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:
three equivalent ratios will be
6:22 , 9:33, 15:55
Answer:
- 0.100
Step-by-step explanation:
Length of the ladder, H = 6 m
Distance at the bottom from the wall, B = 1.3 m
Let the distance of top of the ladder from the bottom at the wall is P
Thus,
from Pythagoras theorem,
B² + P² = H² .
or
B² + P² = 6² ..............(1) [Since length of the ladder remains constant]
at B = 1.3 m
1.3² + P² = 6²
or
P² = 36 - 1.69
or
P² = 34.31
or
P = 5.857
Now,
differentiating (1)
at t = 2 seconds
change in B = 0.3 × 2= 0.6 ft
Thus,
at 2 seconds
B = 1.3 + 0.6 = 1.9 m
therefore,
1.9² + P² = 6²
or
P = 5.69 m
on substituting the given values,
2(1.9)(0.3) + 2(5.69) × 
= 0
or
1.14 + 11.38 × = 0
or
11.38 × = - 1.14
or
= - 0.100
here, negative sign means that the velocity is in downward direction as upward is positive
Answer: ( 2 + 3 ) ( 8d )
Step-by-step explanation:
There are multiple possible answers and I'm not sure if you're missing part of the question.
Correct me if I am incorrect.
-4x-24=8
So you move the -24 to the other side.
-4x=32
Divide both sides by -4.
x=-8
