Answer:
3
Step-by-step explanation:
look at where the points at
A.
that is changge per year
you do change/years=11150000/10=1115000
the mean change ws 1115000 visitors per year
B. it can be negative if the change was massivly negative to offset the 10800000 visitor increase
I believe I can fly, I believe I can’t touch the sky.
Answer:
c
Step-by-step explanation:
the formula is
When we plug that in we get ![\frac{-3-2}{6- (-4)} = \frac{-5}{10} = -\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B-3-2%7D%7B6-%20%28-4%29%7D%20%3D%20%5Cfrac%7B-5%7D%7B10%7D%20%3D%20-%5Cfrac%7B1%7D%7B2%7D)
Hope this helps!
Answer:
The 99% confidence interval for the population mean is 22.96 to 26.64
Step-by-step explanation:
Consider the provided information,
A sample of 49 customers. Assume a population standard deviation of $5. If the sample mean is $24.80,
The confidence interval if 99%.
Thus, 1-α=0.99
α=0.01
Now we need to determine ![z_{\frac{\alpha}{2}}=z_{0.005}](https://tex.z-dn.net/?f=z_%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%3Dz_%7B0.005%7D)
Now by using z score table we find that ![z_{\frac{\alpha}{2}}=2.58](https://tex.z-dn.net/?f=z_%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%3D2.58)
The boundaries of the confidence interval are:
![\mu-z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80-2.58\times \frac{5}{\sqrt{49}}=22.96\\\mu+z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80+2.58\times \frac{5}{\sqrt{49}}=26.64](https://tex.z-dn.net/?f=%5Cmu-z_%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ctimes%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%20%7D%5C%5C24.80-2.58%5Ctimes%20%5Cfrac%7B5%7D%7B%5Csqrt%7B49%7D%7D%3D22.96%5C%5C%5Cmu%2Bz_%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ctimes%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%20%7D%5C%5C24.80%2B2.58%5Ctimes%20%5Cfrac%7B5%7D%7B%5Csqrt%7B49%7D%7D%3D26.64)
Hence, the 99% confidence interval for the population mean is 22.96 to 26.64