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3 years ago

What is the solution? 2/3(x-7)= -2

Mathematics

1 answer:

koban [17]3 years ago

3 0

Answer:

x =4

Step-by-step explanation:

2/3(x-7)= -2

Multiply each side by 3/2 to clear the fractions

3/2 * 2/3(x-7)= -2 *3/2

x-7 = -3

Add 7 to each side

x-7+7 = -3+7

x =4

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3m + 7m - 0.8 < 9.2 someone answer this quick plz

Simora [160]

If you are solving for m
m <1

5 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

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

Then the $k$ -th moment of $X$ is

$M_X^{(k)}(0) = \dfrac{(k+7)!}{7!\,2^k}$

and we obtain the same results as before,

$\Bbb E[X] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=1} = 4$

$\Bbb E[X^2] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=2} = 18$

and the same variance follows.

6 0

2 years ago

Find f(-5) and f(12) for each piecewise function f(x){-4x+3 if x<3

Law Incorporation [45]

<h3>Answers:</h3>

f(-5) = 23
f(12) = 433

==================================================

Explanation:

The given piecewise function is

$f(x) = \begin{cases}-4x+3 \ \text{ if } \ x < 3\\-x^3 \ \text{ if } \ 3 \le x \le 8\\3x^2+1 \ \text{ if } \ x > 8\end{cases}$

At first piecewise functions may be strange confusing things, but they aren't so bad. I like to think of it like this: f(x) is a function that changes its identity based on what the input x is. We have three situations

f(x) = -4x+3 when x < 3
f(x) = -x^3 when $3 \le x \le 8$
f(x) = 3x^2+1 when x > 8

In a sense, we have three different functions but they are combined somehow.

If x is smaller than 3, then we go for the first definition. Or if x is between 3 and 8, then we go for the second definition. Or if x is larger than 8, then we go for the third definition.

-----------------------

f(-5) means f(x) when x = -5. We see that -5 is smaller than 3, so x = -5 makes x < 3 true. We'll use the first definition

f(x) = -4x+3

f(-5) = -4(-5)+3

f(-5) = 20+3

f(-5) = 23

------------------------

Now the input is x = 12. This is larger than 8. In other words, x = 12 makes x > 8 true. We'll use the third definition

f(x) = 3x^2+1

f(12) = 3(12)^2+1

f(12) = 3(144)+1

f(12) = 432+1

f(12) = 433

------------------------

Side notes:

We won't use the second definition since we don't have any x inputs between 3 and 8
To say "less than or equal to" on a keyboard, you can write "<=" without quotes. For example, $x \le 5$ is the same as x<=5

3 0

3 years ago

Exam parallelogram ABCD segments CE and BE have lengths 15-X and 2r what is the value of x

natima [27]

Answer:

The whole triangle is 180° , and means + in math

15-x + 2x= 180

-x+2x=180-15

-x+2x=165

2x=165

2x/2=165+/2

=82.5

4 0

3 years ago

Solve this system of linear equations. Separate

Elanso [62]

Answer:

(-3, -3)

Step-by-step explanation:

1.) Rewrite the second equation so 3y is on one side of the equation:

3y=6+5x

2.) Substitute the given value of 3y (replacing 3y with 6+5x, since we know they equal each other) into the equation 17x=-60-3y

Should end up with this:

17x=-60-(6+5x)

3.) Solve 17x=-60-(6+5x)

Calculate Difference: 17x=-66-5x

Combine Like Terms: 22x = -66

Divided both sides by 22 to isolate and solve for x: -3

So We know x=-3, now we got to find the y value. We can use either the first or second equation to find y value, so lets use the second.

3y=6+5x

1.) We know that x=-3, so we can simply substitute x in the equation

3y=6+5x with -3

3y=6+5(-3)

2.) Solve 3y=6+5(-3)

Combine Like Term: 3y=6+-15

Combine Like Term Even More: 3y= -9

Divide by 3 on both sides to isolate and solve for y: y=-3

So now we know y=-3 and once again we know x=-3, so if we format that

(-3,-3)

^ ^

x y

4 0

3 years ago