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3 years ago

The joint F.D.P of a bivariate VA (X, Y) is given

Mathematics

1 answer:

Nat2105 [25]3 years ago

8 0

Answer:

Step-by-step explanation:

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For what values of y:

Whitepunk [10]

Answer:

$y > \frac{3}{17}$

Step-by-step explanation:

To solve this problem, simply set the binomial, $5y - 1$ as greater than $\frac{3y - 1}{4}$ , and solve for the value of y as you would solve an equation.

Thus:

$5y - 1 > \frac{3y - 1}{4}$

Multiply both sides by 4

$4(5y - 1) > \frac{3y - 1}{4}*4$

$4*5y - 4*1 > 3y - 1$

$20y - 4 > 3y - 1$

Subtract 3y from both sides

$20y - 4 - 3y > 3y - 1 - 3y$

$17y - 4 > -1$

Add 4 to both sides

$17y - 4 + 4 > -1 + 4$

$17y > 3$

Divide both sides by 17

$\frac{17y}{17} > \frac{3}{17}$

$y > \frac{3}{17}$

Therefore, the value of y that will make the binomial $5y - 1$ greater than $\frac{3y - 1}{4}$ is $y > \frac{3}{17}$ .

7 0

3 years ago

Sam read 3 times as many pages in history textbook on Monday night as he read on Tuesday night altogether he read 56 pages how m

AleksAgata [21]

Monday he read 42 pages
Tuesday he read 14 pages

4 0

3 years ago

What is the solution of the system? 10x-2y=24 6x+2y=8 enter your in the box as an ordered pair.

Ivahew [28]

I got a decimal for the 10x-2y=24 I’m trying to figure it out if it’s correct or not

8 0

3 years ago

Factor this expression completely, then place the factors in the proper location on the grid. a^3y + 1

ivolga24 [154]

Answer:

$a^{3y} + 1 = (a^{y}+1 )^{3} - 3a^y(a^{y}+1)\\\\$

Step-by-step explanation:

We are to factorize the expression $a^{3y} + 1$ completely. To do this, we will apply the expression below;

The expression can be rewritten as $a^{3y} + 1^{3}$

To factorize the expression, we need to first factorize $(a^{y}+1 )^{3}$ first

$(a^{y}+1 )^{3} =(a^{y}+1 )(a^{y}+1 )^{2}\\= (a^{y}+1 )((a^y)^{2} } + 2a^{y} +1)\\= (a^y)^{3} +2(a^y)^{2}+a^y+( a^y)^{2}+2a^y+1\\(a^{y}+1 )^{3} = ((a^y)^{3} + 1) +2(a^y)^{2}+a^y+( a^y)^{2}+2a^y\\(a^{y}+1 )^{3} = ((a^y)^{3} + 1) +3(a^y)^{2}+3a^y\\$