Well it all depends on the object you are using but in this case lets say remote control cars. You can have an example like "we drove the car around the building twice. the first time it was -8 than how fast it was supposed to go. the second time it was 9. In which round was the car faster?" I dont think i helped very much but i hope i did
There will be eight marbles on the 9th line
Answer:
- P(≥1 working) = 0.9936
- She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.
Step-by-step explanation:
1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...
... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936
2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.
If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.
This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.
_____
My assessment is that there is significant gain from bringing a backup. (Personally, I might investigate why the probability of failure is so high. I have not had such bad luck with calculators, which makes me wonder if operator error is involved.)
<span>$2,022.92 would be the answer
</span>
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:
![\left[\begin{array}{cc}0.94&0.04\\0.06&0.96\end{array}\right] \left[\begin{array}{c}p_c\\p_s\end{array}\right] = \left[\begin{array}{c}p_c'\\p_s'\end{array}\right]](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%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dp_c%5C%5Cp_s%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dp_c%27%5C%5Cp_s%27%5Cend%7Barray%7D%5Cright%5D)
- 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)
