![P=-(1)/(20)x^2+5x-30](https://img.qammunity.org/2023/formulas/mathematics/college/p115td68w4e1rj2oyjnll097ab6h3s5z42.png)
a) Use the equation above to complete the table: substitute the x in the equation by the given values in the tabe to find the corresponding value of P:
![\begin{gathered} x=0 \\ P=-(1)/(20)(0)^2+5(0)-30 \\ P=-30 \\ \\ x=20 \\ P=-(1)/(20)(20)^2+5(20)-30 \\ P=-20+100-30 \\ P=50 \\ \\ x=30 \\ P=-(1)/(20)(30)^2+5(30)-30 \\ P=-(900)/(20)+150-30 \\ P=-45+150-30 \\ P=75 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/x6gltuni4qdlkdd6q20a0czdewckzluke2.png)
![\begin{gathered} x=50 \\ P=-(1)/(20)(50)^2+5(50)-30 \\ P=-(2500)/(20)+250-30 \\ P=-125+250-30 \\ P=95 \\ \\ x=60 \\ P=-(1)/(20)(60)^2+5(60)-30 \\ P=-(3600)/(20)+300-30 \\ P=-180+300-30 \\ P=90 \\ \\ x=90 \\ P=-(1)/(20)(90)^2+5(90)-30 \\ P=-(8100)/(20)+450-30 \\ P=-405+450-30 \\ P=15 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5ol3bx6zpgjcward3wga4ieqo014ju3a3y.png)
Table:
b) Graph:
c)
i) maximum possible profit is the maximum P in the vertex: 95
ii) the numer of glasses that need to be sold to make the maximum profit is the x value in the vertex: 50
iii) the number of glasses that need to be sold to make a profit of at least 80; x coordinate when P is 80 in the increasing part of the parabola: Approximately 33
iv) the amount of money initially invested by the three students, value of P1when x is 0: it is negative bcause is athe amount invested -30, They invested 30 Swiss Francs
d)Baljeet : Jane : Fiona
1: 2 : 3
The profit is divided into 6 parts, Baljeet receives 1/6, Jane 2/6 and Finona 3/6
If they sold 40 gasses of lemodnade the profit is 90Swiss Francs
![(90)/(6)=15](https://img.qammunity.org/2023/formulas/mathematics/high-school/aloicf2hft06jh77lobaacfd789mmvntbt.png)
1/6 of 90 is 15, as Fiona received 3/6, she has:
![15*3=45](https://img.qammunity.org/2023/formulas/mathematics/college/lpvbw8eh09kq0n1inzqxvh5ta71zi3m546.png)
Fiona's share 45 Swiss Francs