Answer:
The minimum sample size is n = 75 so that the desired margin of error is 5 or less.
Step-by-step explanation:
We are given the following in the question:
Population variance = 484
Standard deviation =
![\sigma^2 = 484\\\sigma =\sqrt{484} = 22](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D%20484%5C%5C%5Csigma%20%3D%5Csqrt%7B484%7D%20%3D%2022)
Confidence level = 0.95
Significance level = 0.05
Margin of error = 5
Formula:
Margin of error =
![E = z\times \dfrac{\sigma}{\sqrt{n}}\\\\n = (\dfrac{z\times \sigma}{E})^2](https://tex.z-dn.net/?f=E%20%3D%20z%5Ctimes%20%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C%5C%5Cn%20%3D%20%28%5Cdfrac%7Bz%5Ctimes%20%5Csigma%7D%7BE%7D%29%5E2)
![z_{critical}\text{ at}~\alpha_{0.05} = 1.96](https://tex.z-dn.net/?f=z_%7Bcritical%7D%5Ctext%7B%20at%7D~%5Calpha_%7B0.05%7D%20%3D%201.96)
Putting values, we get
![E\leq 5\\\\1.96\times \dfrac{22}{\sqrt{n}} \leq 5\\\\\sqrt{n} \geq 1.96\times \dfrac{22}{5}\\\\n \geq (1.96\times \dfrac{22}{5})^2\\\\n \geq 74.373](https://tex.z-dn.net/?f=E%5Cleq%205%5C%5C%5C%5C1.96%5Ctimes%20%5Cdfrac%7B22%7D%7B%5Csqrt%7Bn%7D%7D%20%5Cleq%205%5C%5C%5C%5C%5Csqrt%7Bn%7D%20%5Cgeq%201.96%5Ctimes%20%5Cdfrac%7B22%7D%7B5%7D%5C%5C%5C%5Cn%20%5Cgeq%20%281.96%5Ctimes%20%5Cdfrac%7B22%7D%7B5%7D%29%5E2%5C%5C%5C%5Cn%20%5Cgeq%2074.373)
Thus, the minimum sample size is n = 75 so that the desired margin of error is 5 or less.
Answer:
N= -8
Step-by-step explanation:
a
Can i get a kiss??? x#*||~~
The smallest y can be is y = -1. The largest it can be is y = 1. The value of y can be anything in between. The value of y is a real number.
Answer: Choice D) All real numbers between -1 and 1 including -1 and 1Note: we can write that as
![-1 \le y \le 1](https://tex.z-dn.net/?f=-1%20%5Cle%20y%20%5Cle%201%20)
which in interval notation would look like
![[-1, 1]](https://tex.z-dn.net/?f=%5B-1%2C%201%5D%20)
. The square brackets tell us to include the endpoints.