It's 8/a^6 that's your answer
Answer:
8.2+/-0.25
= ( 7.95, 8.45) years
the 95% confidence interval (a,b) = (7.95, 8.45) years
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
x+/-zr/√n
Given that;
Mean x = 8.2 years
Standard deviation r = 1.1 years
Number of samples n = 75
Confidence interval = 95%
z value(at 95% confidence) = 1.96
Substituting the values we have;
8.2+/-1.96(1.1/√75)
8.2+/-1.96(0.127017059221)
8.2+/-0.248953436074
8.2+/-0.25
= ( 7.95, 8.45)
Therefore the 95% confidence interval (a,b) = (7.95, 8.45) years
Answer: m = $52
Explanation:
Tickets (t) = $8
Popcorn & Soda (s)= $5
Friends = 4
Since all friends bought the same thing the equation will be set as
4(t +s) = m OR 4t +4s = m
Using substitution, you plug each variable
t = 8 and s = 5
4((8) + (5)) = m
4(8+5) = m => 4(13) =m => 52=m
OR
4(8) + 4(5) = m => 32 + 20 = m =>
52 =m
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)
