Question

can anyone help me with this? you don't have to answer for me, just help me through it please.

the biomass of a deciduous forest is 50% carbon. additionally the biomass increases annually at a rate of 2.7x10^5 kg/hectare. calculate the mass of carbon accumulated and stored in 1.0 hectare if the forest in one year. answer to the nearest thousand kilograms

Answer 1

The biomass increases by 2.7*10^5 kg/hectare each year. So 1.0 hectare of forest will accumulate 2.7*10^5 kg in one year. 50% of the biomass is carbon. Half of the accumulated biomass is therefore carbon. How much is 50% of 2.7*10^5 kg, rounded to the nearest 1000 kg?

Answer 2

um if I did this right I got 135000 before rounding

Answer 3

Yes, that is 50% of 2.7*10^5.

Answer 4

is that all I had to do?

Answer 5

Well, you don't know how much biomass was there originally, so you can't really do anything else! You can't figure out the total carbon stored in the hectare, just the amount added in one year.

Answer 6

idk I typed it exactly as it said on my paper. I think they word it too complicated

Answer 7

I mean I get mixed up on what to divide by what. that's my main problem

Answer 8

you've got the figure for the biomass, and 50% of it is carbon. "x% of something" means multiply something by x% which is the same as multiplying by x/100. No division needed.

Answer 9

well I just divided 2.7x10^5 by 2

Answer 10

That is equivalent to taking 50%. But as you expressed confusion about "what to divide by what", I gave you a reliable procedure for calculating percentages. No extra charge :-)

Answer 11

wow... I have a whole packet full of questions I don't know how to do :'(

Answer 12

well, why don't you post a few of them and perhaps two sets of eyes will be better than one...

Answer 13

I just don't want people to feel like I'm not trying

Answer 14

do you want me to post them in the comments or close this question and open a new one

Answer 15

Go ahead and post another one here...

Answer 16

ok. well I tried this next one but it confused me too :/

Answer 17

a population of glasswing butterfly exhibits logistic growth. the carring capacity of the population is 200 butterflies. and \[r_{\max}\] the maximum per capita growth rate of the population is 0.10 butterflies/ (butterflies x month). calculate the maximum population growth rate for the population if the maximum growth occurs when N=K/2. give answer to the nearest tenth

Answer 18

Okay, doing a bit of reading...

Answer 19

K is carrying capacity = 200
\(r_{max}\) is the (maximum) per capita growth rate = 0.10
N is population size = K/2

Answer 20

take all the time you need. I'm also tweeting Zak Bagans so you don't have to reply right away. no hurry

Answer 21

max population growth rate is
\[\frac{dN}{dt} = rN(1-\frac{N}{K})= 0.10(K/2)(1-\frac{(K/2)}{K}) =\]

Answer 22

Plug in the value of K (200) and I believe that is your answer. However, this would be the first such problem I've done, so my usual money-back guarantee is not in force :-)

Answer 23

how in the heck did you get that

Answer 24

I'm more confused now

Answer 25

I went off and read the wikipedia page on logistic growth...

Answer 26

we can't use wikipedia

Answer 27

it might not be true

Answer 28

I'm confident that you aren't expected to just know or invent these formulas, so perhaps doing the reading or viewing of your course materials again will reveal the source.

Given that the section in question has bibliographic references to a number of texts and journal articles, I'm inclined to believe it until it is demonstrated I should not. Did your instructor provide a similar level of references for whatever he or she told you? :-)

Answer 29

yeah there is no "course materials" it's literally a piece of paper with just these questions. that's all this packet is. no references or anything

Answer 30

So, you just showed up, someone handed you a list of problems and said go do them, no lecturing, no reading, no nothing?

Answer 31

You'll forgive my skepticism, but that's unlike any educational experience I've had in my many years...

