CAN SOMEONE HELP WITH THIS QUESTION?✨

Answer:

351.5625

1,440,000

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{7 cm}\underline{General form of an Exponential Function}\\\\$y=Ae^{kt}$\\\\where:\\\phantom{ww}$\bullet$ $A$ is the initial value ($y$-intercept). \\ \phantom{ww}$\bullet$ $k$ is a constant.\\ \phantom{ww}$\bullet$ $t$ is time.\\\end{minipage}}[/tex]

Given:

(Let t = time in minutes).

If the doubling period is 15 minutes, then at t = 135 minutes, y = 180,000:

[tex]\implies 90000=Ae^{120k}[/tex]

[tex]\implies 180000=Ae^{135k}[/tex]

Divide the second equation by the first to eliminate A, and solve for k:

[tex]\implies \dfrac{180000}{90000}=\dfrac{Ae^{135k}}{Ae^{120k}}[/tex]

[tex]\implies 2=\dfrac{e^{135k}}{e^{120k}}[/tex]

[tex]\implies 2=e^{135k} \cdot e^{-120k}[/tex]

[tex]\implies 2=e^{15k}[/tex]

[tex]\implies \ln 2 = \ln e^{15k}[/tex]

[tex]\implies \ln 2 =15k \ln e[/tex]

[tex]\implies \ln 2 =15k[/tex]

[tex]\implies k=\dfrac{1}{15}\ln 2[/tex]

Substitute t = 120, y = 90000 and the found value of k into the formula and solve for A:

[tex]\implies 90000=Ae^{\left(120 \cdot \frac{1}{15}\ln 2\right)}[/tex]

[tex]\implies 90000=Ae^{\left(8\ln 2\right)}[/tex]

[tex]\implies 90000=Ae^{\ln256}[/tex]

[tex]\implies 90000=256A[/tex]

[tex]\implies A=\dfrac{90000}{256}[/tex]

[tex]\implies A=351.5625[/tex]

Therefore, the function that models the scenario is:

[tex]\large\boxed{y=351.5625e^{\left(\frac{1}{15}t \ln 2\right)}}[/tex]

So the initial population at time t = 0 was:

To find the size of the bacteria population after 3 hours, substitute t = 180 into the found formula:

[tex]\implies y=351.5625e^{\left(\frac{1}{15}(180) \ln 2\right)}[/tex]

[tex]\implies y=351.5625e^{\left(12 \ln 2\right)}[/tex]

[tex]\implies y=351.5625e^{\left(\ln 4096\right)}[/tex]

[tex]\implies y=351.5625 \cdot 4096[/tex]

[tex]\implies y=1440000[/tex]

Therefore, the size of the bacterial population after 3 hours was:

The initial population at the time t = 0 is 351.5625. And the size of the bacterial population after 3 hours is 1,440,000.

An exponential function's curve is created by a pattern of data called exponential growth, which exhibits higher increases over time.

If n₀ is the initial size of a population experiencing exponential growth, then the population n(t) at time t is modeled by the function:

n(t) = n₀(e[tex])^{rt}[/tex]

Where r is the relative rate of growth expressed as a fraction of the population.

Given:

Doubling period = 15 minutes

At t = 120 minutes, n(t) = 90,000

If the doubling period is 15 minutes, then at t = 120+15 = 135 minutes,

90000 = n₀(e[tex])^{120r}[/tex]

18000 =  n₀(e[tex])^{135r}[/tex]]

To find the r:

Take ratio of both of the equations,

90000 / 18000 = n₀(e[tex])^{120r}[/tex]  /   n₀(e[tex])^{135r}[/tex]

2 = (e[tex])^{135r}[/tex] . (e[tex])^{-120r}[/tex]

r = 1/15 ln2

Substitute the value of r, t and y.

90000 = n₀(e[tex])^{120r}[/tex]

90000 = 256n₀

n₀ = 351.5652

Now, the function

n(t) = n₀(e[tex])^{rt}[/tex]

n(t) = (351.5652)(e[tex])^{(1/15)(180)(ln2)}[/tex]

n(t) = 1440000

Therefore, the initial population at the time t = 0 is 351.5625. And the size of the bacterial population after 3 hours is 1,440,000.

To learn more about the exponential growth;

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