the data shows the length and weight of a species of fish. based on the data, which is closest to the length…

Explanation:

Step1: Assume linear - relationship

Let the length be $x$ and the weight be $y$. Assume $y = ax + b$. Take two points $(8,1.21)$ and $(10,1.42)$ from the table. First, calculate the slope $a$: $a=\frac{y_2 - y_1}{x_2 - x_1}=\frac{1.42 - 1.21}{10 - 8}=\frac{0.21}{2}=0.105$.

Step2: Calculate the intercept $b$

Substitute $x = 8$, $y = 1.21$ and $a = 0.105$ into $y=ax + b$: $1.21=0.105\times8 + b$. $1.21 = 0.84 + b$. $b=1.21 - 0.84 = 0.37$. So the linear - equation is $y = 0.105x+0.37$.

Step3: Solve for $x$ when $y = 4.2$

Substitute $y = 4.2$ into $y = 0.105x+0.37$: $4.2=0.105x + 0.37$. $0.105x=4.2 - 0.37$. $0.105x = 3.83$. $x=\frac{3.83}{0.105}\approx36.476$. This is wrong. Let's use another method - interpolation.

Let's use linear interpolation. We know that when $x_1 = 16$, $y_1 = 2.24$ and assume another point $(x_2,y_2)$ such that $y_2 = 4.2$. The difference in weight $\Delta y=y_2 - y_1=4.2 - 2.24 = 1.96$. The average increase in weight per unit length from the table: From $x = 8$ to $x = 10$, $\Delta y_1=1.42 - 1.21 = 0.21$, $\Delta x_1 = 2$, rate $r_1=\frac{0.21}{2}=0.105$. From $x = 10$ to $x = 12$, $\Delta y_2=1.63 - 1.42 = 0.21$, $\Delta x_2 = 2$, rate $r_2=\frac{0.21}{2}=0.105$. From $x = 12$ to $x = 14$, $\Delta y_3=1.95 - 1.63 = 0.32$, $\Delta x_3 = 2$, rate $r_3=\frac{0.32}{2}=0.16$. From $x = 14$ to $x = 16$, $\Delta y_4=2.24 - 1.95 = 0.29$, $\Delta x_4 = 2$, rate $r_4=\frac{0.29}{2}=0.145$. The average rate $r=\frac{0.21 + 0.21+0.32 + 0.29}{4\times2}=\frac{1.03}{8}=0.12875$. The additional length $\Delta x=\frac{\Delta y}{r}=\frac{1.96}{0.12875}=15.22$. The length $x=x_1+\Delta x=16 + 15.22\approx28$ inches.

Answer:

D. 28 inches