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3 years ago

Please answer this with the steps included please y-10 = -5/4x + 30/4 for middle school

1 answer:

7 0

<span>y-10 = -5/4x + 30/4 represents a linear function; we know that because both x and y have the power 1.

We could put this equation into one of several forms.

Suppose we wanted to put it into slope-intercept form. Then </span><span>y-10 = -5/4x + 30/4 becomes y = 10 + 30/4 - (5/4)x, or y = 35/2 - (5/4)x. Thus, this line has the slope -5/4 and the y-intercept (0, 35/2).

If this isn't quite what you wanted, then please share the instructions.

</span>

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Which example models a negative integer A.Students in a classroom B. The age of a parent C. Money lost from a pocket D. the heig

Rufina [12.5K]

Answer:

C- Money lost from a pocket

Step-by-step explanation:

It would be C because something is decreasing ( now a negative) from what they did have.

4 0

2 years ago

Brittany and Ashley can paint a room in 7 hours if they work together. If Ashley were to work by himself, it would take him 2 ho

stich3 [128]

Answer:

Step-by-step explanation:

Brittany paints the room in x hours, completing 1/x of the room per hour.

Ashley paints the room in x+2 hours, completing 1/(x+2) of the room per hour.

Working together, they complete 1/x + 1/(x+2) = ⅐ of the room per hour.

7(x+2) + 7x = x(x+2)

14x + 15 = x^2 + 2x

x^2 - 12x - 15 = 0

x = [12 ±√(12²+60)]/2= 6±√51

6-√51 is an extraneous solution.

x = 6+√51 ≈ 13.14

Brittany paints the room in 12.14 hours

3 0

3 years ago

Let X1 and X2 be independent random variables with mean μand variance σ².

My name is Ann [436]

Answer:

a) $E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

b) $Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

Step-by-step explanation:

For this case we know that we have two random variables:

$X_1 , X_2$ both with mean $\mu = \mu$ and variance $\sigma^2$

And we define the following estimators:

$\hat \theta_1 = \frac{X_1 + X_2}{2}$

$\hat \theta_2 = \frac{X_1 + 3X_2}{4}$

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

$E(\hat \theta_i) = \mu , i = 1,2$