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3 years ago

If 4x-3y=12,what is x in the terms of y

Mathematics

2 answers:

Vlada [557]3 years ago

8 0

4x-3y=12 for x
4x=12+3y
x=12+3y/4

4x-3y=12 for y
-3y=12-4x
y=12-4x/-3

faust18 [17]3 years ago

7 0

X= 3/4Y + 3 you have to get X by itself & simplify .

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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.

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

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

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

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