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3 years ago

A line has a slope of -1/3 and includes the points (q, 2) and (6, 3). What is the value of q?

Mathematics

1 answer:

elena55 [62]3 years ago

3 0

Answer:

Step-by-step explanation:

Given the knowledge that the slope is -1/3 and includes the point (6,3), you can subtract one from the y value and add three to the x value, resulting in (9,2). Q is equal to 9.

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(3,10) and (6,13)Put in slope-intercept form

spayn [35]

ANSWER

y = x + 7

EXPLANATION

We want to use the two points to form a linear equation in slope-intercept form.

That is in the form:

y = mx + c

To do this, we first find the slope by using formula:

$m\text{ = }\frac{y_2-y_1}{x_2-x_1}$

where (x1, y1) = (3, 10)

(x2, y2) = (6, 13)

So, we have that:

$\begin{gathered} m\text{ = }\frac{13\text{ - 10}}{6\text{ - 3}} \\ m\text{ = }\frac{3}{3} \\ m\text{ = 1} \end{gathered}$

The slope, m, is 1.

Now, we use the point-slope formula to find the equation. That is:

$\begin{gathered} y-y_1=m(x-x_1) \\ \Rightarrow\text{ y - 10 = 1(x - 3)} \\ y\text{ - 10 = x - 3} \\ =>\text{ y = x - 3 + 10} \\ y\text{ = x + 7} \end{gathered}$

That is the answer.

4 0

1 year ago

William has a lemonade stand .Yesterday he made $17.55in lemonade sales and today he made $14.02 sales. How much does he have no

Zolol [24]

Answer: 31.57

Step-by-step explanation: 17.55 + 14.02 =31.57

5 0

3 years ago

2) X and Y are jointly continuous with joint pdf

Nady [450]

From what I gather from your latest comments, the PDF is given to be

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0\le x,y \le1\\0&\text{otherwise}\end{cases}$

and in particular, f(x, y) = cxy over the unit square [0, 1]², meaning for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (As opposed to the unbounded domain, x ≤ 0 *and* y ≤ 1.)

(a) Find c such that f is a proper density function. This would require

$\displaystyle\int_0^1\int_0^1 cxy\,\mathrm dx\,\mathrm dy=c\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\frac c{2^2}=1\implies \boxed{c=4}$

(b) Get the marginal density of X by integrating the joint density with respect to y :

$f_X(x)=\displaystyle\int_0^1 4xy\,\mathrm dy=(2xy^2)\bigg|_{y=0}^{y=1}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

$f_Y(y)=\displaystyle\int_0^14xy\,\mathrm dx=\begin{cases}2y&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}$

(d) The conditional distribution of X given Y can obtained by dividing the joint density by the marginal density of Y (which follows directly from the definition of conditional probability):

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

(e) From the definition of expectation:

$E[X]=\displaystyle\int_0^1\int_0^1 x\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\boxed{\frac23}$

$E[Y]=\displaystyle\int_0^1\int_0^1 y\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac23}$

$E[XY]=\displaystyle\int_0^1\int_0^1xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac49}$

(f) Note that the density of X | Y in part (d) identical to the marginal density of X found in (b), so yes, X and Y are indeed independent.

The result in (e) agrees with this conclusion, since E[XY] = E[X] E[Y] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)

8 0

3 years ago

Plz help me answer this question

olya-2409 [2.1K]

A. 5 invitations
b. 15 invitations
Comment if you need me to explain it.

8 0

3 years ago

The length of a rectangle is twice its width. The perimeter of the rectangle is 126ft.

umka21 [38]

Alright, so let's make the length x and the width y. 2y=x, and xy=area. 2x+2y=perimeter. Plugging 2y=x into the perimeter equation, we get that 3x=perimeter=126, and x=126/3=42

Length=42ft, width=21ft

8 0

3 years ago