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Student council expects $150 students for a dance and spends $10 on food for each student. The council expects the number of students to increase by 8% each year for the next 5 years and wants to reduce the amount of money spent on food by 4% for each student each year.


Use the drop-down menus to choose or create functions to model:


A. The predicted number of students over time, S(t)

B. The predicted amount spent per student over time,A(t)

C. The predicted total expense for food each year over time, E(t)

Respuesta :

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Answer:

[tex]A) S(t)=150(1.08)^{t-1}[/tex] [tex]B) A(t)=10(0.96)^{t-1}[/tex]

[tex]C) If\:t = 1\\E(t)=150\times10\\ \\If\:t >1\\E(t)=150(1.08)^{t-1}\times10(0.96)^{t-1}[/tex]

Step-by-step explanation:

Considering that The Student Council expects to 150 students for a dance, and spends $10 per student.

If the expectancy is that the number of students increases by  8%

Then since the growth is 8%

we can write the the Predicted Number of Students as a Geometric Sequence,

A.

[tex]a_{1}=150\\a_{2}=150\times 1.08 =162\\a_{3}=162\times 1.08= 174.96\approx 175\\a_{4}=175\times1.08=189\\a_{5}=189\times 1.08=204.12\approx 204\\[/tex]

Therefore,

[tex]S(t)=150(1.08)^{t-1}[/tex]

B) If we spend $10 per student on the 1st year, but the Council wants to reduce the expenditure 4% yearly. Since we want to reduce 4% we must multiply by 96%.

[tex]b_{1}=10\\b_{2}=10*(0.96)=9.6\\b_{3}=9.6*(0.96)=9.22\\b_{4}=9.22*(0.96)=8.85\\b_{5}=8.85*(0.96)=8.50[/tex]

Therefore we can write:

[tex]A(t)=10(0.96)^{t-1}[/tex]

C)

The total expenditure for food must be reduced, by 4%. As long as the public goes increasing by 8% yearly this

So,

[tex]If\:t = 1\\E(t)=150\times10\\ \\If\:t >1\\E(t)=150(1.08)^{t-1}\times10(0.96)^{t-1}[/tex]

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