Question

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Answer 1

This is an easy one.

Answer 2

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

Answer 3

um no not sure

Answer 4

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

Answer 5

Does that make sense on why you would do that?

Answer 6

oh okay yeah that makes sense.

Answer 7

what did you get for your answer?

Answer 8

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

Answer 9

umm something with the 6?

Answer 10

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

Answer 11

does that make sense?

Answer 12

so 37.59 + 1253?

Answer 13

correct. remember its INCREASING, so you will add.

Answer 14

I got 1290.59

Answer 15

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?

Answer 16

37.59 + 1290.59 ?

Answer 17

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

Answer 18

oh okay

Answer 19

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

Answer 20

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?

Answer 21

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

Answer 22

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

Answer 23

that models the deer population actually.

Answer 24

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

Answer 25

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

Answer 26

well that would've helped

Answer 27

ill let hugh explain it.

Answer 28

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

Answer 29

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

Answer 30

these are questions at the end of the lessons

Answer 31

were they talking about continuous compounding ?

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

if that helps at all

Answer 37

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

Answer 38

so your y(0) is your starting population

Answer 39

so y(0)= 1253 then

Answer 40

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

Answer 41

so not 3 .. but ..?

Answer 42

k = 3% or 0.03

Answer 43

good one

Answer 44

and then t is your term :)

Answer 45

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

Answer 46

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

Answer 47

okay

Answer 48

well wait

Answer 49

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

Answer 50

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

Answer 51

y(6) ?

Answer 52

or 1253

Answer 53

and your constant for the rate of growth (k)

Answer 54

yes starting population = y(0)

Answer 55

1253

Answer 56

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

Answer 57

so 1253 e^(0.03*6)

Answer 58

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.

Answer 59

i'm getting 1500

Answer 60

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

Answer 61

looks good..

Answer 62

oh its multipole choice?

Answer 63

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

Answer 64

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

Answer 65

yeah so I guess it's all just approximate

Answer 66

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

Answer 67

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

Answer 68

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

Answer 69

yeah lol

Answer 70

so thanks for helping me it's really appreciated!

Answer 71

no worries.. good luck

Answer 72

thank you :) I understand this more thanks to you