You have the function v(x) = 8(x − 1), where x <span>is the number of days since opening.
</span><span>On the first day, there will be a ceremony with 32 people in attendance.
Therefore, keeping the information above on mind, the </span><span>function that shows total visitors, including the ceremony is:
f(x)=8(x-1)+32
So, as you can see, the answer is:
</span>
f(x)=8(x-1)+32
For a known standard deviation a confidence level of 99% we may consider Z=<span>2.576.
So, the confidence interval would be calculated by:
</span>
![mean-Z\frac{standard~deviation}{ \sqrt{number~of~students~in~the~group} }](https://tex.z-dn.net/?f=mean-Z%5Cfrac%7Bstandard~deviation%7D%7B%20%5Csqrt%7Bnumber~of~students~in~the~group%7D%20%7D%20)
,
![mean+Z\frac{standard~deviation}{\sqrt{number~of~students~in~the~group} }](https://tex.z-dn.net/?f=mean%2BZ%5Cfrac%7Bstandard~deviation%7D%7B%5Csqrt%7Bnumber~of~students~in~the~group%7D%20%7D%20)
⇔
⇔
![15.3-2.576 \frac{8.5}{ \sqrt{463} }](https://tex.z-dn.net/?f=15.3-2.576%20%5Cfrac%7B8.5%7D%7B%20%5Csqrt%7B463%7D%20%7D%20)
,
![15.3+2.576 \frac{8.5}{ \sqrt{463} }](https://tex.z-dn.net/?f=15.3%2B2.576%20%5Cfrac%7B8.5%7D%7B%20%5Csqrt%7B463%7D%20%7D%20)
⇔
⇔
![15.3-1.018](https://tex.z-dn.net/?f=15.3-1.018%20)
,
![15.3+1.018](https://tex.z-dn.net/?f=15.3%2B1.018%20)
⇔
⇔
![14.282](https://tex.z-dn.net/?f=14.282%20)
,
![16.318](https://tex.z-dn.net/?f=16.318%20)
Considering the given data, with a 99% cofidence, it can be said that the true mean of hours of study in a typical week for first-year college students is between 14.282 and 16.318.
Answer:
mmm idgl
Step-by-step explanation: