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3 years ago

Please help lol I have to get this done and not very good with math

Mathematics

1 answer:

Annette [7]3 years ago

7 0

B represents a function......................

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The rate of discount, R, can be determined using the formula, where P is the regular price of an item and D is the discount or a

astraxan [27]

Answer:

B.0.31

Step-by-step explanation:

% of change = amount of change/original price

R=(P-D)÷P

39-27=12

12÷39=.307

≈.31

8 0

3 years ago

I just need help set up

Vilka [71]

Answer:

Step-by-step explanation:

Set up equations (L= Lisa; M = Maria; A = April; J = Jim)

L = 12M

A = 2J

Substitute x for Maria and 6x for Jim

L = 12x

A = 2(6x) = 12x

So, you get:

Maria = x

Jim = 6x

Lisa = 12x

April = 12x

5 0

4 years ago

the ratio of sadia's age to her father's age is 3:6. The sum of their age is 96 .What is sadia's age

max2010maxim [7]

We have,

$a:b=3:6,a+b=96$

Introduce variable $x$ such that $a=3x,b=6x$

The sum $a+b=96$ is therefore $9x=96\implies x=10.\overline{6}$

So,

$a=3\cdot10.\overline{6}=\boxed{32}$ (sadia's age)

$b=6\cdot10.\overline{6}=\boxed{64}$ (father's age)

Hope this helps :)

4 0

3 years ago

Twice a number,x, divided. y 6 is 42

Sonbull [250]

x to the second power over 6=42. X=252

5 0

3 years ago

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