Question

An efficiency expert hired by a manufacturing firm has compiled these data relating workers output to their experience:

Answer 1

Answer:

Q(t) = 300 - 100e^0.0318t

Step-by-step explanation:

Given the relationship between the output as related to the experience t to be Q (t) = 300 - A e^-kt.

From the table, at t = 0, Q(t) = 200, substituting this into he equation to get A we have:

200 = 300 - A e^-k(0).

200 = 300 - A e^-0

200 = 300 - A

A = 300-200

A = 100

Secondly we need to get the constant k by substituting the other condition into the equation. When t = 3, Q(t) = 190

190 = 300 - A e^-k(3)

190 = 300 - 100e^-3k

190-300 = -100e^-3k

-110 = -100e^-3k

e^-3k = -110/-100

e^-3k = 1.1

Taking the natural logarithm of both sides we have;

ln(e^-3k) = ln1.1

-3k = 0.09531

k = 0.09531/-3

k = -0.0318

The function of this form that fits the data can be gotten by substituting A = 100 and k = -0.0318 into the modeled equation given as shown:

Q(t) = 300 - A e^-kt

Q(t) = 300 - 100e^-(-0.0318)t

Q(t) = 300 - 100e^0.0318t