First of all, we have to check out what the yearly decrease really amounted to. Then we'll know if the equation we are using is correct.
If there is a 26% decrease, then there's 74% of the population that remains. So after 1 year, 74% of 426000 people still live in this community. How many people are still there?
P = 426000 * (74/100) ^1 P = 315240 are still left in our community. What about the second year? This is really critical, because if we make a mistake in our theory, it will show up here.
P2 = 315240 * (74/100) P2 = 233277 People still live in this community.
Could we have found P2 directly? Yes. P2 = 426000 * 0.74^2 P2 = 233277 is the answer. So now we know what the answer to the question must be
year one:315240 is the population in the first year year two:233277 is the population in the second year
Now we're ready to begin answering your question. If we start out with P =426000 ((1 - 0.26)^1/12)^(12t) we should be able to get a monthly drop for a yearly %. You better read this over a couple of times to get the idea.
So P = [426000*(.74)^(1/12)]^12t is what you do. By the law of powers, you have done nothing to the actual equation, because the 12s cancel.
Put this in your calculator .74^(1/12) = 0.97522 Ans our equation becomes P = 426000 * 0.97522^(12t) where t is in years.
To check the equation, we will let t = 1 P = 426000 * 0.97522^12 P = 315240
But the acid test is year two. P = 426000 * 0.97522^24 P = 233277 which checks with doing it one step at a time using 0.74
So the answer is P = a * 0.97522^12t b = 0.97522 c = 12*t
In the given scenario, the z-score value is used as the score which is used to shows that how many standard deviation units in this score were uses as the from mean and whenever this score is above or below by the mean, therefore the given statement is true.
