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decay factor formula
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Question:A city of population 2.3 million is expected to experience a 22% decrease in population every 10 years. What is the 10year decay factor?
What is the yearly decay factor? Round the answer to 2 decimal places.
The yearly decay rate? Round the answer to 2 decimal places.
%
Answers:the 10 year decay factor is 78% population = Po*e^at t =years e^at = 0.78 when t =10 10a = ln(0.78)= 0.2485 a = 0.02485 population P = 2.3e6*e^.02485t the 1 year decay rate =(1.0 0.9746) = 2.54%
Answers:the 10 year decay factor is 78% population = Po*e^at t =years e^at = 0.78 when t =10 10a = ln(0.78)= 0.2485 a = 0.02485 population P = 2.3e6*e^.02485t the 1 year decay rate =(1.0 0.9746) = 2.54%
Question:i need to now this an help
Answers:No. It's a plant, which makes it biotic (bio = life), even though it's decaying, it's still a biologic process (think of all the little critters busy decomposing it). BUT... there are abiotic factors involved... for instance, temperature, humidity, and oxygen availability are all abiotic, and they affect the rate of decomposition. So... it's kind of both. Of course, most things are both, which is why we look at both types of factor when studying an organism or ecosystem.
Answers:No. It's a plant, which makes it biotic (bio = life), even though it's decaying, it's still a biologic process (think of all the little critters busy decomposing it). BUT... there are abiotic factors involved... for instance, temperature, humidity, and oxygen availability are all abiotic, and they affect the rate of decomposition. So... it's kind of both. Of course, most things are both, which is why we look at both types of factor when studying an organism or ecosystem.
Question:Find the percent of decrease for each decay factor for the following examples
1) 0.92
2) 0.65
3) 0.04
4) 0.995
5) 0.73
6) 0.18
7) 0.65
8) 0.025
Use a table to graph each function
1) y=18*0.98^x
2) y=36* (1/3)^x
Please help me, i have been searching the internet for help, i forgot my text book in class and cant afford to lose points for not handing in my homework, all help will be appreciated<3
Answers:Your first question is unclear, but I think this is what you want. 1) % decrease = 1  N * 100% % de. = 1  (0 92) * 100% % de. = 0 08 * 100% % de. = 8% 2) % decrease = 1  N * 100% % de. = 1  (0 65) * 100% % de. = 0 35 * 100% % de. = 35% Do the rest in a similar way. Take a range of values for x, and work out the values for y. Plot the points and draw your curve. 1) When x = 2; y = 18 * 0 98^x y = 18 * 0 98^ y = 18 * 1 04 123 282... y = 18 742 19... Pt.(2, 18 74..) When x = 1; y = 18 * 0 98^x y = 18 * 0 98^1 y = 18 * 1 02 0408... y = 18 367 346... Pt.(1, 18 367..) When x = 0; y = 18 * 0 98^x y = 18 * 0 98^ y = 18 * 1 y = 18 Pt.(0, 18) When x = 1; y = 18 * 0 98^x y = 18 * 0 98^1 y = 18 * 0 98 y = 17 64 Pt.(1, 17 64) When x = 2; y = 18 * 0 986^x y = 18 * 0 98 y = 18 * 0 9604 y = 17 2872 Pt.(2, 17 2872) Now plot the points onto the graph. Do likewise with the next question.
Answers:Your first question is unclear, but I think this is what you want. 1) % decrease = 1  N * 100% % de. = 1  (0 92) * 100% % de. = 0 08 * 100% % de. = 8% 2) % decrease = 1  N * 100% % de. = 1  (0 65) * 100% % de. = 0 35 * 100% % de. = 35% Do the rest in a similar way. Take a range of values for x, and work out the values for y. Plot the points and draw your curve. 1) When x = 2; y = 18 * 0 98^x y = 18 * 0 98^ y = 18 * 1 04 123 282... y = 18 742 19... Pt.(2, 18 74..) When x = 1; y = 18 * 0 98^x y = 18 * 0 98^1 y = 18 * 1 02 0408... y = 18 367 346... Pt.(1, 18 367..) When x = 0; y = 18 * 0 98^x y = 18 * 0 98^ y = 18 * 1 y = 18 Pt.(0, 18) When x = 1; y = 18 * 0 98^x y = 18 * 0 98^1 y = 18 * 0 98 y = 17 64 Pt.(1, 17 64) When x = 2; y = 18 * 0 986^x y = 18 * 0 98 y = 18 * 0 9604 y = 17 2872 Pt.(2, 17 2872) Now plot the points onto the graph. Do likewise with the next question.
Question:So here is the formula used:
http://www.freeimagehosting.net/uploads/0ea8b7cc77.gif
the thing is its like it looks extremity complicated to me.
in the exponent is the "t" being multiplied by the 1/5th?
so does that make it a decay function with an exponential line thing because the "t" is in the exponent?
but what does the A(t) represent? the initial amount? or the is it the amount present at time "t"?
so the "1/2" would be the growth factor and the "4" would be the number of times to complete a cycle?
how would i find how much is left after 20 years?
I know this is alot but im sooo confused, things like this are why i hate math. its like hitting a road block.
