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enot [183]
3 years ago
14

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
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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.
6 0
3 years ago
Give the slope-intercept form for the equation.<br><br> 5x+ 2y = -14
irga5000 [103]

Answer:

y=-\frac{5}{2} x-7

Step-by-step explanation:

5x+2y=-14\\\frac{5x}{2} +\frac{2y}{2} =\frac{-14}{2}

\frac{5}{2}x+y=-7

y=-\frac{5}{2} x-7

5 0
2 years ago
Hey guys is this correct?
lutik1710 [3]

Answer:

Step-by-step explanation:

y e s , m e h  f r i e n d

6 0
2 years ago
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