so considering t = 0 to correspond with 2016
1000 = A 3^0 , but 3^0 = 1
A = 1000
b) y(5) = 1000 3^(4/3) = .... Use your calculator
c) 100000 = 1000 3^(t/4)
1000 = 3^(t/4
take log of both sides
log 1000 = log 3^(t/4)
3 = (t/4)log 3
t/4 = 3/log3
t = 12/log3 = appr 25.15 years
iv) y = 1000 3^(t/4)
dy/dt = 1000(3^(t/4)(1/4)(ln3)
= 250ln3 (3^(t/4)
plug in t = 0 for now, and t = 10 for the other case
The number of people expected to have a disease in t years is given by y(t)=A•3^(t/4)
i) if now year(2016) the number of people having disease is 1000, find the value of A?
ii) how many people expected to have the disease in five years?
iii) when are 100,000 people expected to have the disease?
iv) How fast is the number of people with the disease expected to grow now and ten years from now?
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2 answers