Answer: E. All of the above statements are true
Step-by-step explanation:
The mean of sampling distribution of the mean is simply the population mean from which scores were being sampled. This implies that when population has a mean μ, it follows that mean of sampling distribution of mean will also be μ.
It should also be noted that the distribution's shape is symmetric and normal and there are no outliers from its overall pattern.
The statements about the sampling distribution of the sample mean, x-bar that are true include:
• The sampling distribution is normal regardless of the shape of the population distribution, as long as the sample size, n, is large enough.
• The sampling distribution is normal regardless of the sample size, as long as the population distribution is normal. • The sampling distribution's mean is the same as the population mean.
• The sampling distribution's standard deviation is smaller than the population standard deviation.
Therefore, option E is the correct answer as all the options are true.
Answer:
Step-by-step explanation:
![\sf 3a^5-18a^3+6a^2\\\\HCF = 3a^2\\\\Take \ 3a^2 \ common\\\\= 3a^2(a^3-6a+2)\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Csf%203a%5E5-18a%5E3%2B6a%5E2%5C%5C%5C%5CHCF%20%3D%203a%5E2%5C%5C%5C%5CTake%20%5C%203a%5E2%20%5C%20common%5C%5C%5C%5C%3D%203a%5E2%28a%5E3-6a%2B2%29%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
Answer:
(b) 5k+1
Step-by-step explanation:
https://www.symbolab.com/solver/step-by-step/simplify%20%5Cfrac%7B30k%5E%7B2%7D%2B6k%7D%7B5k%2Bk%7D
Answer:
13x-4
Step-by-step explanation:
