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murzikaleks [220]
3 years ago
6

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
You might be interested in
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 = &#10;&#10;5 9 5 0 0 0 0/&#10;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.

8 0
2 years ago
Read 2 more answers
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