# decrease of disease

• Sep 29th 2009, 02:41 PM
xacsempire
decrease of disease
Here is my problem and I just can't get the correct formula.

Suppose http://edugen.wiley.com/edugen/share...=1254263519632 the number of cases of a disease, is reduced by 11% each year.
(a) If there are initially 10,000 cases, express http://edugen.wiley.com/edugen/share...=1254263519632 as a function of http://edugen.wiley.com/edugen/share...=1254263519632 the number of years elapsed.

• Sep 29th 2009, 03:09 PM
pickslides
Some of the characters in your post weren't to clear my friend

Lets say P is the number of people with the disease over t years.

0.89 is the same subtracting 11% so

Try $\displaystyle P = 10,000\times 0.89^t$
• Sep 29th 2009, 03:15 PM
pflo
This is the same as a monetary problem. Imagine investing your money and having it grow at a certain percentage rate each year. The formula for calculating your balance is:
$\displaystyle B=P(1+r)^{t}$
where B is your balance, P is the principal (the amount of your initial investment), r is the interest rate (expressed as a decimal), and t is the time your investment has been growing.

One difference with this problem is that you're not dealing with dollars, but with the number of diseased individuals. This doesn't change anything about the formula though.

Another difference is that the number is not growing, but is decreasing. This just means that the rate is negative.

Therefore, use the same formula:
$\displaystyle B=10,000(1-.11)^{t}$
or
$\displaystyle B=10,000(.89)^{t}$