Answer:
![= \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](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1344%5C%5C84%5C%5C28%5Cend%7Barray%7D%5Cright%5D%20%20%5Cleft%20%5Cbegin%7Barray%7D%7Bccc%7D%7B0%20%5C%20%20%5Cleq%20%20age%20%20%20%5Cleq%20%201%20%7D%5C%5C%7B%201%20%5C%20%20%5Cleq%20%20age%20%20%20%5Cleq%20%202%20%7D%5C%5C%7B2%20%5C%20%20%5Cleq%20%20age%20%20%5Cleq%203%7D%5Cend%7Barray%7D%5Cright)
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]](https://tex.z-dn.net/?f=x%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%262%5C%5C1%261%262%5C%5C1%261%262%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%7B0%20%5C%20%20%5Cleq%20%20age%20%20%20%5Cleq%20%201%20%7D%5C%5C%7B%200%20%5C%20%20%5Cleq%20%20age%20%20%20%5Cleq%20%202%20%7D%5C%5C%7B0%20%5C%20%20%5Cleq%20%20age%20%20%20%5Cleq%203%7D%5Cend%7Barray%7D%5Cright%5D)
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]](https://tex.z-dn.net/?f=L%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%266%263%5C%5C0.75%260%260%20%5C%5C0%260.25%260%5Cend%7Barray%7D%5Cright%5D)
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]](https://tex.z-dn.net/?f=x_2%20%3DLx_1%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%266%263%5C%5C0.75%260%260%20%5C%5C0%260.25%260%5Cend%7Barray%7D%5Cright%5D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%262%5C%5C1%261%262%5C%5C1%261%262%5Cend%7Barray%7D%5Cright%5D)
![= \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](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1344%5C%5C84%5C%5C28%5Cend%7Barray%7D%5Cright%5D%20%20%5Cleft%20%5Cbegin%7Barray%7D%7Bccc%7D%7B0%20%5C%20%20%5Cleq%20%20age%20%20%20%5Cleq%201%20%7D%5C%5C%7B%201%20%5C%20%20%5Cleq%20%20age%20%20%20%5Cleq%20%202%20%7D%5C%5C%7B2%20%5C%20%20%5Cleq%20%20age%20%20%20%5Cleq%20%203%7D%5Cend%7Barray%7D%5Cright)
Answer:
$1.60 for tax, $33.60 all together
Step-by-step explanation:
32 x .05 = 1.6
Step-by-step explanation:
3 meters = 300 cm
anyway, the ratio tells us that the whole ribbon can be seen as the combination of 3 + 2 = 5 equally long parts.
one such part is then
3 m / 5 = 0.6 m or 60 cm
the shorter piece is 2 of these parts long :
2 × 0.6 = 1.2 m or 120 cm
Let the speed of the current be y and the speed of Micah's sailing speed be x. Then 4.48/(x + y) = 0.32
4.48/(x - y) = 0.56
0.32x + 0.32y = 4.48 . . . (1)
0.56x - 0.56y = 4.48 . . . (2)
(1) x 7 => 2.24x + 2.24y = 31.36 . . . (3)
(2) x 4 => 2.24x - 2.24y = 17.92 . . . (4)
(3) - (4) => 4.48y = 13.44
y = 3
From (1), 0.32x + 0.32(3) = 4.48
0.32x = 4.48 - 0.96 = 3.52
x = 3.52/0.32 = 11
Therefore, the speed of the current is 3 miles per hour.
ounces of water
ounces of cocoa
ounces of water
ounces of cocoa (same amount of cocoa as before because pure water is being added)
ounces of water
ounces of cocoa