Answer:
![s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}](https://tex.z-dn.net/?f=%20s%5E2%20%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20%28X_i%20-%5Cbar%20X%29%5E2%7D%7Bn-1%7D)
But we need to calculate the mean with the following formula:
![\bar X = \frac{\sum_{i=1}^n X_I}{n}](https://tex.z-dn.net/?f=%5Cbar%20X%20%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20X_I%7D%7Bn%7D)
And replacing we got:
![\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072](https://tex.z-dn.net/?f=%5Cbar%20X%20%3D%20%5Cfrac%7B%201.04%2B1.00%2B1.13%2B1.08%2B1.11%7D%7B5%7D%3D%201.072)
And for the sample variance we have:
![s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003](https://tex.z-dn.net/?f=%20s%5E2%20%3D%20%5Cfrac%7B%281.04-1.072%29%5E2%20%2B%281.00-1.072%29%5E2%20%2B%281.13-1.072%29%5E2%20%2B%281.08-1.072%29%5E2%20%2B%281.11-1.072%29%5E2%7D%7B5-1%7D%3D%200.00277%5C%20approx%200.003)
And thi is the best estimator for the population variance since is an unbiased estimator od the population variance ![\sigma^2](https://tex.z-dn.net/?f=%5Csigma%5E2)
![E(s^2) = \sigma^2](https://tex.z-dn.net/?f=%20E%28s%5E2%29%20%3D%20%5Csigma%5E2%20)
Step-by-step explanation:
For this case we have the following data:
1.04,1.00,1.13,1.08,1.11
And in order to estimate the population variance we can use the sample variance formula:
![s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}](https://tex.z-dn.net/?f=%20s%5E2%20%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20%28X_i%20-%5Cbar%20X%29%5E2%7D%7Bn-1%7D)
But we need to calculate the mean with the following formula:
![\bar X = \frac{\sum_{i=1}^n X_I}{n}](https://tex.z-dn.net/?f=%5Cbar%20X%20%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20X_I%7D%7Bn%7D)
And replacing we got:
![\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072](https://tex.z-dn.net/?f=%5Cbar%20X%20%3D%20%5Cfrac%7B%201.04%2B1.00%2B1.13%2B1.08%2B1.11%7D%7B5%7D%3D%201.072)
And for the sample variance we have:
![s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003](https://tex.z-dn.net/?f=%20s%5E2%20%3D%20%5Cfrac%7B%281.04-1.072%29%5E2%20%2B%281.00-1.072%29%5E2%20%2B%281.13-1.072%29%5E2%20%2B%281.08-1.072%29%5E2%20%2B%281.11-1.072%29%5E2%7D%7B5-1%7D%3D%200.00277%5C%20approx%200.003)
And thi is the best estimator for the population variance since is an unbiased estimator od the population variance ![\sigma^2](https://tex.z-dn.net/?f=%5Csigma%5E2)
![E(s^2) = \sigma^2](https://tex.z-dn.net/?f=%20E%28s%5E2%29%20%3D%20%5Csigma%5E2%20)
Answer:
Your answer is option C … (-5, -1, 0)
Step-by-step explanation:
Answer:15252.25π^3
Step-by-step explanation:
Radius=r
Circumference=247 ud
Circumference=2 x π x r
247=2πr
Divide both sides by 2π
247/2π=2πr/2π
123.5π=r
Radius=123.5π
Area of circle=π x r x r
Area of circle=π x 123.5π x 123.5π
Area of circle=15252.25π^3
<h2>Volume = 4.19 inches³</h2>
<u>Step-by-step explanation:</u>
radius = 1 inches
π = 3.14
volume of SPHERE = 4/3 × π × radius³
= 4/3 × 3.14 × 1³
= 4.19 inches³
What is your question that your asking?