Question

A school has 200 students and spends $40 on supplies for each student. The principal expects the number of students to increase by 5% each year for the next 10 years and wants to reduce the amount of money spent on supplies by 2% 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 supplies each year over time, E(t) CLEAR

Answer 1

S(t) = 200(1+.05)^t

A(t) = 40(1-.02)^t

E(t) = S(t) X a(t)

Answer 2

`\mathbf{S(t)=200((105)/(100))^(x)}`

`\mathbf{A(t)=40((98)/(100))^(x)}`

`\mathbf{E(t)=S(t) \cdot A(t)=200((105)/(100))^(x) \cdot 40((98)/(100))^(x)=8000((10290)/(10000))^(x)}`

Explanation:

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

Rate of increment is 5% per year.

A function 'S(t)' which gives the number of students in school after 't' years.

S(0) means the initial year when the number of students is 200.

S(0) = 200

S(1) means the number of students in school after one year when the number increased by 5% than previous year which is 200.

S(1) = 200 + 5% of 200 =
`200+(5)/(100)\time200` =
`200(1+(5)/(100))` =
`200((105)/(100))`

S(2) means the number of students in school after two year when the number increased by 5% than previous year which is S(1)

S(2) = S(1) + 5% of S(1) =
`\textrm{S}(1)((105)/(100))` =
`200((105)/(100))((105)/(100))` =
`200((105)/(100))^(2)`

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Similarly
`\mathbf{S(x)=200((105)/(100))^(x)}`

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

Rate of decrements is 2% per year.

A function 'A(t)' which gives the amount spend on each student in school after 't' years.

A(0) means the initial year when the number of students is 40.

A(0) = 40

A(1) means the amount spend on each student in school after one year when the amount decreased by 2% than previous year which is 40.

A(1) = 40 + 2% of 40 =
`40-(2)/(100)\time40` =
`40(1-(2)/(100))` =
`40((98)/(100))`

A(2) means the amount spend on each student in school after two year when the amount decreased by 2% than previous year which is A(1)

A(2) = A(1) + 2% of A(1) =
`\textrm{A}(1)((98)/(100))` =
`40((98)/(100))((98)/(100))` =
`40((98)/(100))^(2)`

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Similarly
`\mathbf{A(x)=40((98)/(100))^(x)}`

The predicted total expense for supplies each year over time, E(t)

Total expense = (number of students) × (amount spend on each student)

E(t) = S(t) × A(t)

`\mathbf{E(t)=S(t) \cdot A(t)=200((105)/(100))^(x) \cdot 40((98)/(100))^(x)=8000((10290)/(10000))^(x)}`

`\mathbf{E(t)=8000((10290)/(10000))^(x)}`

(NOTE : The value of x in all the above equation is between zero(0) to ten(10).)