(a) We are given two points on the curve described by the equation y = 100 - Ae^(-at): (0, 99) and (1, 50).

We can use these points to form two equations and solve for A and a.

From the first point (0, 99), we substitute into the equation and get:

99 = 100 - A * e^(0)
= A = 100 - 99
= A = 1

Substitute A = 1 into the second point (1, 50) we get:

50 = 100 - e^(-a)
= e^(-a) = 50
= -a = ln(50)
= a = -ln(50)

So, A = 1 and a = -ln(50) = -3.912 (to 3 decimal places).

(b) To predict the percentage of consulted books after 5 years, we substitute t = 5 into the equation:

y = 100 - e^(-5*-3.912)
= y = 100 - e^(-19.56)
= y = 100 - 0.0000003
= y = 99.9999997%

So, after 5 years, approximately 100% of the books will have been consulted.

(c) In the long run, as t approaches infinity, the term e^(-at) approaches 0. Therefore, the percentage of consulted books approaches 100 - 0 = 100%. So, in the long run, all the books will have been consulted.

Question

(a) We are given two points on the curve described by the equation y = 100 - Ae^(-at): (0, 99) and (1, 50).

We can use these points to form two equations and solve for A and a.

From the first point (0, 99), we substitute into the equation and get:

99 = 100 - A * e^(0)
=> A = 100 - 99
=> A = 1

Substitute A = 1 into the second point (1, 50) we get:

50 = 100 - e^(-a)
=> e^(-a) = 50
=> -a = ln(50)
=> a = -ln(50)

So, A = 1 and a = -ln(50) = -3.912 (to 3 decimal places).

(b) To predict the percentage of consulted books after 5 years, we substitute t = 5 into the equation:

y = 100 - e^(-5*-3.912)
=> y = 100 - e^(-19.56)
=> y = 100 - 0.0000003
=> y = 99.9999997%

So, after 5 years, approximately 100% of the books will have been consulted.

(c) In the long run, as t approaches infinity, the term e^(-at) approaches 0. Therefore, the percentage of consulted books approaches 100 - 0 = 100%. So, in the long run, all the books will have been consulted.

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