If its not to much here is the other equation im trying to figure out is
http://www.freeimagehosting.net/uploads/a7b8e2cfcd.gif
Answers:A(t) = 4*(1/2)^[ (1/5) * t ] Boy, it just looks even more complicated when I write it like that; doesn't it? It's not as bad as it seems. We can make sense of this if we break it down stepbystep. First off: "n the exponent is the "t" being multiplied by the 1/5th?" Yes, you are correct. Both the (1/5) and the 't' are in the exponent, and so the exponent is the product of (1/5) and 't'. "but what does the A(t) represent? the initial amount? or the is it the amount present at time "t"?" A(t) is a function that describes the amount present at time 't'. By substituting in some value for 't', you get what amount is there at that time. "so the "1/2" would be the growth factor and the "4" would be the number of times to complete a cycle?" Not quite. You're correct that (1/2) is the growth factor, in the sense that the amount of material present is cut in half every five units of time. 4 is the amount that's initially present. To prove it to yourself, try finding A(t) for t = 0. That's: A(0) = 4*(1/2)^[ (1/5) * 0 ] Clearly (1/5)*0 = 0. So you have (1/2) raised to the power of 0 and then times 4. Since anything to the 0 is 1, that means A(0) = 4. So we start out with 4, and we cut the amount in half every 5 units of time. "how would i find how much is left after 20 years?" Simple. A(20) = 4*(1/2)^[ (1/5) * 20) ] You see that (1/5)*20 = 4. So we've got 4 times (1/2) to the fourth. (1/2) to the fourth is 1/16. Multiply that by 4 and you get 1/4. That means 1/4 units of material are left after 20 years. Before I move on to the next part I want to explain what I've been saying about 'every five units of time'. Look at the exponent; you've already noticed that it's (1/5)*t. That means that the exponent is just onefifth of 't'. So if t = 5 then the exponent is 1. If t = 10 then the exponent is 2. Can you see how this means that we lose half the material every five units of time? Now on to the next one. M(t) = 8*(2)^[ (1/6) * t ] This isn't very different from the last one. Of course t is time. Since there is a factor of (1/6) in the exponent, you can see that if t = 6 then the exponent is 1 and if t = 12 then the exponent is 2 and so on. Thus the exponent increases by one for every six units of time that elapse. Since the exponent is on top of a 2, that means the amount of material doubles with every cycle (every six units of time). The leading coefficient is an 8 which tells you that M(0) = 8. Try it if you need to. Good luck. Feel free to drop me an IM if you need more help. Math is not hard once you believe it. =)
Answers:A(t) = 4*(1/2)^[ (1/5) * t ] Boy, it just looks even more complicated when I write it like that; doesn't it? It's not as bad as it seems. We can make sense of this if we break it down stepbystep. First off: "n the exponent is the "t" being multiplied by the 1/5th?" Yes, you are correct. Both the (1/5) and the 't' are in the exponent, and so the exponent is the product of (1/5) and 't'. "but what does the A(t) represent? the initial amount? or the is it the amount present at time "t"?" A(t) is a function that describes the amount present at time 't'. By substituting in some value for 't', you get what amount is there at that time. "so the "1/2" would be the growth factor and the "4" would be the number of times to complete a cycle?" Not quite. You're correct that (1/2) is the growth factor, in the sense that the amount of material present is cut in half every five units of time. 4 is the amount that's initially present. To prove it to yourself, try finding A(t) for t = 0. That's: A(0) = 4*(1/2)^[ (1/5) * 0 ] Clearly (1/5)*0 = 0. So you have (1/2) raised to the power of 0 and then times 4. Since anything to the 0 is 1, that means A(0) = 4. So we start out with 4, and we cut the amount in half every 5 units of time. "how would i find how much is left after 20 years?" Simple. A(20) = 4*(1/2)^[ (1/5) * 20) ] You see that (1/5)*20 = 4. So we've got 4 times (1/2) to the fourth. (1/2) to the fourth is 1/16. Multiply that by 4 and you get 1/4. That means 1/4 units of material are left after 20 years. Before I move on to the next part I want to explain what I've been saying about 'every five units of time'. Look at the exponent; you've already noticed that it's (1/5)*t. That means that the exponent is just onefifth of 't'. So if t = 5 then the exponent is 1. If t = 10 then the exponent is 2. Can you see how this means that we lose half the material every five units of time? Now on to the next one. M(t) = 8*(2)^[ (1/6) * t ] This isn't very different from the last one. Of course t is time. Since there is a factor of (1/6) in the exponent, you can see that if t = 6 then the exponent is 1 and if t = 12 then the exponent is 2 and so on. Thus the exponent increases by one for every six units of time that elapse. Since the exponent is on top of a 2, that means the amount of material doubles with every cycle (every six units of time). The leading coefficient is an 8 which tells you that M(0) = 8. Try it if you need to. Good luck. Feel free to drop me an IM if you need more help. Math is not hard once you believe it. =)
From Youtube
Algebra 2  Exponential Growth and Decay :Free Math Help at Brightstorm! www.brightstorm.com How to use the exponential growth and decay formula.
Exponential Decay Formula Proof (can skip, involves Calculus) :Showing that N(t)=Ne^(kt) describes the amount of a radioactive substance we have at time T. For students with background in Calculus. Not necessary for intro chemistry class.