Answer:
The population of the city in 2002 is 469,280 while the population of the suburb is 730,720.
Step-by-step explanation:
- 6% of the city's population moves to the suburbs (and 94% stays in the city).
- 2% of the suburban population moves to the city (and 98% remains in the suburbs).
The migration matrix is given as:
![A= \left \begin{array}{cc} \\ C \\S \end{array} \right\left[ \begin{array}{cc} C&S\\ 0.94&0.06 \\0.02&0.98 \end{array} \right]](https://tex.z-dn.net/?f=A%3D%20%5Cleft%20%5Cbegin%7Barray%7D%7Bcc%7D%20%20%5C%5C%20C%20%5C%5CS%20%5Cend%7Barray%7D%20%5Cright%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bcc%7D%20%20C%26S%5C%5C%200.94%260.06%20%5C%5C0.02%260.98%20%5Cend%7Barray%7D%20%5Cright%5D)
The population in the year 2000 (initial state) is given as:
![\left[ \begin{array}{cc} C&S\\ 500,000&700,000 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bcc%7D%20%20C%26S%5C%5C%20500%2C000%26700%2C000%20%20%5Cend%7Barray%7D%20%5Cright%5D)
Therefore, the population of the city and the suburb in 2002 (two years after) is:
![S_0A^2=\left \begin{array}{cc} [500,000&700,000]\\& \end{array} \right\left \begin{array}{cc} \end{array} \right\left[ \begin{array}{cc} 0.94&0.06 \\0.02&0.98 \end{array} \right]^2](https://tex.z-dn.net/?f=S_0A%5E2%3D%5Cleft%20%5Cbegin%7Barray%7D%7Bcc%7D%20%5B500%2C000%26700%2C000%5D%5C%5C%26%20%20%5Cend%7Barray%7D%20%5Cright%5Cleft%20%5Cbegin%7Barray%7D%7Bcc%7D%20%5Cend%7Barray%7D%20%5Cright%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bcc%7D%200.94%260.06%20%5C%5C0.02%260.98%20%5Cend%7Barray%7D%20%5Cright%5D%5E2)
![A^{2} = \left[ \begin{array}{cc} 0.8848 & 0.1152 \\\\ 0.0384 & 0.9616 \end{array} \right]](https://tex.z-dn.net/?f=A%5E%7B2%7D%20%3D%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bcc%7D%200.8848%20%26%200.1152%20%5C%5C%5C%5C%200.0384%20%26%200.9616%20%5Cend%7Barray%7D%20%5Cright%5D)
Therefore:
![S_0A^2=\left \begin{array}{cc} [500,000&700,000]\\& \end{array} \right\left \begin{array}{cc} \end{array} \right \left[ \begin{array}{cc} 0.8848 & 0.1152 \\ 0.0384 & 0.9616 \end{array} \right]\\\\=\left[ \begin{array}{cc} 500,000*0.8848+700,000*0.0384& 500,000*0.1152 +700,000*0.9616 \end{array} \right]\\\\=\left[ \begin{array}{cc} 469280& 730720 \end{array} \right]](https://tex.z-dn.net/?f=S_0A%5E2%3D%5Cleft%20%5Cbegin%7Barray%7D%7Bcc%7D%20%5B500%2C000%26700%2C000%5D%5C%5C%26%20%20%5Cend%7Barray%7D%20%5Cright%5Cleft%20%5Cbegin%7Barray%7D%7Bcc%7D%20%5Cend%7Barray%7D%20%5Cright%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bcc%7D%200.8848%20%26%200.1152%20%5C%5C%200.0384%20%26%200.9616%20%5Cend%7Barray%7D%20%5Cright%5D%5C%5C%5C%5C%3D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bcc%7D%20500%2C000%2A0.8848%2B700%2C000%2A0.0384%26%20500%2C000%2A0.1152%20%2B700%2C000%2A0.9616%20%5Cend%7Barray%7D%20%5Cright%5D%5C%5C%5C%5C%3D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bcc%7D%20469280%26%20730720%20%5Cend%7Barray%7D%20%5Cright%5D)
Therefore, the population of the city in 2002 is 469,280 while the population of the suburb is 730,720.
The answer is C. x<span> ≤ </span>3
Answer:
The average velocity of the ball at the given time interval is -122.3 ft/s
Step-by-step explanation:
Given;
velocity of the ball, v = 34 ft/s
height of the ball, y = 34t - 26t²
initial time, t₀ = 3 seconds
final time, t = 3 + 0.01 = 3.01 seconds
At t = 3 s
y(3) = 34(3) - 26(3)² = -132
The average velocity of the ball in ft/s is given as;
Therefore, the average velocity of the ball at the given time interval is -122.3 ft/s
Here, as the sides are equal, middle line would be exact half then the longer (bottom) side of the triangle.
48 = 2(3x)
6x = 48
x = 48/6
x = 8
In short, Your Answer would be Option D
Hope this helps!
Subtract the total of the 3 fish she is buying from the amount of money she has:
85 - (12 + 13 + 14) = 85 - 39 = 46
She has $46 to spend on crabs.
Divide the amount she has to spend by the price on one crab:
46 / 8 = 5.75
Since it isn't a whole number you need to round down, since she cant buy 0.75 of a crab.
She has enough money to buy 5 crabs.
