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- anonymous

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- schrodinger

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- anonymous

This is an easy one.

- anonymous

Ok first you try finding the deer population in the first year. Do you know what you would do?

- anonymous

um no not sure

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## More answers

- anonymous

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

- anonymous

Does that make sense on why you would do that?

- anonymous

oh okay yeah that makes sense.

- anonymous

what did you get for your answer?

- anonymous

Ok good, now since the deer population is INCREASING do you have an idea on what you would do with that number?

- anonymous

umm something with the 6?

- anonymous

I'll get to the 6 later. Since we did the first year, we will add 37.59 to the ORIGINAL 1253 deer.

- anonymous

does that make sense?

- anonymous

so 37.59 + 1253?

- anonymous

correct. remember its INCREASING, so you will add.

- anonymous

I got 1290.59

- anonymous

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?

- anonymous

37.59 + 1290.59 ?

- anonymous

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

- anonymous

oh okay

- anonymous

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

- anonymous

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?

- anonymous

or is Rstones method okay? because they will both work.

- anonymous

this is 8th year and I think they want me to write a function that represents the deer population.

- anonymous

that models the deer population actually.

- anonymous

do you know if they want you to use e like eulers number?

- anonymous

yeah I think they want that exponential function in there because that's what the lesson is about

- anonymous

well that would've helped

- anonymous

ill let hugh explain it.

- anonymous

its cool man, you can do it if you like, didn't mean to hijack your help

- anonymous

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

- anonymous

these are questions at the end of the lessons

- anonymous

were they talking about continuous compounding ?

- anonymous

I barely know what any of this means, I missed the class

- anonymous

umm nope I do think it just needs to be written as exponential function to get the answer

- anonymous

ok 1 sec... lol.. dusting off my cobwebs...

- anonymous

my book has this function y=y(0)e^kt

- anonymous

if that helps at all

- anonymous

When critters and bacteria multiply they use a compounding exponential function.. and the euler number.
the function would look like.
yah awesome... thats it..

- anonymous

so your y(0) is your starting population

- anonymous

so y(0)= 1253 then

- anonymous

good,
the k is the rate at which they grow.. as a decimal of 1..

- anonymous

so not 3 .. but ..?

- anonymous

k = 3% or 0.03

- anonymous

good one

- anonymous

and then t is your term :)

- anonymous

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

- anonymous

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

- anonymous

okay

- anonymous

well wait

- anonymous

to be a function... then you would want a function of t... so...

- anonymous

f[t] = y(0) E^(k t)
and you would plug in your value for y(0)

- anonymous

y(6) ?

- anonymous

or 1253

- anonymous

and your constant for the rate of growth (k)

- anonymous

yes starting population = y(0)

- anonymous

1253

- anonymous

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

- anonymous

so 1253 e^(0.03*6)

- anonymous

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.

- anonymous

i'm getting 1500

- anonymous

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

- anonymous

looks good..

- anonymous

oh its multipole choice?

- anonymous

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

- anonymous

f[6] = 1253 E^(0.03 (6)) = 1500.11

- anonymous

yeah so I guess it's all just approximate

- anonymous

wow that's sneaky hey ... yeah I'd go with the closest value..

- anonymous

In the question is subtly uses "about how many dear" so I guess it doesn't have to be exact.

- anonymous

thats like 1252 deer hey.. I guess one died.

- anonymous

yeah lol

- anonymous

so thanks for helping me it's really appreciated!

- anonymous

no worries.. good luck

- anonymous

thank you :) I understand this more thanks to you

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