly or forgetting the y-intercept when writing the final equation.

This problem involves modeling exponential growth with a non-unit time factor. Since the culture doubles every 3 hours, after t hours it has gone through t/3 doubling periods. Starting with 200 cells, the model is N = 200(2)^(t/3), which correctly accounts for the 3-hour doubling time. For example, after 6 hours (2 doubling periods), N = $200(2)^2$ = 800 cells. Students often write N = $200(2)^t$, forgetting to adjust for the doubling period. When the growth period doesn't match the time unit, always include the appropriate fraction in the exponent. Test your model at easy values like t = 3 to verify correctness.

This problem models exponential decay of battery charge over time. Since the battery loses 12% of its remaining charge each hour, it retains 88% (100% - 12% = 88%), giving a decay factor of 0.88. The exponential model B = $100(0.88)^h$ correctly represents this, starting at 100% when h = 0. A linear model like B = 100 - 12h would incorrectly subtract 12% of the original charge each hour, eventually going negative. The key insight is that percentage-based changes are multiplicative, not additive. When a quantity changes by a constant percentage, always use an exponential model with base (1 ± rate).

This problem requires finding an equation from given data points showing plant growth over time. Checking the points (0,6), (1,9), (2,12), (3,15), we see the height increases by 3 cm each week - a constant difference indicating linear growth. The equation h = 3t + 6 correctly models this, with slope 3 (growth rate) and y-intercept 6 (initial height). To verify, check that each point satisfies the equation: when t = 2, h = 3(2) + 6 = 12 ✓. Students might be tempted by exponential models, but constant differences always indicate linear relationships. Calculate differences between consecutive y-values first to determine the model type.

This question tests interpretation of parameters in a linear model for subscriber growth. In S = 12000 + 800t, the coefficient 800 represents the rate of change - specifically, the increase in subscribers per month. The constant term 12000 is the initial number of subscribers at launch (when t = 0). Students often confuse these roles or misinterpret 800 as a percentage rather than an absolute increase. In linear models of the form y = mx + b, m always represents the rate of change per unit of the independent variable. When solving context problems, always identify what each parameter means in real-world terms.

This problem asks for an exponential growth equation modeling compound interest. Since the account grows by 6% each month, the growth factor is 1 + 0.06 = 1.06, and we multiply the initial amount by this factor raised to the power of months: B = 500(1.06)^m. Linear models like B = 500 + 0.06m would only add 6% of the original $500 each month, not 6% of the current balance. The key distinction is that exponential growth compounds - each month's 6% is calculated on the new, larger balance. Always use exponential models for percentage-based growth or decay.

This question tests understanding of exponential decay model parameters, specifically what each number represents in context. In the equation V = $22000(0.82)^t$, the number 22000 is the coefficient that appears when t = 0, making it the initial value of the car. The base 0.82 represents the decay factor (1 - 0.18 = 0.82, since the car loses 18% of its value each year). A common mistake is thinking 22000 represents the value after one year, but that would be $22000(0.82)^1$ = 18040. When interpreting exponential models, always evaluate at t = 0 to find the initial value.

We need to model bacteria that start at 500 cells and double every 3 hours. Since the population doubles (multiplies by 2), this is exponential growth with base 2. However, doubling happens every 3 hours, not every hour, so we need to adjust the exponent. After h hours, the number of 3-hour periods is h/3, giving us N = 500(2)^(h/3). We can verify: at h=0, $N=500(2)^0$=500; at h=3, $N=500(2)^1$=1000; at h=6, $N=500(2)^2$=2000. A common error is using N = $500(2)^h$, which would mean doubling every hour instead of every 3 hours. When the growth period doesn't match the time unit, divide the time variable by the growth period in the exponent.

This question asks about the meaning of 0.85 in the exponential decay model V(t) = $24000(0.85)^t$. In exponential models of the form y = $a(b)^t$, the base b represents the decay factor when 0 < b < 1. A base of 0.85 means the car retains 85% of its value each year (or loses 15%). To verify: if the car keeps 85% of its value, then after 1 year it's worth 24000(0.85) = 0.85 × 24000, which is 85% of the original. The common error is thinking 0.85 means losing 85%, but remember that the base represents what remains, not what is lost. For percent problems, convert the remaining percentage to decimal form for the base.