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1 year ago

11

A rectangular swimming pool has a length of 50 meters, a width of 25 meters, and a depth of 2 meters. How much paint would be ne

eded to cover the sides and bottom of the pool?

1 answer:

sweet-ann [11.9K]1 year ago

5 0

Answer:

2500m³

Step-by-step explanation:

A = length x width x height

50m x 25m x 2 m

=2500m³

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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 $\lambda=1$ .

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2 years ago

Determine whether the Mean Value Theorem applies to the function f(x)=2x^1/5 on the interval [-32, 32]

Anettt [7]

Answer

given,

$f(x) = 2 x^{1/5}$

interval = [-32, 32]

differentiating the given equation

$f'(x) = \dfrac{d}{dx}(2 x^{1/5})$

$f'(x) = \dfrac{2}{5x^{4/5}}$

hence, the above solution is not defined at x = 0

and x = 0 lie in the given interval i.e. [-32, 32]

so, at x = 0 the function is not differentiable.

Hence, mean value theorem does not apply to the given function.

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2 years ago

A scientist used 2 gallons of liquid for every 3 hours he works. He uses __ of a gallon each hour he works.

tiny-mole [99]

He uses 2/3 of a gallon each hour he works.

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2 years ago

Choose the missing exponent to create a polynomial?

lions [1.4K]

3.

The answers 2 and 4 are incorrect because inserting that ad the missing exponent would create only a trinomial.

-9 is incorrect as it would create something more than a polynomial. 7 is incorrect as, if doing a certain type of division, would result in a larger polynomial.

Hope this helps!

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3 years ago

Chad needed to buy a new part for his bicycle, and thought that the parts were sold in metric units. When he arrived at the stor

svp [43]

Answer:

<h2>9.1 inches</h2>

Step-by-step explanation:

1cm = 0.3937007874 inches.

As there are 2.54cm in an inch, to convert your cm figure to inches you need to divide your figure by 2.54.

9.05512 -> 9.1 inches

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2 years ago