Suppose a certain type of fertilizer has an expected yield per acre of mu 1 with variance sigma 2, whereas the expected yield for a second type of fertilizer is mu 2 with the same variance sigma 2. let s1^2 and s2^2 denote the sample variances of yields based on sample sizes n1 and n2, respectively, of the two fertilizers. show that the pooled (combined) estimator cap sigma 2 = (n1 - 1)s1^2 + (n2 - 1)s2^2/n1 + n2 - 2 is an unbiased estimator of sigma 2.

See the proof below.

Step-by-step explanation:

For this case we just need to apply properties of expected value. We know that the estimator is given by:

And we want to proof that

So we can begin with this:

And we can distribute the expected value into the temrs like this:

And we know that the expected value for the estimator of the variance s is , or in other way so if we apply this property here we have:

And we know that so using this we can take common factor like this:

And then we see that the pooled variance is an unbiased estimator for the population variance when we have two population with the same variance.

a. 3/4 ÷ 1/8 = 6

step-by-step explanation:

in the question, the number line represents 3/4 divided into 6 equal units. each arrow represents a unit and there are a total of six arrows.

the size of each unit is 1/8 as shown.

showing that 1/8 divides 3/4 into 6 divisions:

3/4 ÷ 1/8

= 3/4 x 8/1

= 3/1 x 8/4

= 3 x 2

= 6

hence, the number line represents six divisions given by 3/4 ÷ 1/8.

i used to know but i learned it a long time ago sorry

step-by-step explanation: