Answer:
The nomber of people can attend a dinner that costs $131. 50 is 14.
Step-by-step explanation:
With the information provided, it can be said that the total cost of a dinner would be equal to the initial fee plus the result of multiplying the price per person for the number of people, which can be expressed as:
c=16+8.25x, where:
c is the total cost of he dinner
x is the number of people
Now, you can replace c with 131.50 that is the cost of a dinner and solve for x to find the number of people that can attend:
131.50=16+8.25x
131.50-16=8.25x
115.50=8.25x
x=115.50/8.25
x=14
According to this, the answer is that the nomber of people can attend a dinner that costs $131.50 is 14.
Answer: Karen has 20, Trey has 15, Dante has 40
Step-by-step explanation:
Karen=x+5 Trey=x Dante=2x+10
x+x+5+2x+10=75
4x+15=75
4x=60
x=15
Answer:
City @ 2017 = 8,920,800
Suburbs @ 2017 = 1, 897, 200
Step-by-step explanation:
Solution:
- Let p_c be the population in the city ( in a given year ) and p_s is the population in the suburbs ( in a given year ) . The first sentence tell us that populations p_c' and p_s' for next year would be:
0.94*p_c + 0.04*p_s = p_c'
0.06*p_c + 0.96*p_s = p_s'
- Assuming 6% moved while remaining 94% remained settled at the time of migrations.
- The matrix representation is as follows:
- In the sequence for where x_k denotes population of kth year and x_k+1 denotes population of x_k+1 year. We have:
![\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_k = x_k_+_1](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%20x_k%20%3D%20x_k_%2B_1)
- Let x_o be the populations defined given as 10,000,000 and 800,000 respectively for city and suburbs. We will have a population x_1 as a vector for year 2016 as follows:
![\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_o = x_1](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%20x_o%20%3D%20x_1)
- To get the population in year 2017 we will multiply the migration matrix to the population vector x_1 in 2016 to obtain x_2.
![x_2 = \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right]\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] x_o](https://tex.z-dn.net/?f=x_2%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%20x_o)
- Where,
![x_o = \left[\begin{array}{c}10,000,000\\800,000\end{array}\right]](https://tex.z-dn.net/?f=x_o%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D10%2C000%2C000%5C%5C800%2C000%5Cend%7Barray%7D%5Cright%5D)
- The population in 2017 x_2 would be:
![x_2 = \left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right]\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] \left[\begin{array}{c}10,000,000\\800,000\end{array}\right] \\\\\\x_2 = \left[\begin{array}{c}8,920,800\\1,879,200\end{array}\right]](https://tex.z-dn.net/?f=x_2%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0.94%260.04%5C%5C0.06%260.96%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D10%2C000%2C000%5C%5C800%2C000%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5Cx_2%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%2C920%2C800%5C%5C1%2C879%2C200%5Cend%7Barray%7D%5Cright%5D)
The first one is (1,10)
The second one is (1,5)
The third one is (0.5,10)
I hope this helps
The percentage of change from 45 to 56 is 24.4%
To find this, we first need to find the difference between our two values. This difference would be 11.
The next step would be to divide 11 by 45. We then move the decimal point over two spots to the right which gives us our answer.