This expression is correct. It shows the price John pays for the jeans on the left side of the + symbol (30n), and the amount of tax he has to pay on the right side of the symbol (0.08(30n)).
Hope this helps!
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:
We are given an area and three different widths and we need to determine the corresponding length and perimeter.
The first width that is provided is 4 yards and to get an area of 100 we need to multiply it by 25 yards. This would mean that our length is 25 yards and our perimeter would be 2(l + w) which is 2(25 + 4) = 58 yards.
The second width that is given is 5 yards and in order to get an area of 100 yards we need to multiply by 20 yards. This would mean that our length is 20 yards and our perimeter would be 2(l + w) which is 2(20 + 5) = 50 yards.
The final width that is given is 10 yards and in order to get an area of 100 yards we need to multiply by 10. This would mean that our length is 10 yards and our perimeter would be 2(l + w) which is 2(10 + 10) = 40 yards.
Therefore the field that would require the least amount of fencing (the smallest perimeter) is option C, field #3.
<u><em>Hope this helps!</em></u>
The answer is 20 im almost sure
Best is to draw a sketch of the three points.
Next step is to find the distances BC, CD, DB.
The perimeter is the sum of the three distances.
The distances are found using the distance formula:
D=sqrt((y2-y1)^2+(x2-x1)^2)
order of (x1,y1), (x2,y2) is not important.
BC=sqrt((3- -3)^2+(5-3)^2)=sqrt(6^2+2^2)=sqrt(40)=5.324
CD=sqrt((-1-3)^2+(0-5)^2)=sqrt(4^2+5^2)=sqrt(41)=6.403
DB=sqrt((-3- -1)^2+(3-0)^2)=sqrt(2^2+3^2)=sqrt(13)=3.606
So perimeter = BC+CD+DB=16.333 (approximately)
