Give the starting value a , the growth rate r , and the continuous growth rate k .

Question

   Q= 6^-(t-5)/3

Answer 1

q= 6^-(t-5)/3

time t =5 q=1

t=20 q= 1/6

in 15 units of time, say minutes, the population decayed from 1 to 1/6 = a negative growth rate of 5/6 = r = about -17%

and about -1% per minute

initial population is when time t =0

q = 6^-(-5)/3 = about 19.8