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This is an easy one.
Ok first you try finding the deer population in the first year. Do you know what you would do?
um no not sure
Well as you said, the rate of increase is 3%. All you would do for the first year, is to multiply the original amount (1253) by 3% or .03
Does that make sense on why you would do that?
oh okay yeah that makes sense.
what did you get for your answer?
Ok good, now since the deer population is INCREASING do you have an idea on what you would do with that number?
umm something with the 6?
I'll get to the 6 later. Since we did the first year, we will add 37.59 to the ORIGINAL 1253 deer.
does that make sense?
so 37.59 + 1253?
correct. remember its INCREASING, so you will add.
I got 1290.59
yes! OK we just finished year 1, now since its increasing by 3% every year, you must do the same thing we just did for the next year. what do you think that is?
37.59 + 1290.59 ?
You have the right idea. You will take 1290.59 and take 3% of that number.
you take 3% of the previous year, and add that to the new population.
what level of schooling is it though? Is it possible they want you to use an exponential function? likeP= A^(r t) where P is population A is starting amount r is the rate t is the term?
or is Rstones method okay? because they will both work.
this is 8th year and I think they want me to write a function that represents the deer population.
that models the deer population actually.
do you know if they want you to use e like eulers number?
yeah I think they want that exponential function in there because that's what the lesson is about
well that would've helped
ill let hugh explain it.
its cool man, you can do it if you like, didn't mean to hijack your help
sorry I mean it didn't specifically say to use that, but since that was what some of the lesson talked about... I didn't know if it needed to be used
these are questions at the end of the lessons
were they talking about continuous compounding ?
I barely know what any of this means, I missed the class
umm nope I do think it just needs to be written as exponential function to get the answer
ok 1 sec... lol.. dusting off my cobwebs...
my book has this function y=y(0)e^kt
if that helps at all
When critters and bacteria multiply they use a compounding exponential function.. and the euler number. the function would look like. yah awesome... thats it..
so your y(0) is your starting population
so y(0)= 1253 then
good, the k is the rate at which they grow.. as a decimal of 1..
so not 3 .. but ..?
k = 3% or 0.03
and then t is your term :)
term? the 6? that's the only one left
that's it.. a term of 6 years.
to be a function... then you would want a function of t... so...
f[t] = y(0) E^(k t) and you would plug in your value for y(0)
and your constant for the rate of growth (k)
yes starting population = y(0)
So you should have something that looks like f(t) = 1253 e^(0.03 t)
so 1253 e^(0.03*6)
you ahve the correct equation.. and if you need to make it a formula instead.. then you keep the t, and set a parameter in your function for it.
i'm getting 1500
I have a choice that is 1456. That's the closest to that
oh its multipole choice?
yeah there's a few to pick from 1044 9134 1777 1496
f = 1253 E^(0.03 (6)) = 1500.11
yeah so I guess it's all just approximate
wow that's sneaky hey ... yeah I'd go with the closest value..
In the question is subtly uses "about how many dear" so I guess it doesn't have to be exact.
thats like 1252 deer hey.. I guess one died.
so thanks for helping me it's really appreciated!
no worries.. good luck
thank you :) I understand this more thanks to you