Answer 32

yes. it's for my AP bio class next year. summer assignment. I literally walked in and he said "here's your assignment, due the first day of school. have a good break"

Answer 33

Perhaps you're expected to get the text to be used and read up about such things, or visit the library... Anyhow, looking at other sources on the internet, they seem to construct the same formula.

https://www.youtube.com/watch?v=rXlyYFXyfIM
http://personal.kenyon.edu/holdenerj/Math108Spring2007/classnotes/logisticgrowth/lesson4logisticgrowth.htm
http://mathworld.wolfram.com/LogisticEquation.html
https://home.comcast.net/~sharov/PopEcol/lec5/logist.html

Answer 34

ok so what's the little d supposed to represent?

Answer 35

oh, that's calculus notation, dN/dt is the rate of change of N with respect to t.

Answer 36

my interpretation is that dN/dt is what they are asking for when they ask for the maximum population growth rate

Answer 37

I have \[\frac{ dN }{ dt }=r _{\max}^N (\frac{ K-N }{ K })\]

Answer 38

I could of course be incorrect. I'll ask my ex-girlfriend with the PhD in population biology if I run into her, but as we live on opposite sides of the continent, it's not very likely :-)

Yes, that's the same.
\[\frac{K-N}{K} = 1-\frac{N}{K}\]

Answer 39

um I kinda need this done by next month

Answer 40

how do I know what N is

Answer 41

because K is the carrying capacity which is 200

Answer 42

I think this is the key: "if the maximum growth occurs when N=K/2."

Answer 43

but N is supposed to be the population size

Answer 44

oh... hold on a sec

Answer 45

Yes. Why can't it equal K/2?

Btw, the youtube video I sent you looks pretty good.

Answer 46

brain function in process... think it's registering

Answer 47

so for this question the population is supposed to be 200/2 (K/2). so the maximum growth rate would be 10?

Answer 48

I get 5, not 10.

Answer 49

wait I did 100x .10. then went back and did \[\frac{ 200-100 }{ 100}\]x .10 and got .05

Answer 50

\[\frac{dN}{dt} = rN(1-\frac{N}{K})= 0.10(K/2)(1-\frac{(K/2)}{K}) = 0.1(100)(1-\frac{1}{2}) = 5\]

Answer 51

wtf is all that

Answer 52

denominator of that fraction should be 200, not 100

Answer 53

oops. my bad

Answer 54

I think you're missing a factor of 100 in there somewhere, because I'm confident that the little butterflies are going to be able to add more than 5/100ths of a butterfly per month to the population

Answer 55

I still got .05

Answer 56

show me your formula and work. mine is right there, and correct, I believe.

Answer 57

this is why I hate math

Answer 58

I told you what I did

Answer 59

no, you didn't. show me the entire formula you used, with the numbers you put in it, just like I did.

Answer 60

I'm getting frustrated

Answer 61

I haven't met anyone who doesn't at some point. How are you going to deal with it? Learning how to do so may be more valuable in the long run than any handful of facts you learn in AP biology.

Answer 62

the question asks for the population growth and my paper gives me a different equation for that.

Answer 63

Okay, what equation does it give?

Answer 64

yeah I want to work delivering babies. I seriously doubt this stuff will help

Answer 65

\[\frac{ dN }{ dt }= r_{\max}N \]

Answer 66

but the dn/dt stuff doesn't matter cause it's set equal to rmaxN so as long as I have that it doesn't matter

Answer 67

I don't think that equation shows logistic growth, from what I understand about it. Read the links I sent you, watch the video, make your own decision.

Answer 68

but the question itself doesn't ask for logistic growth, it asks for population growth

Answer 69

"a population of glasswing butterfly exhibits logistic growth"

Answer 70

logistic growth is a particular kind of population growth

Answer 71

In any case, it's time for me to go investigate the biology of eating a large cheeseburger :-) Good luck with the problems!

Answer 72

XD

Answer 73

ooh a chili cheese burger sounds nice...

Answer 74

I'm getting extremely pissed off right now.

Answer 75

I got it