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2 years ago

F(x) = -3x^2 Find f(-2)

Mathematics

1 answer:

MAXImum [283]2 years ago

6 0

9514 1404 393

Answer:

f(-2) = -12

Step-by-step explanation:

Put -2 where x is and do the arithmetic.

$f(x)=-3x^2\\\\f(-2)=-3(-2)^2=-3\cdot4=\boxed{-12}$

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Answer:

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

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 $\sigma^2$

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

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 $\sigma^2$

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