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[size=150]Suppose The Population Of Deer In A Region Was 3,500 In The Year 2000. Since Then The Population Has Grown By 3.5 % Annually. What Will The Approximate Population Be In The Year 2020? Use This Equation Y= a*b^x When a Is The Starting Amount; b Is The Growth Rate; And x Is The Time Period.[/size]
 Mar 6, 2014
 #1
avatar+64 
0
Hm, i think you should try to use your equation as below:
(3.5% of 3500 = 3500*0.035). So, b=0.035:

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When a is the starting amount, b=0.035 and x is the time period.
Try it yourself, by just using the right numbers and if you need more help, ask me here! Good luck!
Dms
 Mar 6, 2014
 #2
avatar+118723 
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Guest:

[size=150]Suppose The Population Of Deer In A Region Was 3,500 In The Year 2000. Since Then The Population Has Grown By 3.5 % Annually. What Will The Approximate Population Be In The Year 2020? Use This Equation Y= a*b^x When a Is The Starting Amount; b Is The Growth Rate; And x Is The Time Period.[/size]



Thanks Dms,

Let the year 2000 be year 0 x=0
and the year 2020 be year 20 x=20

As Dms said, a=3500, and b=0.035

so you have the equation

Y = 3500 * 0.035 x

Now just plug the numbers in
 Mar 6, 2014
 #3
avatar+64 
0
Melody:


Thanks Dms,

Let the year 2000 be year 0 x=0
and the year 2020 be year 20 x=20

As Dms said, a=3500, and b=0.035

so you have the equation

Y = 3500 * 0.035x

Now just plug the numbers in


Nice, but your equation for Y is wrong. I'll give you an example:
y=3500*0.035^20
So, y<a ? Remember that already exist 3500 deers in the year 2000. This is why you should use the equation:

0


Dms
 Mar 7, 2014
 #4
avatar+118723 
0
Yes, of course. Thank you.
Melody
 Mar 7, 2014

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