All categories

3 years ago

Find "x"

1 answer:

4 0

1. −2x−4=20+6x
Step 1: Simplify both sides of the equation.
−2x−4=20+6x
−2x+−4=20+6x
−2x−4=6x+20
Step 2: Subtract 6x from both sides.
−2x−4−6x=6x+20−6x
−8x−4=20
Step 3: Add 4 to both sides.
−8x−4+4=20+4
−8x=24
Step 4: Divide both sides by -8.
−8x/−8=24/−8
x=−3
Answer:
x=−3

3. (x/4)*(4)=(9)*(4)
x=36
Answer:
x=36

4.
2(5x−1)=3x+5
Step 1: Simplify both sides of the equation.
2(5x−1)=3x+5
(2)(5x)+(2)(−1)=3x+5(Distribute)
10x+−2=3x+5
10x−2=3x+5
Step 2: Subtract 3x from both sides.
10x−2−3x=3x+5−3x
7x−2=5
Step 3: Add 2 to both sides.
7x−2+2=5+2
7x=7
Step 4: Divide both sides by 7.
7x/7=7/7
x=1
Answer:
x=1

5.
8−2(3x−4)+2x=5(x+6)+4
Step 1: Simplify both sides of the equation.
8−2(3x−4)+2x=5(x+6)+4
8+(−2)(3x)+(−2)(−4)+2x=(5)(x)+(5)(6)+4(Distribute)
8+−6x+8+2x=5x+30+4
(−6x+2x)+(8+8)=(5x)+(30+4)(Combine Like Terms)
−4x+16=5x+34
−4x+16=5x+34
Step 2: Subtract 5x from both sides.
−4x+16−5x=5x+34−5x
−9x+16=34
Step 3: Subtract 16 from both sides.
−9x+16−16=34−16
−9x=18
Step 4: Divide both sides by -9.
−9x/−9=18/−9
x=−2
Answer:
x=−2

You might be interested in

Y^2 − x^2 = 100 is that a function or no?

Ugo [173]

Answer:

No, it is not an equation

Step-by-step explanation:

See attached graph:

7 0

3 years ago

Read 2 more answers

A random variable X has a gamma density function with parameters α= 8 and β = 2.

DerKrebs [107]

I know you said "without making any assumptions," but this one is pretty important. Assuming you mean $\alpha,\beta$ are shape/rate parameters (as opposed to shape/scale), the PDF of $X$ is

$f_X(x) = \dfrac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\beta x} = \dfrac{2^8}{\Gamma(8)} x^7 e^{-2x}$

if $x>0$ , and 0 otherwise.

The MGF of $X$ is given by

$\displaystyle M_X(t) = \Bbb E\left[e^{tX}\right] = \int_{-\infty}^\infty e^{tx} f_X(x) \, dx = \frac{2^8}{\Gamma(8)} \int_0^\infty x^7 e^{(t-2) x} \, dx$

Note that the integral converges only when $t$ .

Define

$I_n = \displaystyle \int_0^\infty x^n e^{(t-2)x} \, dx$

Integrate by parts, with

$u = x^n \implies du = nx^{n-1} \, dx$

$dv = e^{(t-2)x} \, dx \implies v = \dfrac1{t-2} e^{(t-2)x}$

so that

$\displaystyle I_n = uv\bigg|_{x=0}^{x\to\infty} - \int_0^\infty v\,du = -\frac n{t-2} \int_0^\infty x^{n-1} e^{(t-2)x} \, dx = -\frac n{t-2} I_{n-1}$

Note that

$I_0 = \displaystyle \int_0^\infty e^{(t-2)}x \, dx = \frac1{t-2} e^{(t-2)x} \bigg|_{x=0}^{x\to\infty} = -\frac1{t-2}$

By substitution, we have

$I_n = -\dfrac n{t-2} I_{n-1} = (-1)^2 \dfrac{n(n-1)}{(t-2)^2} I_{n-2} = (-1)^3 \dfrac{n(n-1)(n-2)}{(t-2)^3} I_{n-3}$

and so on, down to

$I_n = (-1)^n \dfrac{n!}{(t-2)^n} I_0 = (-1)^{n+1} \dfrac{n!}{(t-2)^{n+1}}$

The integral of interest then evaluates to

$\displaystyle I_7 = \int_0^\infty x^7 e^{(t-2) x} \, dx = (-1)^8 \frac{7!}{(t-2)^8} = \dfrac{\Gamma(8)}{(t-2)^8}$

so the MGF is

$\displaystyle M_X(t) = \frac{2^8}{\Gamma(8)} I_7 = \dfrac{2^8}{(t-2)^8} = \left(\dfrac2{t-2}\right)^8 = \boxed{\dfrac1{\left(1-\frac t2\right)^8}}$

The first moment/expectation is given by the first derivative of $M_X(t)$ at $t=0$ .

$\Bbb E[X] = M_x'(0) = \dfrac{8\times\frac12}{\left(1-\frac t2\right)^9}\bigg|_{t=0} = \boxed{4}$

Variance is defined by

$\Bbb V[X] = \Bbb E\left[(X - \Bbb E[X])^2\right] = \Bbb E[X^2] - \Bbb E[X]^2$

The second moment is given by the second derivative of the MGF at $t=0$ .

$\Bbb E[X^2] = M_x''(0) = \dfrac{8\times9\times\frac1{2^2}}{\left(1-\frac t2\right)^{10}} = 18$

Then the variance is

$\Bbb V[X] = 18 - 4^2 = \boxed{2}$

Note that the power series expansion of the MGF is rather easy to find. Its Maclaurin series is

$M_X(t) = \displaystyle \sum_{k=0}^\infty \dfrac{M_X^{(k)}(0)}{k!} t^k$

where $M_X^{(k)}(0)$ is the $k$ -derivative of the MGF evaluated at $t=0$ . This is also the $k$ -th moment of $X$ .

Recall that for $|t|$ ,

$\displaystyle \frac1{1-t} = \sum_{k=0}^\infty t^k$

By differentiating both sides 7 times, we get

$\displaystyle \frac{7!}{(1-t)^8} = \sum_{k=0}^\infty (k+1)(k+2)\cdots(k+7) t^k \implies \displaystyle \frac1{\left(1-\frac t2\right)^8} = \sum_{k=0}^\infty \frac{(k+7)!}{k!\,7!\,2^k} t^k$