All categories

3 years ago

I WILL GIVE BRAINLIEST FOR ANSWER WITH EXPLANATION!!!! (trolls get banned)

Mathematics

2 answers:

BARSIC [14]3 years ago

8 0

The answer is C. :)

natka813 [3]3 years ago

8 0

Hii
It’s C!
If you need an explanation tell me <8
Have a nice day byee

You might be interested in

If x + y = 2 and x - y = 6, solve for x and y.

anyanavicka [17]

Answer:

Step-by-step explanation:

Givens

x + y = 2

x - y = 6

Solution

Just add the two equations together. The y's will drop out.

x + y = 2

<u>x - y = 6 </u> Add

2x = 8 Divide by 2

2x/2 = 8/2 Combine

x = 4 Substitute x = 4 into the top equation

x + y = 2

4 + y = 2 Subtract 4 from both sides

4-4+y=2-4 Combine

y = - 2

Answer

x = 4

y = -2

7 0

1 year ago

I'm struggling a little bit can someone help?? A cell phone costs $850. You have a coupon for 15% off. How much will you pay for

Greeley [361]

Answer:

722.50

Step-by-step explanation:

not exactly sure its right but i did use a calculator so I'm pretty sure it's right

3 0

3 years ago

Read 2 more answers

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 $\lambda=1$ .