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2 years ago

Please help! Correct answer only!

Mathematics

1 answer:

melisa1 [442]2 years ago

5 0

Answer:

$7.50

Step-by-step explanation:

Since there is a 50% chance that the card is odd, the expected payoffs would be $15 * 50%=0.5*15=7.5. Hope this helps!

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Pls help me help me pls

grin007 [14]

Answer:

Point B

Step-by-step explanation:

Lines given:

y= 2/5x - 1/2
y= -1/3x +2/3

x coordinate of intersection of given lines is:

2/5x - 1/2= -1/3x+2/3
2/5x+1/3x= 2/3 +1/2
6/15x+5/15x= 4/6+3/6
11/15x=7/6
x= 7*15/11*6
x= 35/22

For this value of x, the value of y is:

y= 2/5*35/22 - 1/2
y=14/22- 1/2
y= 3/22

As per graph point (35/22, 3/22) is point B

4 0

3 years ago

I need help finding x please?

enot [183]

$\large\mathfrak{{\pmb{\underline{\blue{To\:find}}{\blue{:}}}}}$

The value of $x$ .

$\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}$

$\longrightarrow{\green{x\:=\: 25° }}$

$\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}$

We know that,

$\sf\purple{Sum\:of\:angles\:on\:a\:straight\:line\:=\:180°}$

➪ 125° + $x$ + 30° = 180°

➪ $x$ + 155° = 180°

➪ $x$ = 180° - 155°

➪ $x$ = 25°

Therefore, the value of $x$ is 25°.

Now, the three angles of the triangle are 125°, 25° and 30°.

$\large\mathfrak{{\pmb{\underline{\pink{To\:verify}}{\pink{:}}}}}$

✒ 125° + 25° + 30° = 180°

✒ 180° = 180°

✒ L. H. S. = R. H. S.

$\boxed{Hence\:verified.}$

$\huge{\textbf{\textsf{{\orange{My}}{\blue{st}}{\pink{iq}}{\purple{ue}}{\red{35}}{\green{ヅ}}}}}$

3 0

2 years ago

Read 2 more answers

At a local Brownsville play production, 420 tickets were sold. The ticket

Dahasolnce [82]

Answer:

jsdcjdvnjkdnjnjdanskcbanknqnjfkrbgiyrwhgondfkv

Step-by-step explanation:

8 0

2 years ago

Consider right triangle MNO below which expression represents the length of side MN

azamat

Answer:

I can help you via Watsapp

5 0

2 years ago

G with the definition of covariance, prove cov[(ay − b),(cy − d)] = accov(x, y ), where x, y are random variables and a, b, c, d

____ [38]

By definition of covariance,

$\mathrm{Cov}(X,Y)=\mathbb E[(X-\mathbb E[X])(Y-\mathbb E[Y])]$

$\mathrm{Cov}(X,Y)=\mathbb E[XY-\mathbb E[X]Y-X\mathbb E[Y]+\mathbb E[X]\mathbb E[Y]]=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]$

We have

$\mathbb E[(aX-b)(cY-d)]=\mathbb E[acXY-adX-bcY+bd]$

$=ac\mathbb E[XY]-ad\mathbb E[X]-bc\mathbb E[Y]+bd$

$\mathbb E[aX-b]=a\mathbb E[X]-b$

$\mathbb E[cY-d]=c\mathbb E[Y]-d$

$\mathbb E[aX-b]\mathbb E[cY-d]=ac\mathbb E[X]\mathbb E[Y]-ad\mathbb E[X]-bc\mathbb E[Y]+bd$

Putting everything together, we find the covariance reduces to

$\mathrm{Cov}(aX-b,cY-d)=ac(\mathbb E[XY]-\mathbb E[X]\mathbb E[Y])=ac\mathrm{Cov}(X,Y)$

as desired.

5 0

2 years ago