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Citrus2011 [14]

2 years ago

8

Solve for z. -4z+1 = bz+c

1 answer:

blagie [28]2 years ago

3 0

Answer:

$\large\boxed{z=\dfrac{1-c}{4+b}}$

Step-by-step explanation:

$-4z+1=bz+c\qquad\text{subtract 1 from both sides}\\\\-4z=bz+c-1\qquad\text{subtract}\ bz\ \text{from both sides}\\\\-4z-bz=c-1\qquad\text{change the signs}\\\\4z+bz=1-c\qquad\text{distribute}\\\\(4+b)z=1-c\qquad\text{divide both sides by}\ (4+b)\neq0\\\\z=\dfrac{1-c}{4+b}$

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Answer:

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But we need to calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_I}{n}$

And replacing we got:

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And for the sample variance we have:

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And thi is the best estimator for the population variance since is an unbiased estimator od the population variance $\sigma^2$

$E(s^2) = \sigma^2$

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:

$s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

But we need to calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_I}{n}$

And replacing we got:

$\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072$

And for the sample variance we have:

$s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003$

And thi is the best estimator for the population variance since is an unbiased estimator od the population variance $\sigma^2$

$E(s^2) = \sigma^2$

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