1. ## maths Help

I have this done half way but not sure of how the remaining should be handled, pleaes help me out.
The equationP = P0 (1-e-0.05T) has been developed to simulate the spread of a virus and relates the number of persons infected (P) in a population P0 , after a period of T days after it has first been detected.
Use this model to determine:
(i) The number of persons in a town with a population of 10,000 that would be expected to be infected five days after the first detection of the virus
(ii) The time in terms of number of days before 50% of the population is infected
(iii) The time in days before 95% of the population is infected.
Here is my steps:
P = 10,000 (1 – e-0.0(5)(5))
P = 10,000 (1 – e-0.025)
P = 10,000 (1- 1/e0.025)

I was stocked at how to convert 1/e0.025.

Ola

2. Originally Posted by Ola
I have this done half way but not sure of how the remaining should be handled, pleaes help me out.
The equationP = P0 (1-e-0.05T) has been developed to simulate the spread of a virus and relates the number of persons infected (P) in a population P0 , after a period of T days after it has first been detected.
Use this model to determine:
(i) The number of persons in a town with a population of 10,000 that would be expected to be infected five days after the first detection of the virus
(ii) The time in terms of number of days before 50% of the population is infected
(iii) The time in days before 95% of the population is infected.
Here is my steps:
P = 10,000 (1 – e-0.0(5)(5))
P = 10,000 (1 – e-0.025)
P = 10,000 (1- 1/e0.025)

I was stocked at how to convert 1/e0.025.

Ola
1. I assume that the given equation reads:

$P(T)=P_0\left(1-e^{-0.05 \cdot T} \right)$

If so then

(i) would be: $P(5)=10,000 \left(1-e^{-0.05 \cdot 5} \right) = 10,000 \left(1-e^{-0.25} \right)
$

and now use your calculator. Keep in mind that there aren't any decimal fractions of persons.

(ii) would be: $\frac12 P_0 = P_0 \left(1-e^{-0.05 \cdot T} \right)$

Solve for T.

(iii) should be done in exactly the same way as (ii).

3. ## Maths - Exponential

Hi,

Very grateful for your help, pls can you help me to have a look at the solutions I finally arrived at just to be sure I am on the right track:

(i)
P=Po(1-e^-0.05(5))
P = 10,000(1-^-0.25)
P = 10000 (1-0.7788)
P = 10,000(0.2212)
P = 2,212

(ii)
1/2 of 10,000 (1-e^0.05(T))
= 1/2(10,000) (1 - 0.95123T)
= 5,000 (1 - 0.95123T
= 5,000 - 4756.15T
4756.15T = 5,000
T = 5,000/4756.15
T = 1.05127
T = 1 day (aproximately)

(iii)
95% of 10,000 (1 - e^0.05(T))
= 9,500 (1 - 0.95123T)
= 9,500 - 9,036.68T
9036.68T = 9,500
T = 9,500/9036.68
T = 1.051271 day
T = 1 day (aproximately)

4. Originally Posted by Ola
Hi,

Very grateful for your help, pls can you help me to have a look at the solutions I finally arrived at just to be sure I am on the right track:

(i)
P=Po(1-e^-0.05(5))
P = 10,000(1-^-0.25)
P = 10000 (1-0.7788)
P = 10,000(0.2212)
P = 2,212

(ii)
1/2 of 10,000 (1-e^0.05(T))
= 1/2(10,000) (1 - 0.95123T)
= 5,000 (1 - 0.95123T
= 5,000 - 4756.15T
4756.15T = 5,000
T = 5,000/4756.15
T = 1.05127
T = 1 day (aproximately) <<<<< if you have 2000 infections in 5 days it would take a little bit longer to have 5000 infections

(iii)
95% of 10,000 (1 - e^0.05(T))
= 9,500 (1 - 0.95123T)
= 9,500 - 9,036.68T
9036.68T = 9,500
T = 9,500/9036.68
T = 1.051271 day
T = 1 day (aproximately) <<<<<<< No
to (ii):

50% of 10,000 = 5,000 . Plug in this value for P(T) and solve for T:

$5000 = 10000 \left(1-e^{-0.05 \cdot T} \right)~\implies~\dfrac12= 1-e^{-0.05 \cdot T}~\implies~\dfrac12= e^{-0.05 \cdot T}
$

Now use logarithms to solve for T.

$\ln\left(\frac12\right)=-0.05\cdot T~\implies~T=\dfrac{\ln\left(\frac12\right)}{0.05} \approx 13.86\ days$

(iii) has to be done in exactly the same way.

5. I accept with information:I have this done half way but not sure of how the remaining should be handled, pleaes help me out.
The equationP = P0 (1-e-0.05T) has been developed to simulate the spread of a virus and relates the number of persons infected (P) in a population P0 , after a period of T days after it has first been detected.
Use this model to determine:
(i) The number of persons in a town with a population of 10,000 that would be expected to be infected five days after the first detection of the virus
(ii) The time in terms of number of days before 50% of the population is infected
(iii) The time in days before 95% of the population is infected.
Here is my steps:
P = 10,000 (1 – e-0.0(5)(5))
P = 10,000 (1 – e-0.025)
P = 10,000 (1- 1/e0.025)

I was stocked at how to convert

6. Originally Posted by cinlymcre
...

I was stocked at how to convert 1/e0.025.

1. I don't know why you want to do this ...

2. According to my first reply the term should be: $1-e^{-0.25}$

3. Since $0.25 = \dfrac14$ the term becomes:

$1-e^{-\frac14} = 1-\dfrac1{e^{\frac14}} = 1-\dfrac1{\sqrt[4]{e}}$

7. Hi,
I know it will be a surprise to you. My background in maths is terrible very sorry for the mistake and that is the reason why I have come to somebody like yourself to rescue me. I appreciate your efforts.
Ola.