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3 years ago

Write an answer in the form of rooted ax+b=x+c

1 answer:

5 0

Lets solve for a
ax+b=x+c
step 1:add -b to both sides
ax+b-b=c+x+-b
step 2:divide both sides
ax=-b+c+x
ax/x=-b+c+x/x
a=-b+c+x/x
answer:
a=-b+c+x/x

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Suppose a certain type of fertilizer has an expected yield per acre of mu 1 with variance sigma 2, whereas the expected yield fo

mart [117]

Answer:

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:

$S^2_p= \frac{(n_1 -1) S^2_1 +(n_2 -1) S^2_2}{n_1 +n_2 -2}$

And we want to proof that $E(S^2_p)= \sigma^2$

So we can begin with this:

$E(S^2_p)= E(\frac{(n_1 -1) S^2_1 +(n_2 -1) S^2_2}{n_1 +n_2 -2})$

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

$E(S^2_p)= \frac{(n_1 -1) E(S^2_1) +(n_2 -1) E(S^2_2)}{n_1 +n_2 -2}$

And we know that the expected value for the estimator of the variance s is $\sigma$ , or in other way $E(s) = \sigma$ so if we apply this property here we have:

$E(S^2_p)= \frac{(n_1 -1 )\sigma^2_1 +(n_2 -1) \sigma^2_2}{n_1 +n_2 -2}$

And we know that $\sigma^2_1 = \sigma^2_2 = \sigma^2$ so using this we can take common factor like this:

$E(S^2_p)= \frac{(n_1 -1) +(n_2 -1)}{n_1 +n_2 -2} \sigma^2 =\sigma^2$

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.

8 0

3 years ago

Use intercepts to graph the linear equation -1/2x+y=-4

ololo11 [35]

Do you need to switch it around or do something else?

6 0

3 years ago

Read 2 more answers

All ratios that are equivalent to14 to 20

gulaghasi [49]

Answer:

7/10, 14/20, 21/30, 28/40, 35/50, 42/60, 49/70, 56/80, 63/90, 70/100, 77/110, 84/120, 91/130, 98/140, 105/150, 112/160, 119/170, 126/180, 133/190, 140/200,

Step-by-step explanation:

8 0

3 years ago

Tue 05/28<br> 11. Find the common ratio of the sequence.<br> 2, -10,50.-250,<br> 0-5<br> .

Montano1993 [528]

Answer:

-5

Step-by-step explanation:

Geometric sequences have common ratios.

You can see if a sequence is geometric by computing:

$\frac{\text{term}}{\text{previous term}}$ .

If you get the same result per consecutive pair, then there is a common ratio and the sequence is geometric.

Let's see:

$\frac{-250}{50}=\frac{50}{-10}=\frac{-10}{2}$

So if this equations holds (if all 3 fractions equal the same thing), then is geometric and we have also have computed the common ratio.

The first fraction results in -250/50=-5.

The second fraction results in 50/-10=-5.

The third fraction results in -10/2=-5.

All three fractions are equal and so the equation holds so it is geometric and the common ratio is -5.

5 0

3 years ago

Find h(-1) if h(x) =- 3|3x – 6|– 3?

Dahasolnce [82]

Answer:

Step-by-step explanation:

h(-1) = - 3|3(-1) - 6| - 3

h(-1) = - 3|-3 - 6| - 3

h(-1) = - 3|-9| - 3

h(-1) = - 3(9) - 3

h(-1) = - 27 - 3

h(-1) = - 30

6 0

3 years ago