Respuesta :
Answer:
greater by about 50 bacteria
hope this helps,
T
Step-by-step explanation:
Populations can be modeled by exponential functions and logarithm functions.
- Kirk's function is: [tex]\mathbf{f(x) = (10)2^{\frac{x}{20}}}[/tex]
- Kirk’s model predict the population will be greater than about 50 bacteria after 60 minutes than Jana’s model
Kirk's graph is an exponential function, and it is represented as:
[tex]\mathbf{y = ab^x}[/tex]
From the graph (see attachment);
x = 0, when y =10.
So, we have:
[tex]\mathbf{10 = ab^0}[/tex]
[tex]\mathbf{10 = a\times 1}[/tex]
[tex]\mathbf{10 = a}[/tex]
Rewrite as:
[tex]\mathbf{a = 10 }[/tex]
Substitute [tex]\mathbf{a = 10 }[/tex] in [tex]\mathbf{y = ab^x}[/tex]
So, we have:
[tex]\mathbf{y = 10(b)^x}[/tex]
x = 20, when y =20.
[tex]\mathbf{20 = 10(b)^{20}}[/tex]
Divide both sides by 10
[tex]\mathbf{2 = b^{20}}[/tex]
Take 20th root of both sides
[tex]\mathbf{b = 2^{\frac{1}{20}}}[/tex]
Substitute [tex]\mathbf{b = 2^{\frac{1}{20}}}[/tex] in [tex]\mathbf{y = 10(b)^x}[/tex]
[tex]\mathbf{y = 10(2^{\frac{1}{20}})^x}[/tex]
[tex]\mathbf{y = (10)2^{\frac{x}{20}}}[/tex]
Represent as function
[tex]\mathbf{f(x) = (10)2^{\frac{x}{20}}}[/tex]
So, Kirk's function is: [tex]\mathbf{f(x) = (10)2^{\frac{x}{20}}}[/tex]
When x = 60, we have:
[tex]\mathbf{f(60) = (10)2^{\frac{60}{20}}}[/tex]
[tex]\mathbf{f(60) = (10)2^3}[/tex]
[tex]\mathbf{f(60) = (10)8}[/tex]
[tex]\mathbf{f(60) = 80}[/tex] ---- Kirk's model
For Jana's model, we have:
[tex]\mathbf{y = 12log(x + 1) + 10}[/tex]
Substitute 60 for x
[tex]\mathbf{y = 12log(60 + 1) + 10}[/tex]
[tex]\mathbf{y = 12log(61) + 10}[/tex]
Using a calculator, we have:
[tex]\mathbf{y \approx 31}[/tex] --- Jana's model
Calculate the difference between both models
[tex]\mathbf{d = 80 - 31}[/tex]
[tex]\mathbf{d = 49}[/tex]
Approximate
[tex]\mathbf{d \approx 50}[/tex]
Hence, Kirk’s model predict the population will be greater than about 50 bacteria after 60 minutes than Jana’s model
Read more about population models at:
https://brainly.com/question/8993571
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