word problem: disease

[kellyanne]

suppose that in any given year, the number of cases of a disease is reduced by 20%. if there are 10,000 cases today, how many years will it take (a) to reduce the number of cases to 1000 and (b) to eliminate the disease; that is, reduce the number of cases to less than 1?

 

[tkhunny]

x = time from today, in years

\(\displaystyle Cases(x) = 10000*0.8^{x}\)

\(\displaystyle Cases(0) = 10000*0.8^{0} = 10000\)

You do the rest. Please show your work and more helpful responses can be had.

 

[stapel]

suppose that in any given year, the number of cases of a disease is reduced by 20%. if there are 10,000 cases today, how many years will it take (a) to reduce the number of cases to 1000 and (b) to eliminate the disease; that is, reduce the number of cases to less than 1?

Use the compound-interest formula you've memorized, with a negative "interest" rate of -0.2.

Eliz.

 

[TchrWill]

suppose that in any given year, the number of cases of a disease is reduced by 20%. if there are 10,000 cases today, how many years will it take (a) to reduce the number of cases to 1000 and (b) to eliminate the disease; that is, reduce the number of cases to less than 1?

Comsider the sequence of reductions to be a geometric progression with first term a = 10,000 =, common factor = r = 0.8 and last term L = 1000.

The nth trerm of a geometric progression is given by L = ar^(n - 1)

Therefore, 10,000(.8)^(n - 1) = 1000 or (.8)^(n - 1) = .10

I estimate that 1000 cases will be reached between the 10th and 11th year, based on year 1 being 10,000(.8) = 6400, which I will let you figure out exaclty,

I estimate that the less than 1 case will be reached between the 41st and 42nd years which you can figure out.