Answer: Ksh 6144
Step-by-step explanation:
On Thursday, a fruit vendor bought 1948 oranges on a Thursday and sold 750. The number of oranges left will be: = 1948 - 750 = 1198
On n Friday, he sold 240 more oranges than on Thursday. This means the number of oranges sold is: = 240 + 750 = 990.
The number for f oranges left will then be: = 1198 - 990 = 208
On Saturday, he bought 560 more
oranges, the total oranges that the vendor has now will be:
= 208 + 560 = 768
Since sold all the oranges he had at a price of Ksh. 8 each, the amount if money made will be:
= Ksh 8 × 768
= Ksh 6144
The vendor made Ksh 6144
It is A. 2^12.
The question asks for the value of 8 to the x-power. Divided by 2 to the y-power. the trick to solving this is to rewrite 8^x as 2^3x. so you will have 2^3x divided by 2^y. since the bases are now the same , you can subtract the exponents to get the result of 2^3x-y. since it is given the 3x-y is 12. your final answer is 2^12
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.
