To find out how much Simon will have to pay for 30 cases of water, we need to estimate the cost using the given price data for different quantities of cases. We have price information for 15, 20, and 35 cases of water as shown in the table:
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Number of cases} & \text{Price in dollars} \\
\hline
15 & 66.00 \\
\hline
20 & 88.00 \\
\hline
35 & 154.00 \\
\hline
\end{array}
\][/tex]
Simon wants to buy 30 cases, which is between the quantities given for 20 cases and 35 cases. To estimate the price for 30 cases, we use a method called linear interpolation, which assumes that the price increases at a constant rate between the given points.
From the table, we see:
- For 20 cases, the price is \[tex]$88.00.
- For 35 cases, the price is \$[/tex]154.00.
Calculate the increase in price per additional case between 20 and 35 cases:
[tex]\[
\text{Increase in price} = 154.00 - 88.00 = 66.00
\][/tex]
[tex]\[
\text{Number of additional cases} = 35 - 20 = 15
\][/tex]
[tex]\[
\text{Price increase per case} = \frac{66.00}{15} = 4.40
\][/tex]
To find the price for 30 cases, we calculate the number of additional cases from the closest known quantity (20 cases) and multiply by the price increase per case:
[tex]\[
\text{Number of additional cases from 20 to 30 cases} = 30 - 20 = 10
\][/tex]
[tex]\[
\text{Price for 30 cases} = \text{Price for 20 cases} + (10 \times \text{Price increase per case})
\][/tex]
[tex]\[
\text{Price for 30 cases} = 88.00 + (10 \times 4.40) = 88.00 + 44.00 = 132.00
\][/tex]
Now we see that the interpolated price for 30 cases is \[tex]$132.00. We compare this result to the possible given options:
- \$[/tex]99.00
- \[tex]$119.00
- \$[/tex]121.00
- \[tex]$132.90
The closest option to our interpolated price (\$[/tex]132.00) is \$132.90.
Therefore, Simon will most likely have to pay:
[tex]\[
\boxed{132.90}
\][/tex]
