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dmitriy555 [2]
2 years ago
7

While growing bacteria in a laboratory Petri dish, a scientist notices the number of bacteria over time. The observations are sh

own in the table.
Number of bacteria
Number of minutes
from initial state
0 4
5 128
10 4096
15 131072

What function can be used to describe the number (n) of bacteria after 2 minutes?
Mathematics
1 answer:
9966 [12]2 years ago
8 0

Answer:

The function that can be used to describe the number (n) of bacteria after 2 minutes is;

P = 4 \cdot e^{\left(\dfrac{ln(32)}{5} \times 2\right)} \approx 4 \cdot e^{\left(0.693\times 2\right)}

Step-by-step explanation:

The data in the table are presented as follows;

Number of bacteria; 4, 128, 4,096, 131,072

Number of minutes from initial state; 0, 5, 10, 15

The general equation for population growth is presented as follows;

P = P_0 \cdot e^{r\cdot t}

Where;

P = The population after 't' minutes

P₀ = The initial population

r = The population growth rate

t = The time taken for the growth in population numbers

At t =  minutes. we have;

4 = P_0 \cdot e^{r\times0} = P_0

∴ P₀ = 4

At t = 5, we have;

128 = 4 \cdot e^{r\times 5}

\therefore  e^{r\times 5} = \dfrac{128}{4} = 32

ln\left(e^{r\times 5}\right) = ln(32)

∴ r × 5 = ㏑(32)

r = ln(32)/5 ≈ 0.693

The number (n) of bacteria after 2 minutes is therefore;

P = 4 \cdot e^{\left(\dfrac{ln(32)}{5} \times 2\right)} \approx 4 \cdot e^{\left(0.693\times 2\right)}

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anzhelika [568]

Answer:

(c) Probability that a failure is due to loose keys = 0.2376

(d) Probability that a failure is due to improperly connected or poorly welded wires = 0.078

Step-by-step explanation:

The Whole probability scenario is given for Computer Keyboard failures.

(a) Let F be the event of failure due to faulty electrical connects, P(F) = 0.12

 M be the event of failure due to mechanical defects, P(M) = 0.88

 LK be the event of mechanical defect due to loose keys, P(LK/M) = 0.27

 IA be the event of mechanical defect due to improper assembly, P(IA/M)   =0.73

 DW be the event of electrical connects due to defective wires,P(DW/F) = 0.35

 IC be the event of electrical connects due to improper connections,

  P(IC/F) = 0.13 .

PWW be the event of electrical connects due to poorly welded wires,

  P(PWW/F) = 0.52

(b)                                     <u> </u><u>Keyboard failures</u>

<h2>                              /               \</h2>

           <u> </u><u> Faulty electrical connects   </u>            <u>Mechanical Defects </u>          

                      P(F) = 0.12                                             P(M) = 0.88

<h2>        /            |             \                  /            \</h2>

<u><em>Defective wires</em></u>  <u><em>Improper</em></u>        <u><em>Poorly</em></u>                  <u><em>Loose Keys</em></u>      <u><em>Improper</em></u><em> </em>

P(DW/F)=0.35   <u><em>Connections</em></u>   <u><em>Welded wires</em></u>      P(LK/M)=0.27   <em> </em><u><em>Assembly</em></u>

                           P(IC/F)=0.13     P(PWW/F)=0.52                            P(IA/M)=0.73              

This is the required tree diagram.

(c) Probability that a failure is due to loose keys is given by:

  P(LK) =P(LK/M) * P(M) {This means mechanical failure is due to loose  

                                               keys}

    P(LK) = 0.27 * 0.88 = 0.2376 .

(d) Probability that a failure is due to improperly connected or poorly welded

     wires is given by P(IC \bigcup PWW) ;

 P(IC \bigcup PWW) = P(IC) + P(PWW) - P(IC \bigcap PWW) { Here P(IC \bigcap PWW) = 0 }

 P(IC) = P(IC/F) * P(F)  = 0.13 * 0.12 = 0.0156

 P(PWW) = P(PWW/F) * P(F) = 0.52 * 0.13 = 0.0676

Therefore, P(IC \bigcup PWW) = 0.0156 + 0.0676 - 0 = 0.078 .

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