Options for the first box: inverse and direct Options: 275, 50, 5,000, 13,750 Options for the third box: $275.00, $137.50, $550.00
![Options for the first box inverse and direct Options 275 50 5000 13750 Options for the third box 27500 13750 55000 class=](https://us-static.z-dn.net/files/dc8/c1ecc2acfd47d8f07f8c675ebd21950c.png)
• The proportional situation represents an Inverse Variation
,• The constant of variation, k = 13750
,• If all 100 students participate in the fundraiser, each will contribute $137.5
Explanation:From the given information, we notice that the more students involved in the fundraiser, the less the amount each student needs to contribute.
This is an INVERSE VARIATION
Let s represent the number of students who participated in the fundraiser, and t represents the amount needed to be contributed by each student, we have:
[tex]\begin{gathered} s\propto\frac{1}{t} \\ \\ \Rightarrow s=\frac{k}{t} \\ \\ OR \\ st=k \end{gathered}[/tex]To find k, we use the information that s = 50 when t = 275
So,
[tex]\begin{gathered} k=50\times275 \\ =13750 \end{gathered}[/tex]From the above, we have the formula:
[tex]st=13750[/tex]If 100 students participate in the fundraiser, we have:
[tex]\begin{gathered} 100t=13750 \\ t=\frac{13750}{100}=137.5 \end{gathered}[/tex]Each student needs to contribute $137.5