a) y =10000(0.11)^t
b) y = 10000(0.11)^6 and you solve it
c) 1000 = 10000(0.11)^t and I would divide both sides by 10000 first then either uses exponents or take the log of both sides to solve
Suppose y , the number of cases of a disease, is reduced by 11% per year.
(a) If there are initially 10,000 cases, express y as a function of t , the number of years elapsed.
(b) How many cases will there be 6 years from now?
(c) How long does it take to reduce the number of cases to 1000?
4 answers
it is reduced by .11/yr so multiply by 0.89^t
It is reduced by 11% leaving 89%
so y = 10000(.89)^t
b) y = 10000(.89)^6
c)
10000(.89)^t = 1000
.89^t = .1
t log .89 = log .1
t = -1/-.0506099.. = appr 19.76 years
so y = 10000(.89)^t
b) y = 10000(.89)^6
c)
10000(.89)^t = 1000
.89^t = .1
t log .89 = log .1
t = -1/-.0506099.. = appr 19.76 years
Thanks math gang! Time for a cup of afternoon tea. I was doing the math in my head and .89 didn't feel just right. LOL! Thanks again for the correction : )
1 answer