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3 years ago

7

A clothing company dyes cotton clothes in a huge container. They pour liquid dye into the container at a constant rate. How many

ounces of dye do they pour into the container in 22 minutes

1 answer:

AURORKA [14]3 years ago

6 0

Answer:

Sorry, but I will need more info to solve this problem. Like how many oz they pour in for each load or how often they do a load.

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If X is a r.v. such that E(X^n)=n! Find the m.g.f. of X,Mx(t). Also find the ch.f. of X,and from this deduce the distribution of

astraxan [27]

$M_X(t)=\mathbb E(e^{Xt})$
$M_X(t)=\mathbb E\left(1+Xt+\dfrac{t^2}{2!}X^2+\dfrac{t^3}{3!}X^3+\cdots\right)$
$M_X(t)=\mathbb E(1)+t\mathbb E(X)+\dfrac{t^2}{2!}\mathbb E(X^2)+\dfrac{t^3}{3!}\mathbb E(X^3)+\cdots$
$M_X(t)=1+t+t^2+t^3+\cdots$
$M_X(t)=\displaystyle\sum_{k\ge0}t^k=\frac1{1-t}$

provided that $|t|$ .

Similarly,

$\varphi_X(t)=\mathbb E(e^{iXt})$
$\varphi_X(t)=1+it+(it)^2+(it)^3+\cdots$
$\varphi_X(t)=(1-t^2+t^4-t^6+\cdots)+it(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=(1+it)(1-t^2+t^4-t^6+\cdots)$
$\varphi_X(t)=\dfrac{1+it}{1+t^2}=\dfrac1{1-it}$

You can find the CDF/PDF using any of the various inversion formulas. One way would be to compute

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{e^{itx}\varphi_X(-t)-e^{-itx}\varphi_X(t)}{it}\,\mathrm dt$

The integral can be rewritten as

$\displaystyle\int_0^\infty\frac{2i\sin(tx)-2it\cos(tx)}{it(1+t^2)}\,\mathrm dt$

so that

$F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt$

There are lots of ways to compute this integral. For instance, you can take the Laplace transform with respect to $x$ , which gives

$\displaystyle\mathcal L_s\left\{\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt\right\}=\int_0^\infty\frac{1-s}{(1+t^2)(s^2+t^2)}\,\mathrm dt$
$=\displaystyle\frac{\pi(1-s)}{2s(1+s)}$

and taking the inverse transform returns

$F_X(x)=\dfrac12+\dfrac1\pi\left(\dfrac\pi2-\pi e^{-x}\right)=1-e^{-x}$

which describes an exponential distribution with parameter $\lambda=1$ .

6 0

3 years ago

Difference between relative frequencies and percentages.

4vir4ik [10]

Answer:

is a measure of the number of times that an event occurs.

Step-by-step explanation:

A frequency count is a measure of the number of times that an event occurs. The above equation expresses relative frequency as a proportion. It is also often expressed as a percentage. Thus, a relative frequency of 0.50 is equivalent to a percentage of 50%.

3 0

3 years ago

What is the speed of a sailboat thats is traveling 100 meters in 120 seconds

lys-0071 [83]

Given:

100m = distance

120s = time

Asked:

Speed = ?

Formula:

Speed = d/t

Solution:

Speed = 100m/120s

Speed = 0.83333333333

Speed = 0.83m/s

Final Answer:

The speed of the sailboat is 0.83m/s

7 0

3 years ago

Work the following area application problem.

Nookie1986 [14]

SInce you need 1.5 feet of overhang, add 3 feet to each axis dimension 1.5 on each side):

Minor Axis = 18 + 3 = 21 feet

Major axis = 25 + 3 = 28 feet

The area of an ellipse is found by multiplying half the minor axis by half the major axis by PI.

1/2 minor axis = 21 / 2 = 10.5

1/2 major axis = 28 / 2 = 14

Using 3.14 for PI

Area = 10.5 x 14 x 3.14 = 147 x 3.14 = 461.6 sq ft

7 0

3 years ago

Read 2 more answers

An article presents a study of health outcomes in women with symptoms of heart disease. In a sample of 115 women whose test resu

My name is Ann [436]

Answer:

z(s) is in the rejection region we reject H₀ μ₁ = μ₂ and support the claim that at CI 95 % the means of the two groups differs

Step-by-step explanation:

Sample 1:

Sise sample n₁ = 115

μ₁ = 169,9 mmHg

σ₁ = 24,8 mmHg

Sample 2:

Sise sample n₂ = 235

μ₂ = 163,3 mmHg

σ₂ = 25,8 mmHg

We can develop a test hypothesis for differences in means to investigate if the mean peak systolic blood pressure differs between these two groups

We will choose CI = 95 % then significance level α = 5 %

α = 0,05 α/2 = 0,025

z(c) for 0,025 is from z-table z(c) = 1,96

Test Hypothesis:

Null Hypothesis H₀ μ₁ = μ₂

Alternative Hypothesis Hₐ μ₁ ≠ μ₂

The alternative hypothesis tells us that the test is a two-tail test.

z(s) = ( μ₁ - μ₂ ) / √ σ₁²/n₁ + σ₂²/n₂

z(s) = ( 169,9 -163,3 ) / √ (24,8)² /115 + ( 25,8)²/235

z(s) = 6,6 / √5,35 + 2,83

z(s) = 6,6 / 2,86

z(s) = 2,30

Comparing |z(c)| and |z(s)|

z(s) > z(c)

z(s) is in the rejection region we reject H₀ μ₁ = μ₂ and support the claim that at CI 95 % the means of the two groups differs

6 0

3 years ago