The table shows the rate at which water is being pumped into a swimming pool. Time (min) 2 4 8 12 Amount (gal) 44 88 176 264 5 Use the unit rate and the amount of water pumped after 12 minutes to find how much water will have 1 been pumped into the pool after 13 minutes. Complete the explanation of your reasoning. 2 The unit rate is 1 gal/min. So 15 minutes after 12 minutes, an additional gallons will be pumped in. A total of 1 gallons will have been pumped into the pool after 13= minutes. 2

The table shows the rate at which water is being pumped into a swimming pool Time min 2 4 8 12 Amount gal 44 88 176 264 5 Use the unit rate and the amount of wa class=

Respuesta :

The unit rate is 22 gal/min. So, you have

[tex]\begin{gathered} \frac{22\text{ gal}}{1\text{ min}}=\frac{x\text{ gal}}{1\frac{1}{2}\text{ min}} \\ \text{ Multiply }by\text{ }1\frac{1}{2}\text{ min on both sides of the equation} \\ \frac{22\text{ gal}}{1\text{ min}}\cdot1\frac{1}{2}\text{ min}=\frac{x\text{ gal}}{1\frac{1}{2}\text{ min}}\cdot1\frac{1}{2}\text{ min} \\ 22\cdot\frac{3}{2}\text{ gal }=x\text{ } \end{gathered}[/tex][tex]\begin{gathered} \text{Because } \\ 1\frac{1}{2}=\frac{1\cdot2+1}{2}=\frac{2+1}{2}=\frac{3}{2} \end{gathered}[/tex]

Then, you have

[tex]\begin{gathered} 22\cdot\frac{3}{2}\text{ gal }=x\text{ gal} \\ \frac{66}{2}\text{ gal }=x \\ 33\text{ gal = x} \end{gathered}[/tex]

So, a minute and a half after 12 minutes, an additional 33 gallons will be pumped in. A total of 297 gallons

[tex]264\text{ gal + 33 gal = }297\text{ gal}[/tex]

will have been pumped into the pool after 13 and a half minutes.