$= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right] \left \begin{array}{ccc}{0 \ \leq age \leq 1 }\\{ 1 \ \leq age \leq 2 }\\{2 \ \leq age \leq 3}\end{array}\right$

i.e after the first year ;

there 1344 members in the first age class

84 members for the second age class; and

28 members for the third age class

Step-by-step explanation:

We can deduce that the age distribution vector x represents the number of population members for each age class; Given that in each class of age there are 112 members present.

The current age distribution vector is as follows:

$x = \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right] \left[\begin{array}{ccc}{0 \ \leq age \leq 1 }\\{ 0 \ \leq age \leq 2 }\\{0 \ \leq age \leq 3}\end{array}\right]$

Also , the age transition matrix is as follows:

$L = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right]$

After 1 year ; the age distribution vector will be :

$x_2 =Lx_1 = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right] \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right]$

$= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right] \left \begin{array}{ccc}{0 \ \leq age \leq 1 }\\{ 1 \ \leq age \leq 2 }\\{2 \ \leq age \leq 3}\end{array}\right$

6 0

2 years ago

If 3/4 of an inch of rain fell in 1/2 hour, explain how to get the unit rate

bearhunter [10]

Answer:

Search to find the detailed rainfall collections if you want more detailed rainfall intensity data. If all available is monthly average in mm, take the monthly average and divide by number of hours in that month. So if you had 300 mm in April, take 360 mm/(30 days * 24 hours) = 0.5 mm/hour average monthly intensity. so that's it

4 0

2 years ago