1. Zoe stacked the most brownies. She stacked 27 brownies out of the 50 brownies.
2. 15
3. 43 + 3n ≥ 70, so n ≥ 9
4. D. 58x + 13 + 58y
5. The answer would be ≤4
6. 9.94 + 10.01p ≤ 70, so p ≤ 6
7. C 40 89
8. D. x+(x+5)=77
Hope I helped you! Please consider marking my answer the brainliest :)
Answer:
<h2>Adults and Seniors tendo to by more Unlimited Meals tickets</h2>
Step-by-step explanation:
The relation between the park guest and the type to ticket is that Adults and Seniors tendo to by more Unlimited Meals tickets, because they don't play that much in the park, children do.
Therefore, there's a relation that aroun Unlimited Meals tickets and Adults-Seniors costumers: they tend to get these tickest more than children.
Answer:
I think it is 0,2
Step-by-step explanation:
Because you need to substitute the ordered pairs into the linear equation:
so,
2(0)+3(2) = 6
6=6
The two side, so equation are identical, so the correct answer is (0,2).
I hope this helps!
A tenth is 1/10.
5 tenths is 5/10.
5/10 is a fraction that can be made simpler, by dividing by 5, so that the fraction becomes 1/2.
1/2 is a half, so if you have a half a dollar, you have 50 cents.
Hope I helped. :)
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)