suppose a population of 50 crickets doubles in size every 4 months. how many crickets will there be after 5 years?
Ok so there is 12 months in a year correct?
First of all, Welcome to OpenStudy :D
yes
Ok so then 5x12 is 60 so divide 60 into 4
thats 15
but my answer choices is in the thousands
Shoot my dyslexia got the best of me
Lemme redo that
We should know: 1 year = 12 months Given: Doubles every 4 months So: 4 * 3 =12 Therefore: 1 year, there is 3 sets of 4 months. Hence: 4 * 3 = 12 Then: 2 years have 6 sets of 4 months. 3 years have 9 sets of 4 months 4 years have 12 sets of 4 months 5 years have 15 sets of 4 months Good so far?
\i did all that but i confused myself
We start with a population of 50 crickets So then: \[50 \times 2^{n}\] \[50 \times 2^{15}\] whats 2^15? after u get that, multiply what you got from that by 50. Try it and let me see what you get. :)
50 X 2^15 is 1638400
yes, that's correct. is that one of your answer choices?
^ yes good.
So: \[1,638,400\] Is that an answer choice? @roxxy251
You were on the right path @dmezzullo :)
That stuff is easy for me but dyslexia dont help
it was on my quiz, i go it wrong by puttin 1500 buti believe that was 1 of my answer choices,im doing school online, and ihova another question also
Another question from ur quiz?
@roxxy251 just to clarify, are these problems that you got wrong on the quiz?
yes these the problem i had wrong on the quiz
okay. I'm not interested in helping someone take a quiz, but happy to explain after the fact.
yep ^
what was the next one you got wrong? if you can explain what your wrong answer was, and how you got it, that would be great — wouldn't have to worry that you were trying to pull a fast one.
again, yep ^
and we promise not to laugh :-)
suppose an investment of $8,200 doubles in value every 7 years. how much is the investment worth after 28 years
how many doublings is that?
4
right. and what is 2*2*2*2?
16
and 16 * 8200?
yep ^
131200
its kinda the same concept as your 1st question
so how did you get it wrong?
@roxxy251 what did you put as your answer; which was wrong according to you?
86,000 i used the wrong formula and how you putting it make it seemd a whole lot easier, thanks
well, I try :-)
r u very good n math
lol yes @whpalmer4 is quite good; very good, in math :3
Alot of ppl here are!
okay im having problems with exponential growth and decay
okay, let's have one of those problems.
u still there
I am.
this is home work now, and i still dont understand this at all . but the question is, a boat costs 16,600 an decreases in value per 14% per year. how much will the boat be worth after 11 years
Okay. present value is $16,600. Decreases by 14% each year. 14% = 14% * 1/100% = 0.14, right? If we start with 1, and subtract 14%, that's 1-0.14 = 0.86. After 1 year, the boat is worth 0.86 * purchase price. After 2 years, the boat is worth how much?
14276
ugh i feel dumb
i hate math
No, it's worth that much after 1 year. At the end of 1 year, it has decreased by 14%, which is the same as saying it is worth 86% of its previous value. Each passing year, we'll multiply by 86% again. After n years, the value will be the starting value * 0.86^n — do you see why that is?
no ,n itsmaking me very upset
Okay, do you see why multiplying by 0.86 is the same as decreasing 14%?
nope
what do you think it means to decrease by 14%?
less than
how much?
If I have 100 dollars, and it decreases by 14%, how much do I have left?
im not really good with percentages,
percent just means "out of 100" 14 percent means 14 out of 100.
okay so basically divide 14 over a 100 or subtract 14and 100
if we decrease something by 14%, it means we chop it up into 100 equal pieces, and get rid of 14 of them.
that leaves us with 86 of those pieces
so decreasing by 14% over and over means keeping only 86% over and over...
okay now w/ that problem i suppose to do what
well, you wanted to find out how much the boat is worth after some number of years of depreciation at 14% per year. the value of the boat is going to be the starting value (what ou paid for it) * 0.86 after one year, starting value * 0.86 * 0.86 after two years, starting value * 0.86 * 0.86 * 0.86 after three years, and so on: \(16,600*0.86^n\) where \(n\) is the number of years.
in this case, we want to find how much it is worth after n=11 years. your calculator probably has a y^x button, so you enter 0.86, press y^x, then enter 11, then press * 16600 =
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