1. Proportional
2. Not Proportional
3. Not Proportional
4. Proportional
5. Proportional
<h2>52</h2>
Step-by-step explanation:
Let the number of fruit trees planted additionally be ![n](https://tex.z-dn.net/?f=n)
Initially it is given that there are
trees.
Number of trees after planting
additional trees is ![n+20](https://tex.z-dn.net/?f=n%2B20)
Let the yield due to each tree after planting
additional trees be ![y](https://tex.z-dn.net/?f=y)
Initially it is given that ![y=252](https://tex.z-dn.net/?f=y%3D252)
Yield due to each tree after planting
trees is ![y=252-(3\times n)](https://tex.z-dn.net/?f=y%3D252-%283%5Ctimes%20n%29)
![\text{total yield}=\text{yield for each tree}\times\text{total number of trees}](https://tex.z-dn.net/?f=%5Ctext%7Btotal%20yield%7D%3D%5Ctext%7Byield%20for%20each%20tree%7D%5Ctimes%5Ctext%7Btotal%20number%20of%20trees%7D)
![\text{total yield}=(252-3n)(20+n)](https://tex.z-dn.net/?f=%5Ctext%7Btotal%20yield%7D%3D%28252-3n%29%2820%2Bn%29)
=![252\times 20-192n-3n^{2}](https://tex.z-dn.net/?f=252%5Ctimes%2020-192n-3n%5E%7B2%7D)
To maximise yield,we take that value of
for which ![\frac{d\text{total yield}}{dn}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctext%7Btotal%20yield%7D%7D%7Bdn%7D)
![=0](https://tex.z-dn.net/?f=%3D0)
=![192-6n](https://tex.z-dn.net/?f=192-6n)
So,
and ![n=32](https://tex.z-dn.net/?f=n%3D32)
So,32 additional trees has to be planted to maximise yield.
So,there should be 52 trees in total
Total Kenilworth students = 4 boys + 3 girls = 7.
Total students = 9+3+7+6+4+5=34.
So, the probability of a student from Kenilworth being called upon is 7/34=0.20588, or 20.588%