A researcher measures the mass (in mg) of a culture after days. The data are shown below. | (days) | (mg) | | --- | --- | | 0 | 500 | | 2 | 760 | | 4 | 1160 | | 6 | 1760 | | 8 | 2680 | Based on the pattern in the data, which function is the most appropriate model for as a function of ?

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SAT Two-Variable Data Question 21 | Free SAT Prep | Aniko

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Question 21·Hard·Two-Variable Data: Models and Scatterplots

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A researcher measures the mass $m$ (in mg) of a culture after $t$ days. The data are shown below.

$t$ (days)	$m$ (mg)
0	500
2	760
4	1160
6	1760
8	2680

Based on the pattern in the data, which function is the most appropriate model for $m$ as a function of $t$ ?

For model-selection questions with a table of $(x,y)$ data, first check whether the changes in $y$ are roughly constant (linear) or whether the ratios $y_{next}/y_{current}$ are roughly constant (exponential). Once you see a nearly constant ratio, estimate that ratio, identify the initial value from $x=0$ , and write an exponential of the form $a\cdot b^{x/k}$ if the data are spaced every $k$ units. Finally, compare this form to the answer choices and quickly test one or two points to confirm the best match.

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Linear vs exponential

Look at how much the mass increases every 2 days. Are those increases staying the same, or are they changing?

Try ratios instead of differences

Instead of subtracting, divide each new mass by the previous one (for example, $760\div 500$ ). Do these quotients stay about the same?

Connect the ratio to an equation

Once you know the approximate multiplication factor every 2 days and the starting mass at $t=0$ , think about how to write an exponential function that uses $t$ (in days) in the exponent.

Desmos Guide

Enter the data as a table

In Desmos, click the "+" icon and choose "Table." Enter the $t$ values (0, 2, 4, 6, 8) in the first column and the corresponding $m$ values (500, 760, 1160, 1760, 2680) in the second column. You will see the data points plotted.

Graph the linear model option

In a new expression line, type y = 500 + 160x. Compare this line to the plotted data points. Check whether the line passes close to all the points or if it misses by a lot, especially at larger $t$ values.

Graph the exponential model options with base 1.52

On separate lines, type y = 500(1.52)^(x) and y = 500(1.52)^(x/2). Compare each curve to the data points. Look for which curve goes through or very close to all the points in the table.

Graph the exponential decay option

On another line, type y = 500(0.76)^(x/2). Since this has a base less than 1, note how the curve behaves compared to the increasing data points. Decide whether it can reasonably model the given masses.

Choose the best-fitting model

After viewing all four graphs together with the data points, decide which function’s graph best passes through or closely follows the plotted points. That function is the most appropriate model.

Step-by-step Explanation

Check if the data look linear (constant difference)

Compute the changes in mass over each 2-day interval:

From $t=0$ to $t=2$ : $760-500=260$
From $t=2$ to $t=4$ : $1160-760=400$
From $t=4$ to $t=6$ : $1760-1160=600$
From $t=6$ to $t=8$ : $2680-1760=920$

These differences are not constant; they are increasing. That means a linear function like $500+160t$ is not a good model. We should look for an exponential pattern (constant ratio) instead.

Check for a constant ratio (exponential pattern)

Now compute the ratios (new mass divided by previous mass) for each 2-day interval:

From $0$ to $2$ days: $760\div 500 = 1.52$
From $2$ to $4$ days: $1160\div 760 \approx 1.53$
From $4$ to $6$ days: $1760\div 1160 \approx 1.52$
From $6$ to $8$ days: $2680\div 1760 \approx 1.52$

These ratios are all very close to $1.52$ . That suggests an exponential model where the mass is multiplied by about $1.52$ every 2 days.

Translate the pattern into a model form (without selecting yet)

At $t=0$ , the mass is $500$ mg, so the model should have an initial value of $500$ . Because the culture is multiplied by about $1.52$ every 2 days, the exponent should count how many 2-day periods have passed; in $t$ days, that is $t/2$ . Therefore, we are looking for an exponential model with initial value $500$ and a growth factor of about $1.52$ applied once per 2 days (exponent $t/2$ ). We will match this description to an answer choice in the next step.

Match the model to the answer choices

From the answer choices, the only function with initial value $500$ and growth factor $1.52$ applied once per 2 days (exponent $t/2$ ) is

\boxed{m(t)=500(1.52)^{t/2}}.

This is the most appropriate model for the data.

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Step-by-step Explanation

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