Answer:
200 ml
Step-by-step explanation:
It's proportions basically.
frst off, 2% of 100 ml is 2 ml, so the 100 ml solution has 2 ml of Minoxidil.
Eventually we want 6/100 to be the porportion. so the total is going to be 100 (from the initial 100 ml) plus however much is needed x. And we want the amount of Minoxidil to be 2 + 8 percent of x. so this gets us the equation
6/100 = (2 + .08x)/(100+x) Then we solve
(100 + x) 6/100 = 2 + .08x
(100 + x) 6 = 200 + 8x
600 + 6x = 200 + 8x
400 = 2x
x = 200
so they should add 200 ml
You can check too. there is 100 + 200 ml of solution total and 2 + 16 ml of Minoxidil so that's 18/300 = 6/100
Answer:
It is A, 65
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)
•First you multiply parentheses by 3 24+3n-15=24
•then u cancel equal terms 3n-15=0
•Move the constant to the right
3n-15=0
3n=15
•Divide both sides by 3
Answer is n=5 ☺️