This graph describes Mina's displacement from her house during a walk on a straight path. What is the average velocity on the interval 4<t<18 4leqslant tleqslant 18 800. Displacement (m) 700 800 500 400 300 200 100 Time (min) n 12

To calculate the **average velocity** of Mina on the interval \( 4 \leq t \leq 18 \), we use the formula for average velocity:<br /><br />\[<br />v_{\text{avg}} = \frac{\Delta x}{\Delta t}<br />\]<br /><br />where:<br />- \( \Delta x \) is the change in displacement (final displacement - initial displacement),<br />- \( \Delta t \) is the change in time (final time - initial time).<br /><br />### Step 1: Identify the displacement at \( t = 4 \) and \( t = 18 \)<br />From the graph:<br />- At \( t = 4 \, \text{min} \), the displacement is \( x(4) = 700 \, \text{m} \).<br />- At \( t = 18 \, \text{min} \), the displacement is \( x(18) = 300 \, \text{m} \).<br /><br />### Step 2: Calculate \( \Delta x \) and \( \Delta t \)<br />- \( \Delta x = x(18) - x(4) = 300 - 700 = -400 \, \text{m} \),<br />- \( \Delta t = 18 - 4 = 14 \, \text{min} \).<br /><br />### Step 3: Compute the average velocity<br />Substitute into the formula:<br /><br />\[<br />v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{-400}{14} \approx -28.57 \, \text{m/min}.<br />\]<br /><br />### Final Answer:<br />The average velocity on the interval \( 4 \leq t \leq 18 \) is approximately:<br /><br />\[<br />\boxed{-28.57 \, \text{m/min}}<br />\]