Answer:
<h2>
Q(t) = 300 - 100e^0.0318t</h2>
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
