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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

Does that make sense on why you would do that?

oh okay yeah that makes sense.

what did you get for your answer?

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

37.59 + 1290.59 ?

You have the right idea. You will take 1290.59 and take 3% of that number.

oh okay

you take 3% of the previous year, and add that to the new population.

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

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

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

good one

and then t is your term :)

term? the 6? that's the only one left

that's it.. a term of 6 years.

okay

well wait

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)

y(6) ?

or 1253

and your constant for the rate of growth (k)

yes starting population = y(0)

1253

So you should have something that looks like
f(t) = 1253 e^(0.03 t)

so 1253 e^(0.03*6)

i'm getting 1500

I have a choice that is 1456. That's the closest to that

looks good..

oh its multipole choice?

yeah there's a few to pick from
1044
9134
1777
1496

f[6] = 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.

yeah lol

so thanks for helping me it's really appreciated!

no worries.. good luck

thank you :) I understand this more thanks to you