I believe it would be under 8 percent since 9 is lower
1. x^2+x-6
2. x^2-8x-48
3. x^2+7x+12
4. x^2+13x+42
5. x^2+17x+72
6. x^2-25
7. x^2-9
8. x^2+20x+100
9. x^2+10x-24
10. x^2+x-12
Answer:
B. 0.03110
Step-by-step explanation:
Given
Probability of Hit = 60%
Required
Determine the probability that he misses at 6th throw
Represent Probability of Hit with P
![P = 60\%](https://tex.z-dn.net/?f=P%20%3D%2060%5C%25)
Convert to decimal
![P = 0,6](https://tex.z-dn.net/?f=P%20%3D%200%2C6)
Next; Determine the Probability of Miss (q)
Opposite probabilities add up to 1;
So,
![p + q = 1](https://tex.z-dn.net/?f=p%20%2B%20q%20%3D%201)
![q = 1 - p](https://tex.z-dn.net/?f=q%20%3D%201%20-%20p)
Substitute 0.6 for p
![q = 1 - 0.6](https://tex.z-dn.net/?f=q%20%3D%201%20-%200.6)
![q = 0.4](https://tex.z-dn.net/?f=q%20%3D%200.4)
Next,is to determine the required probability;
Since, he's expected to miss the 6th throw, the probability is:
![Probability = p^5 * q](https://tex.z-dn.net/?f=Probability%20%3D%20p%5E5%20%2A%20q)
![Probability = 0.6^5 * 0.4](https://tex.z-dn.net/?f=Probability%20%3D%200.6%5E5%20%2A%200.4)
![Probability = 0.031104](https://tex.z-dn.net/?f=Probability%20%3D%200.031104)
Hence;
<em>Option B answers the question</em>
Answer:
The degrees of freedom are given by;
![df =n-1= 5-1=4](https://tex.z-dn.net/?f=%20df%20%3Dn-1%3D%205-1%3D4)
The significance level is 0.1 so then the critical value would be given by:
![F_{cric}= 7.779](https://tex.z-dn.net/?f=%20F_%7Bcric%7D%3D%207.779)
If the calculated value is higher than this value we can reject the null hypothesis that the arrivals are uniformly distributed over weekdays
Step-by-step explanation:
For this case we have the following observed values:
Mon 25 Tue 22 Wed 19 Thu 18 Fri 16 Total 100
For this case the expected values for each day are assumed:
![E_i = \frac{100}{5}= 20](https://tex.z-dn.net/?f=%20E_i%20%3D%20%5Cfrac%7B100%7D%7B5%7D%3D%2020)
The statsitic would be given by:
![\chi^2 = \sum_{i=1}^n \frac{(O_i-E_i)^2}{E_i}](https://tex.z-dn.net/?f=%20%5Cchi%5E2%20%3D%20%5Csum_%7Bi%3D1%7D%5En%20%5Cfrac%7B%28O_i-E_i%29%5E2%7D%7BE_i%7D)
Where O represent the observed values and E the expected values
The degrees of freedom are given by;
![df =n-1= 5-1=4](https://tex.z-dn.net/?f=%20df%20%3Dn-1%3D%205-1%3D4)
The significance level is 0.1 so then the critical value would be given by:
![F_{cric}= 7.779](https://tex.z-dn.net/?f=%20F_%7Bcric%7D%3D%207.779)
If the calculated value is higher than this value we can reject the null hypothesis that the arrivals are uniformly distributed over weekdays
Answer:
0.35, 0.40
Step-by-step explanation: