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# Followup from CPhill's answer & another question that someone else can answer

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185
12

So for this problem:

CPhill said that for #4, After 15 years...there will be   1000 / [ 1 + 9e^(-.4 * 15) ]  ≈  978  bluegills. What I'm wondering, is what exactly (or how exactly) he put the function in the calculator to get 978. Because when I put the function in my calculator, I didn't get approx 978.

Also, can somebody tell me how I would figure out #3 (What is the point of maximum growth?)?

Many thanks to anyone who helps!!!

Mar 7, 2019

#1
+19675
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Hmmmm.... I put it in my calculator and got  978.18 bluegills.

Mar 7, 2019
#2
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But what exactly did you put into ur calc? Because my worksheet requires me to show my work. So I need to know what you or CPhill put in his/your calc.

Guest Mar 7, 2019
#12
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I put in    1000 / (1+9e^(-.4 *15))

Guest Mar 7, 2019
#3
+716
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How long ago was your original post posted? So i can look and see if i can answer.

Mar 7, 2019
#4
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March 6, 2019, heres the link: https://web2.0calc.com/questions/sumbody-help-me-with-this-question#r3

Guest Mar 7, 2019
#5
+716
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thank you

ilovepuppies1880  Mar 7, 2019
#6
+716
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im sorry this is to advanced for me.

ilovepuppies1880  Mar 7, 2019
#7
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That's alright... I appreciate you doing your best to help

Guest Mar 7, 2019
#8
+1

Use your formula above and plug in 15 for t as follows:

P(t) = 1000/(1 + 9 e^(-(0.4×15)))

= 1000/(1 + 9* e^-6)

= 1000/(1 + 9* 0.00247875....)

= 1000/(1 + 0.022308767)

= 1000/(1.022308767)

= 978.178

Guest Mar 7, 2019
#9
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Thanks, guest!

So for #3, the point of maximum growth would be 1000 right?

Guest Mar 7, 2019
#10
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YES!!.

Guest Mar 7, 2019
#11
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Thanks soooooo much, guest!!!!!

Guest Mar 7, 2019