The intensity, Io, of a light source is reduced to I after passing through d meters of fog according to the formula I = Ioe−0.12d. If 1,200 lumens of light are present initially, what is an equivalent expression written as a percentage rate of lumens lost?
Hint: Find the value of e−0.12 on your calculator. I = 1,200ed I = 1,200(0.887)d I = 1,200e0.88d I = 1,200(0.88)d
@jim_thompson5910 @jagr2713
@Jaynator495
@SolomonZelman can you help with a few questions?
jim can you help with a few of these?
what is the value of `e^(-0.12)` ? use this calculator if you don't have a calculator with you http://web2.0calc.com/
0.88692043671
so b yes?
Yes correct \[\Large \color{red}{e^{-0.12}} \approx \color{red}{0.88692043671712}\] ------------------------------------------------ \[\Large I = I_0 e^{-0.12d}\] \[\Large I = I_0\left(e^{-0.12}\right)^{d}\] \[\Large I = I_0\left(\color{red}{e^{-0.12}}\right)^{d}\] \[\Large I = I_0\left(\color{red}{0.88692043671712}\right)^{d}\] \[\Large I = I_0*0.887^{d}\] Then we plug in the initial value \(\Large I_0 = 1200\) to get this final answer \[\Large I = 1200*0.887^{d}\]
knew it thank you! i need a few more done you have some time?
I'll help with 2 more
great
An automobile is depreciating at 13% per year, every year. A $52,000 car depreciating at this rate can be modeled by the equation V(t) = 52,000(0.87)t. What is an equivalent equation for this vehicle at a daily depreciation and what is it worth (rounded to the nearest thousand dollar) 5 years after purchase? V(t) = 52,000(0.87)365t, $26,000 V(t) = 52,000(0.9996)t, $52,000 V(t) = 52,000(0.87)start fraction t over 365 end fraction, $52,000 V(t) = 52,000(0.99962)365t, $26,000
13% per 52k?
`An automobile is depreciating at 13% per year` how much is it depreciating per day? hint: 365 days in a year
wait is 52k what it is after 5 years? so then i have to split that by 5 correct? and that should give me the per year value?
I don't know what you mean
:o 365 in 52 k?
what is 0.13/365 equal to?
0.00035616438 yes?
one moment
ok take your time.
I don't agree with these answer choices. Can you take a screenshot of the full problem?
mm i can probably attach a file give me a moment.
ok I'll be right back
give me a minute need to re-download my photo album.
hey sorry man can I just send you an email of it?
@jim_thompson5910
@SolomonZelman do you perhaps have a solution for this i think jim has removed himself from this chat for a bit?
ok I'm back now
the "attach file" button doesn't work?
uh no this is a photo from a preview gallary i cant attach this i dont believe.
well the answer is either B or C. The issue is that both answer choices say that the car is worth $52,000 after 5 years, but that's the starting value. The value after 5 years should be much less. So I'm thinking there's a typo which is why I wanted to look directly at the problem in the form of a screenshot.
sorry @jim_thompson5910 my internet went dead ok i checking your stuff now i don see a typo so ill tell you if i find something.
V(t) = 52,000(0.9996)t, $52,000 V(t) = 52,000(0.87)start fraction t over 365 end fraction, $52,000 difference one is fraction of t over 365 one is just t and in the parenthesis the decimals are different too.
If there's an answer choice that has V(t) = 52,000(0.9996)^t; $26,000 then it would work since plugging t = 5 into V(t) = 52,000(0.9996)^t gives roughly 26,000
but I'm guessing there's a typo that the teacher made
there is an answer that says that lol
thats what D is
well there's a 365 in there though
so second guess is b because it doesn't have the 365?
yeah but it has the 52000 instead of 26000
if only there was a way to combine the answer choices
lol so what would you suggest? I'm not just putting down anything. I would like to know which one it is and how we got to it?
i say your correct about D but why is 265 is set back... i mean you did ask me to use that when i divided.
well it's definitely V(t) = 52,000(0.9996)^t because 0.13/365 = 0.00035616438357 then 1-0.00035616438357 = 0.99964383561643
if 365 is what got you this : (0.9996) then why is it just ^t and not ^t365 after it was used to find that decimal right?
when I plug t = 5 into the original V(t) I get roughly 25917.8878764000 which rounds to $26,000 but when I plug t = 5 into choice D, I get 27143.2375407055 roughly which rounds to $27,000.
So this means D is incorrect yes?
yeah D is a bit off
this problem is just odd, idk why it's like this
so b is the winner? e.e
what about A though?
yeah I guess it's close enough, just that last part is off
ok well i just have one more left now...
A doesn't match either
my last question is - In a recent stock market downturn, the value of a $1,000 stock is decreasing at 1.6% per month. This situation can be modeled by the equation A(t) = 1,000(0.984)12t, where A(t) is the final amount and t is time in years. Assuming the trend continues, what is the equivalent annual devaluation rate of this stock (rounded to the nearest tenth of a percent) and what is it worth (rounded to the nearest dollar) after 1 year? 82.4% and $824.00 19.2% and $808.00 17.6% and $824.00 80.8% and $808.00
what is 0.984^12 equal to?
0.82402650513
now subtract that from 1
-0.17597349487?
I was thinking 1-0.82402650513947 = 0.17597349486051
multiply 0.17597349486051 by 100% to get 0.17597349486051*100% = 17.597349486051% that rounds to roughly 17.6%
824 dollars correct?
so 1.6% per month ----> 17.6% per year
yeah after 1 year, the stock is worth $824
k got it.. i know i can do the rest but i doubt they will be right so instead of helping the time can you just check and see if my other two are correct if so then I'm done if not then i have to correct them.
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