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3 years ago

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Which set of ordered pairs (x, y) could represent a linear function of x?? A ((-2,8), (0,4) , (2, 3) , (4,2)) B ((1,2), (1,3),(1

,4), (1,5)) C ((-2,7) , (0,12),(2,17),(4,22)) D ((3,5), 4,7) , (3, 9) , (5,11))

1 answer:

Harlamova29_29 [7]3 years ago

4 0

Answer:

A linear function is when it goes up the same amount each time. The first one does because it increases by 2 each time, so A

Step-by-step explanation:

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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.

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

b)

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

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

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

$= \dfrac{\dfrac{3}{8}}{\dfrac{5}{8}}$

$= \dfrac{3}{8}}\times {\dfrac{8}{5}}$

$= \dfrac{3}{5}}$

= 0.6

(c) The required probability is:

$P(y_2 \ge 0.75 \ y_1 = 0.50) = \dfrac{P(y_2 \ge 0.75 . y_1 =0.50)}{P(y_1 = 0.50)}$

$= \dfrac{\int \limits ^1_{0.75} (y_2 +0.50) \ dy_2}{(0.50 + \dfrac{1}{2})}$

$= \dfrac{0.34375}{1}$

= 0.34375

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3 years ago

Type this equation in standard form. 3x = 6y + 8

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Standard form : Ax + By = C

3x = 6y + 8...subtract 6y from both sides
3x - 6y = 8 <===

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3 years ago

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While there is no drop-down menu options for me to choose. I will show you how to solve this.

$\mathrm{Subtract\:the\:numbers:}\:-16-2=-18 \ \textgreater \ 4k-6=-2k-18$

$\mathrm{Add\:}6\mathrm{\:to\:both\:sides} \ \textgreater \ 4k-6+6=-2k-18+6 \ \textgreater \ Simplify$

$4k=-2k-12 \ \textgreater \ \mathrm{Add\:}2k\mathrm{\:to\:both\:sides} \ \textgreater \ 4k+2k=-2k-12+2k$

Simplify again.

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3 years ago

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Step-by-step explanation:

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3 years ago