Question 22·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

Each table input rises by $2$, so $x/2$ rises by $1$ per two-hour step. The shift $c$ sits outside the power, so $b$ multiplies $y-c$, not the full value of $y$. Ratios of the listed $y$ values won't give you $b$.

Put the given pairs in a Desmos table and type $y_1\sim ab^{x_1/2}+c$. The $\sim$ tells Desmos to run a regression, fitting the constants to the rows. Under PARAMETERS, it gives $a=6$, $b=2$, and $c=-1$. Include the shift in what you type; the full outputs don't double.

At $x=7$, rewrite the exponent as three whole steps and half another: $\frac{7}{2}=3+\frac{1}{2}$. A half power is a square root, so split the power: $2^{7/2}=2^3\cdot2^{1/2}$. Simplify the whole-number power: $2^{7/2}=8\sqrt{2}$. Don't round the square root when the choices are exact.

Type $f(x)=ab^{x/2}+c$, then $f(7)$. Desmos shows $f(7)\approx66.882$; $f(7)$ means the model's output at hour $7$. Use the power from the previous step for the exact value: $f(7)=6(8\sqrt{2})-1$. Multiply the coefficients: $f(7)=48\sqrt{2}-1$. The predicted concentration index after $7$ hours is $48\sqrt{2}-1$. Choice B.

Subtract consecutive outputs. These differences let you work around the unknown fixed shift:

$D_2=47-23=24$. The changes double even though the outputs themselves don't.

For the first two rows, subtract the model outputs; the fixed $c$ cancels:

$D_0=a(b-1)$. For the next pair, cancel the same shift:

$D_1=ab(b-1)$. So the second change contains one extra factor of $b$.

Because $D_1$ is $b$ times $D_0$, divide the changes:

$b=2$. This is the factor for the exponential part every two hours, not every hour.

The first change is $D_0=a(b-1)$. Since $b=2$, the factor $b-1$ is $1$, so $a$ equals that first change:

At $x=0$, the power is $b^0=1$, so the first row gives: $5=a+c$. Substitute $a=6$: $5=6+c$. Subtract $6$: $-1=c$. The negative shift explains why the listed outputs don't double.

Put the constants and $x=7$ into the model: $y(7)=6\cdot2^{7/2}-1$. Split the exponent into three whole steps and a half step: $y(7)=6\cdot2^3\sqrt{2}-1$. Multiply the coefficients: $y(7)=48\sqrt{2}-1$. So the seven-hour concentration index is $48\sqrt{2}-1$. Choice B.