Given:
The parallel sides of a trapezium are 28.7 cm and 22.3 cm.
The distance between parallel sides is 16 cm.
To find:
The area of a trapezium.
Solution:
The area of the trapezium is:

Where,
are parallel sides and
is the vertical distance between the parallel sides.
Putting
in the above formula, we get




Therefore, the area of the trapezium is 408 square cm.
5.5hours to drive264 miles
7 hours to drove (264÷5.5)× 7 =....
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.
Answer:
1.5 km
Step-by-step explanation:
15 cm * 10 000 = 150 000 cm = 1.5 km