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3 years ago

A store sells snacks by weight. A six ounce bag of nuts costs 3.60 will a two ounce bag cost more than or less than 1.25

Mathematics

1 answer:

densk [106]3 years ago

6 0

5.30 it is correct hope this works brah

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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

Use stokes' theorem to evaluate c f · dr where c is oriented counterclockwise as viewed from above. f(x, y, z = xyi + 5zj + 7yk,

Helga [31]

The intersection can be parameterized by

$C:=\mathbf r(t)=\begin{cases}x(t)=6\cos t\\y(t)=6\sin t\\z(t)=5-6\cos t\end{cases}$

with $0\le t$ .

By Stoke's theorem, the integral of $\mathbf f(x,y,z)=xy\,\mathbf i+5z\,\mathbf j+7y\,\mathbf k$ along $C$ is equivalent to

$\displaystyle\int_C\mathbf f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r(t)=\iint_S\nabla\times\mathbf f\,\mathrm dS$

where $S$ is the region bounded by $C$ . The line integral reduces to

$\displaystyle\int_0^{2\pi}(36\sin t\cos t\,\mathbf i+(25-30\cos t)\,\mathbf j+42\sin t\,\mathbf k)\cdot(-6\sin t\,\mathbf i+6\cos t\,\mathbf j+6\sin t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}(54(\cos3t-\cos t)-30(3\cos2t-5\cos t+3)+(126-126\cos2t)\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}(36+96\cos t-216\cos2t+54\cos3t)\,\mathrm dt$
$=72\pi$

4 0

3 years ago

What is 5.95 × 106 written in standard form? A. 595,000 B. 59,500 C. 5950 D. 5,950,000

tiny-mole [99]

The answer is option D "<span>5,950,000".

</span> $5.95 * 10^6 = 

5 9 5 0 0 0 0/
1 2 3 4 5 6$

Which would equal 5,950,000.

Hope this helps!

7 0

3 years ago

The work of a student to solve the equation 2(5y – 2) = 12 + 6y is shown below: Step 1: 2(5y – 2) = 12 + 6y Step 2: 7y – 4 = 12

kirza4 [7]

2(5y - 2) = 12 + 6y
7y - 4 = 12 + 6y ⇒ wrong, it has to be: 10y - 4 = 12 + 6y

so the correct answer is B

7 0

3 years ago

Read 2 more answers

The scatter plot shows the relationship between pages read per week and age. What is the range of the cluster shown in the scatt

Katen [24]

Answer:

part 1 of the question is d ,and part 2 of the question is b.