ACT · Math · Modeling
Pasar de una situación real a un modelo (lineal o exponencial) y usarlo para predecir.
Traducir situaciones reales a modelos lineales o exponenciales y usarlos. En el examen real estas preguntas también cuentan en otra categoría.
Pierde el 15 % al año: 24 $000(0{,}85)t$.
Intenta responder antes de abrir la solución. Las preguntas están en el idioma del examen.
A taxi company charges a fixed fee of $3.50 plus $2.25 for each mile driven. What is the total charge for an 8-mile ride?
Respuesta correcta: B. $21.50
The model is $C=3.50+2.25m$. For $m=8$: $3.50+2.25(8)=3.50+18.00$ = \$21.50.
\$18.00 leaves out the fixed fee. \$23.75 charges for 9 miles. \$$46.00=(3.50+2.25)\times8$ incorrectly charges the fixed fee for every mile.
A car was bought for \$24,000. Its value decreases by 15% each year. Which of the following expressions models the value of the car, in dollars, $t$ years after it was bought?
Respuesta correcta: A. $24{,}000(0.85)t$
Losing 15% each year means keeping 85% of the previous year’s value, so the value is multiplied by 0.85 every year: $24{,}000(0.85)t$.
$24{,}000(0.15)t$ keeps only 15% each year. $24{,}000(1.15)t$ models a 15% increase. $24{,}000-0.15t$ subtracts 15 cents per year, a linear decrease that does not represent a percent change.
The population of a species of frog in a wetland was 2,000 at the start of a study and decreased by 8% each year after that. Based on this model, what is the first whole number of years after the start of the study at which the population is below 1,000?
Respuesta correcta: D. $9$
The model is $P(t)=2{,}000(0.92)t$. We need $0.92t<0.5$. Checking: $0.928\approx0.513$ ($P\approx1{,}026$, still above 1,000) and $0.929\approx0.472$ ($P\approx943$). The first whole year is 9. (With logarithms: $t>\frac{\ln0.5}{\ln0.92}\approx8.3$.)
$7$ treats the decrease as linear: 8% of 2,000 is 160 per year, and $2{,}000-7(160)=880$. $8$ rounds $8.3$ down instead of up. $6$ is $1{,}000\div160$ rounded down.