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ISSN: 2155-6180
Journal of Biometrics & Biostatistics
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More on the Robust Solution for Epidemiology: Nineteenth-Century Quebec

Paul TE Cusacki*

Brealey Drive, Peterborough, 77432, Infrobright, Canada

*Corresponding Author:
Cusack P
Brealey Drive, Peterborough, 77432
Infrobright, Canada
Tel: (506) 214-331
E-mail: [email protected]

Received Date: March 02, 2017; Accepted Date: April 22, 2017; Published Date: April 25, 2017

Citation: Cusacki PTE (2017) More on the Robust Solution for Epidemiology: Nineteenth-Century Quebec. J Biom Biostat 8: 342. doi:10.4172/2155-6180.1000342

Copyright: © 2017 Cusacki PTE. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

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Abstract

Here we consider the Robust Solution as applied to the cholera epidemic in Lower Canada (Quebec) in 1832. We find that the mathematics from that procedure provides the mathematical foundation or the study. The rate of growth of the virus must be kept below 14% to terminate the spread of the disease.

Keywords

Energy; Time; Density; The bell normal; The golden mean parabola

Introduction

This paper is an examination of the mathematics already wellestablished as the Robust solution. We use this solution applied to the cholera epidemic, particularly in what is toddy, Quebec-Montreal and Quebec City. The data was found in Bilson’s book, A Darkened house, Cholera in nineteenth Century Canada. Some figures come from the Saint John Cholera epidemic 1854 [1-3]. We begin there.

In Saint John, 1854,

1103 deaths from Cholera/pop. 30,000=1/e=e^-t=E

In Quebec 1832-33:

3451/X=1/e

X=1269=rho=density ~ 4/Pi

1269=78.8 deaths/1000.

In Montreal, there was a cholera death rate of 74/1000. In Quebec City, Twas a cholera death rate of 82/1000. Average (74+82)/1000=41/1000 [4,5].

Now rho/c=126.9/2.9979=0.4235 ~ Pi-e=0./4233.=Resistance to Disease=Rd

[Deaths/1000]=rho

Rho/c=cuz

Rho/c* Pi=Space s

And, from Astro-theology mathematics:

The cross-product vector is:

S=|E||t|cos theta

Resistance to death=(Vd) (Cycle)cos (Cycle)

=e*(40% of a cycle) cos (1 rad)

=58.75%

58.75%=Pi=54.18%

1-54.18%=45.82% ~ 45.7=death rate in the entire province of Quebec

1-58.75%=41.25%

Cf Average Death rate above=41/1000.

Re=rho v/Nu

(127(0.8415)/0.27=395

395=S.D.=Sqrt [(1/N) SUM (X-Xbar/S.D)]Let S.D.=Re=395, and solving:

1560 N=G X^3-X^2-X

2/3((0.4)^3-(0.4^2)-0.4-1560 N=0

N=1

S.D.=Sqrt {1/1*(X-Xbar)^2

S=0.6

Sigma=0.3

1-0.3<=Mew<=1+0.3

Standard Normal:

Phi=1/Sqrt(s*2Pi) e^-1/2 (X-0.7)/0.3)^2

X=0, 1.4 for mew=0.7

X=t=1+t

1.4=1+t

T=0.4=1 rad=1/2Pi

Mew=1.4+/-1]/2

Mew bar=1.2

1.2 * c^2=1.08=Z score for 85.99% 1/85.99=116.29=Mass no of elements in the periodic table.

1-0.8599=0.14=14% minimum profit to sustain growth.

Finally,

E=Mc^2

=(1/e)((2.997929)²

=3.3063

E=1/t

t=1/E=1/3.3=0.302

Root for the Bell Normal.

Φ=1/√(σ2π) e-1/2 [(X-1.30/1.30)]²

Root X=t=3.02

1/c=Mc²

1=Mc³

=99.125=1/1.009 ~ 1.01=E

E=Φ

Roots X=0, 1.4

0 ≤ X ≤ 1.4

1-0.8599=0.1401

ℤ=1.08

1/81.99=116.29=Mass of final element in periodic table.

Y=e-t cos t

116.29=e-t cos t

Y=1/e=E

Refer to Figure 1.

t²=²-1=0

dE/dt=2t-1=1

2=1

biometrics-biostatistics-time

Figure 1: Energy vs. time.

E=1/e-t

E=1/e-1=2.718=Y

At t=0, Ln t=0

t=1

At t=π

Ln π+cuz=2.568 ~ π/2=t/2

t/2=1/2=Emin=-1.25

And

Ln π+0.4233=1/et

G ~ 6.54=1/et

et=0.1529=1-sin 1=Moment

1-0.1529=0.8471

sin-1(0.8471)=57.89°=1.01 rads=E.

Conclusion

Like every other two pole problem (infected or not infected) the two-pole solution works to solve problems in epidemiology. Keeping the growth rate below 14% will terminate a pandemic.

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