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Dmitry_Shevchenko [17]

3 years ago

8

The point-slope form of the equation of the line that passes through (–5, –1) and (10, –7) is y + 7 = (x – 10). What is the stan

dard form of the equation for this line? 2x – 5y = –15 2x – 5y = –17 2x + 5y = –15 2x + 5y = –17

2 answers:

taurus [48]3 years ago

7 0

The answer is 2x +5y=-15

ra1l [238]3 years ago

3 0

2x-5y=-17!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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F (x) = 1 – 4x<br> Calculate the value of f (2)

Irina-Kira [14]

Answer:

Step-by-step explanation:

F (x) = 1 – 4x

Calculate the value of f (2)

Subsitution 1-4(2)=1-8=-7

The answer of f(2)=-7

I hope this answers ur question

3 0

3 years ago

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

Garcia's Grill offers 77 side dishes, 55 types of steak, and 88 toppings. How many different smothered steak dinners can be made

svet-max [94.6K]

Answer:

Therefore the total number of ways that the customer can order the smothered steak dinner is 9800.

Step-by-step explanation:

There are 7 side dishes, and types of stake and 8 toppings.

The customer will have a smothered steak dinner.

The smothered steak dinner will consist of steak and 3 different toppings and 4 $\binom{8}{3} = \frac{8!}{3! \times 5!} = 56$ different side dishes.

The number of ways steak can be selected is = $\binom{5}{1} = \frac{5!}{1! 4!} =5$

The number of ways toppings can be chosen is = $\binom{8}{3} = \frac{8!}{3! 5!} = 56$

The number of ways side dishes can be chosen = $\binom{7}{4} = \frac{7!}{4!3!} = 35.$

Therefore the total number of ways that the customer can order the smothered steak dinner is

= 5 × 56 × 35 = 9800.

7 0

3 years ago

In a class full of men and women, 5/9 of the class are women. What is the ratio of men to women in its simplest form?

andrew-mc [135]

Answer:

4/5

Step-by-step explanation:

5x are women

9x are full class

9x - 5x = 4x

4x are men

$\frac{4x}{5x} = \frac{4}{5}$ is the ratio of men to women in its simplest form

Hope this helps ^-^

3 0

3 years ago

Solve 2x-8=-2-8. How do I do this

faust18 [17]

Is that negative 2 minus 8? if so put that into the calculator and add 8 to both sides and divide by 2 after that. simple algebra

5 0

3 years ago

Read 2 more answers