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3 years ago

I've been trying to get an answer for this all day and don't know where to begin what is the answer like I have no clue

Mathematics

1 answer:

Shtirlitz [24]3 years ago

6 0

so, we have two 54x18 rectangles, so their perimeter is simply all those units added together, 54+54+54+54+18+18+18+18 = 288.

we know the circle's diameter is 1.5 times the width, well, the width is 18, so the diameter of the circle must be 1.5*18 = 27.

$\bf \stackrel{\textit{circumference of a circle}}{C=d\pi }~~ \begin{cases} d=diameter\\[-0.5em] \hrulefill\\ d=27 \end{cases}\implies C=27\pi \implies C=\stackrel{\pi =3.14}{84.78} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{perimeter of the rectangles}}{288}~~~~+~~~~\stackrel{\textit{perimeter of the circle}}{84.78}~~~~=~~~~372.78$

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Exercise 7.3.5 The following is a Markov (migration) matrix for three locations        1 5 1 5 2 5 2 5 2 5 1 5 2 5 2 5 2

fredd [130]

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Both get the same results that is,

$\left[\begin{array}{ccc}140\\160\\200\end{array}\right]$

Step-by-step explanation:

Given :

$\bf M=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]$

and initial population,

$\bf P=\left[\begin{array}{ccc}130\\300\\70\end{array}\right]$

a) - After two times, we will find in each position.

$P_2=[P].[M]^2=[P].[M].[M]$

$M^2=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]\times \left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]$

$=\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]$

$\therefore\;\;\;\;\;\;\;\;\;\;\;P_2=\left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right] \times\left[\begin{array}{ccc}130\\300\\70\end{array}\right] = \left[\begin{array}{ccc}140\\160\\200\end{array}\right]$

b) - With in migration process, 500 people are numbered. There will be after a long time,

$After\;inifinite\;period=[M]^n.[P]$

$Then,\;we\;get\;the\;same\;result\;if\;we\;measure [M]^n=\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]$

$=\left[\begin{array}{ccc}140\\160\\200\end{array}\right]$

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3 years ago

I've been trying to get an answer for this all day and don't know where to begin what is the answer like I have no clue ​

I've been trying to get an answer for this all day and don't know where to begin what is the answer like I have no clue