The rule for the translation is like a function. If X=2 and Y=-1, we just substitute those values in the rule of (x,y) => (x-1, y+3)

(2,-1)=>(1, 2) are the new coordinates. You went 1 to the left on the X axis and 3 up on the Y axis.

8 0

3 years ago

) Consider the following probability density function of the random variable X. f(x) = k(2x + 3x2 ), 0 ≤ x ≤ 2. (i) Determine th

julia-pushkina [17]

I assume $f(x)=0$ otherwise. If $f(x)$ is indeed a proper PDF, then its integral over the support of $X$ is 1.

$\displaystyle \int_{-\infty}^\infty f(x) \, dx = k \int_0^2 (2x + 3x^2) \, dx = 1$

Compute the integral.

$\displaystyle \int_0^2 (2x + 3x^2) \, dx = (x^2 + x^3)\bigg|_{x=0}^{x=2} = (2^2 + 2^3) - (0^2 + 0^3) = 12$

Then

$12k = 1 \implies \boxed{k=\dfrac1{12}}$

8 0

3 years ago

Factor by grouping 3n^3 - 12n^2 + 2n - 8 (show the work pls..)

densk [106]

Answer:

The answer will be ( n - 4 ) ( 3n² + 2 ).

Step-by-step explanation:

1) Factor out common terms in the first two terms, then in the last two terms.

${3n}^{2} (n - 4) + 2(n - 4)$

2) Factor out the common term n - 4.

$(n - 4)( {3n}^{2} + 2)$

Therefor, the answer is ( n - 4 ) ( 3n² + 2 ).

8 0

3 years ago