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3 years ago

Write a equation for the graph of a linear function that had a slope -1/3and y-intercept

Mathematics

1 answer:

kiruha [24]3 years ago

4 0

Answer:

Round answers to three decimal places.

What are the total number of ways to choose ten mentors from the group of seniors, juniors, and sophomores?

What is the probability that the mentors will be five seniors, three juniors, and two sophomores?

Step-by-step explanation:

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Which lists all the integer solutions of the inequality |x| < 3?

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The answer is c to your problemo

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4 years ago

Solve this using the elimination method<br> 4x - 3y = 22<br> 2x + 8y = 30

Inga [223]

4x-3y=22
-4x-16y=-60
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3 years ago

Read 2 more answers

Please help with 3 math questions ( 50 points )

Mariulka [41]

Answer:

Step-by-step explanation:

1. Reflection over the x-axis: (x, y) → (x, -y)

ΔABC with A(5, 1), B(-2, 6) C(-3, 4), after a reflection along the x-axis becomes A'(5, -1), B'(-2, -6) C'(-3, -4)

2. Counterclockwise rotation of 90° about the origin: (x, y) → (-y, x)

ΔXYZ with X(1, 2), Y(9, 7) Z(10, 1), after a counterclockwise rotation of 90° about the origin the coordinates are X'(-2,1) Y'(-7,9) Z'(-1,10)

Parallelogram WXYZ with W(2, -3), X(-5, -1), Y(-1, 2), Z(6, 0), has the image after a translation of (x, y) → (x – 5, y + 2) as

W'(2-5, -3+2), X'(-5-5, -1+2), Y'(-1-5, 2+2), Z'(6-5, 0+2)

W'(-3, -1), X'(-10, 1), Y'(-6, 4), Z'(1, 2)

The coordinates of W'(-3, -1)

See Graphs...

7 0

3 years ago

Let Y1 and Y2 denote the proportion of time during which employees I and II actually performed their assigned tasks during a wor

Lemur [1.5K]

Answer:

Step-by-step explanation:

From the information given:

The joint density of $y_1$ and $y_2$ is given by:

$f_{(y_1,y_2)} \left \{ {{y_1+y_2, \ \ 0\ \le \ y_1 \ \le 1 , \ \ 0 \ \ \le y_2 \ \ \le 1} \atop {0, \ \ \ elsewhere \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \right.$

a)To find the marginal density of $y_1$ .

$f_{y_1} (y_1) = \int \limits ^{\infty}_{-\infty} f_{y_1,y_2} (y_1 >y_2) \ dy_2$

$=\int \limits ^{1}_{0}(y_1+y_2)\ dy_2$

$=\int \limits ^{1}_{0} \ \ y_1dy_2+ \int \limits ^{1}_{0} \ y_2 dy_2$

$= y_1 \ \int \limits ^{1}_{0} dy_2+ \int \limits ^{1}_{0} \ y_2 dy_2$

$= y_1[y_2]^1_0 + \bigg [ \dfrac{y_2^2}{2}\bigg]^1_0$

$= y_1 [1] + [\dfrac{1}{2}]$

$= y_1 + \dfrac{1}{2}$

i.e.

$f_{(y_1}(y_1)}= \left \{ {{y_1+\dfrac{1}{2}, \ \ 0\ \ \le \ y_1 \ \le , \ 1} \atop {0, \ \ \ elsewhere \ \\ \ \ \ \ \ \ \ \ } \right.$

The marginal density of $y_2$ is:

$f_{y_1} (y_2) = \int \limits ^{\infty}_{-\infty} fy_1y_1(y_1-y_2) dy_1$

$= \int \limits ^1_0 \ y_1 dy_1 + y_2 \int \limits ^1_0 dy_1$

$=\bigg[ \dfrac{y_1^2}{2} \bigg]^1_0 + y_2 [y_1]^1_0$

$= [ \dfrac{1}{2}] + y_2 [1]$

$= y_2 + \dfrac{1}{2}$

i.e.

$f_{(y_1}(y_2)}= \left \{ {{y_2+\dfrac{1}{2}, \ \ 0\ \ \le \ y_1 \ \le , \ 1} \atop {0, \ \ \ elsewhere \ \\ \ \ \ \ \ \ \ \ } \right.$

$P\bigg[y_1 \ge \dfrac{1}{2}\bigg |y_2 \ge \dfrac{1}{2} \bigg] = \dfrac{P\bigg [y_1 \ge \dfrac{1}{2} . y_2 \ge\dfrac{1}{2} \bigg]}{P\bigg[ y_2 \ge \dfrac{1}{2}\bigg]}$

$= \dfrac{\int \limits ^1_{\frac{1}{2}} \int \limits ^1_{\frac{1}{2}} f_{y_1,y_1(y_1-y_2) dy_1dy_2}}{\int \limits ^1_{\frac{1}{2}} fy_1 (y_2) \ dy_2}$

$= \dfrac{\int \limits ^1_{\frac{1}{2}} \int \limits ^1_{\frac{1}{2}} (y_1+y_2) \ dy_1 dy_2}{\int \limits ^1_{\frac{1}{2}} (y_2 + \dfrac{1}{2}) \ dy_2}$