Respuesta :
Answer:
A) 2033%
B) 75 bacteria
C) n(t) = 75(1 + 1.83%)^t R = 1.83%
D) 81.4 bacteria
E) 348.83 seconds
Step-by-step explanation:
Given that the count in a culture of bacteria was 600 after 2 hours and 38,400 after 6
A) growth (600 - 0)/2 = 300 bacteria/s
38400/6 = 6400 bacteria/s
Relative growth=(6400-300)/300 ×100
= 6100/300 × 100
= 2033%
B) initial size of the culture
Using exponential equation
P = I( 1 + R)^t
Where I = initial size
R = rate
600 = I(1 + R)^2
Log both sides
Log600 = log I(1+R)^2
2.778 = logI + 2log(1 +R)
2log (1+R) = 2.778 - logI ..... (1)
Also,
38400 = I(1 + R)^6
Log both sides
Log 38400 = logI + 6log(1+R)
6Log(1+R) = 4.584 - logI .... (2)
Divide equation 2 by 1
6/2 = (4.584 - logI)/(2.778 - logI)
Cross multiply
16.668 - 6logI = 9.168 - 2logI
6logI - 2logI = 16.668 - 9.168
4logI = 7.5
LogI = 7.5/4
LogI = 1.875
I = 74.98 = 75 bacteria
C) A function that models the number of bacteria n(t) after t hours.
If I = 75 bacteria
Then n(t) = 75(1 + R)^t
600 = 75(1+R)^2
8 = (1+R)^2
Log both sides
Log8 = 2log(1+R)
0.903/2 = log(1+R)
0.45 = log(1+R)
1 + R = 2.83
R = 1.83%
The model function is therefore
n(t) = 75(1 + 1.83%)^t
D) the number of bacteria after 4.5 hours
n(t) = 75(1.02)^4.5
n(t) = 81.4 bacteria
E) After how many hours will the number of bacteria reach 75,000
n(t) = 75(1.02)^t
75000 = 75(1.02)^t
1000 = 1.02^t
Log both sides
Log 1000 = tlog 1.02
3 = 0.0086t
t = 348.83 seconds
The bacteria has an exponential growth rate and the population of the
bacteria increases rapidly with time.
- (a) The relative rate of growth is [tex]\underline{\dfrac{ln(64)}{4}}[/tex].
(b) The initial size of the culture is 75 bacteria.
(c) The function that models the number of bacteria n(t) is; [tex]\underline{n(t) = 75 \cdot e^{1.04 \cdot t}}[/tex]
(d) The number of bacteria after 4.5 hours is approximately 8,100 bacteria.
(e) The number of hours after which the bacteria will reach 75,000 is approximately 6.64 hours.
Reasons:
The count in the culture of bacteria after 2 hours = 600
The count after 6 hours = 38,400
(a) The relative rate of growth, k is given by the formula;
[tex]y = \mathbf{C \cdot e^{k \cdot t}}[/tex]
Therefore, we get;
[tex]600 = C \cdot e^{k \times 2}[/tex]
[tex]38,400= C \cdot e^{k \times 6}[/tex]
Which gives;
[tex]\dfrac{38,400}{600} = \mathbf{\dfrac{C \cdot e^{k \times 6}}{C \cdot e^{k \times 2}}}[/tex]
[tex]64= \dfrac{e^{k \times 6}}{e^{k \times 2}} = e^{k \times 6- k \times 2} = e^{4\cdot k}[/tex]
[tex]\mathbf{e^{4\cdot k}} = 64[/tex]
[tex]ln\left(e^{4\cdot k}) = ln(64)[/tex]
4·k = ㏑(64)
[tex]The \ relative \ rate \ of \ growth, \, \underline{ k = \dfrac{ln(64)}{4}}[/tex]
(b) The initial size of the culture, C, is given by the relation;
[tex]C = \mathbf{\dfrac{y}{e^{k \cdot t}}}[/tex]
Therefore, we get;
[tex]C = \dfrac{600}{e^{\dfrac{ln(64)}{4} \times 2}} = \mathbf{ 75}[/tex]
The initial size of the culture, C = 75
(c) The function is [tex]y = C \cdot e^{k \cdot t}[/tex]
Where:
y = n(t)
C = n₀
k = r
We get;
[tex]n(t) = \mathbf{n_0 \cdot e^{r \cdot t}}[/tex]
n₀ = C = 75
[tex]r = k = \dfrac{ln(64)}{4} \approx 1.04[/tex]
Which gives the function as follows; [tex]\underline{n(t) = 75 \cdot e^{1.04 \cdot t}}[/tex]
(d) The number of bacteria after 4.5 hours is [tex]n(4.5) = 75 \cdot e^{1.04 \times 4.5}[/tex] ≈ 8,100 bacteria
(e) At n(t) = 75,000, we have;
[tex]n(t) = 75,000 = \mathbf{75 \cdot e^{1.04 \times t}}[/tex]
[tex]t = \dfrac{ln\left(\dfrac{75,000}{75} \right)}{1.04} \approx \mathbf{6.64}[/tex]
The time at which the bacteria population will reach 75,000, t ≈ 6.64 hours.
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