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3 years ago

Y-x=15 y-x=5 using elimination

Mathematics

1 answer:

Galina-37 [17]3 years ago

3 0

Answer:

No solution

Step-by-step explanation:

To eliminate is to get rid of one of the variables.

You can choose to either add each term in the equations or subtract each term in the equation.

For a variable to be eliminated, there must one pair that have the same constant with it. Each equation already has the same constant with a variable.

Try adding them.

. y - x = 15

2y - 2x = 20 No variables were eliminated.

Try subtracting.

. y - x = 15

0y - 0x = 10 All variables were eliminated.

. 0 = 10 This is false.

This system of equations cannot be solved.

Graphically, these two lines would have the same slope and are parallel. The solution to a system is same as the point of intersection. Parallel lines never meet, never intersect, therefore there is no solution.

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3 years ago

PLZZZZZZZ HELPP Which ordered pair is the solution to the system of equations? {y=2x−15} {4x+3y=−5} (5, −5) (4, −7) (0, 6) (7

Sophie [7]

solution is (4, - 7 )

given the equations

y = 2x - 15 → (1)

4x + 3y = - 5 → (2)

substitute y = 2x - 15 into (2)

4x + 3(2x - 15) = - 5

4x + 6x - 45 = - 5

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10x = 40 ( divide both sides by 10 )

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2 years ago

We have two fair three-sided dice, indexed by i = 1, 2. Each die has sides labeled 1, 2, and 3. We roll the two dice independent

Bogdan [553]

Answer:

(a) P(X = 0) = 1/3

(b) P(X = 1) = 2/9

(d) P(X = 3) = 0

(a) P(Y = 0) = 0

(b) P(Y = 1) = 1/3

Step-by-step explanation:

Given:

- Two 3-sided fair die.

- Random Variable X_1 denotes the number you get for rolling 1st die.

- Random Variable X_2 denotes the number you get for rolling 2nd die.

- Random Variable X = X_2 - X_1.

Solution:

- First we will develop a probability distribution of X such that it is defined by the difference of second and first roll of die.

- Possible outcomes of X : { - 2 , -1 , 0 ,1 , 2 }

- The corresponding probabilities for each outcome are:

( X = -2 ): { X_2 = 1 , X_1 = 3 }

P ( X = -2 ): P ( X_2 = 1 ) * P ( X_1 = 3 )

: ( 1 / 3 ) * ( 1 / 3 )

: ( 1 / 9 )

( X = -1 ): { X_2 = 1 , X_1 = 2 } + { X_2 = 2 , X_1 = 3 }

P ( X = -1 ): P ( X_2 = 1 ) * P ( X_1 = 3 ) + P ( X_2 = 2 ) * P ( X_1 = 3)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 2 / 9 )

( X = 0 ): { X_2 = 1 , X_1 = 1 } + { X_2 = 2 , X_1 = 2 } + { X_2 = 3 , X_1 = 3 }

P ( X = -1 ):P ( X_2 = 1 )*P ( X_1 = 1 )+P( X_2 = 2 )*P ( X_1 = 2)+P( X_2 = 3 )*P ( X_1 = 3)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 3 / 9 ) = ( 1 / 3 )

( X = 1 ): { X_2 = 2 , X_1 = 1 } + { X_2 = 3 , X_1 = 2 }

P ( X = 1 ): P ( X_2 = 2 ) * P ( X_1 = 1 ) + P ( X_2 = 3 ) * P ( X_1 = 2)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 2 / 9 )

( X = 2 ): { X_2 = 1 , X_1 = 3 }

P ( X = 2 ): P ( X_2 = 3 ) * P ( X_1 = 1 )

: ( 1 / 3 ) * ( 1 / 3 )

: ( 1 / 9 )

- The distribution Y = X_2,

P(Y=0) = 0

P(Y=1) = 1/3

P(Y=2) = 1/ 3

- The probability for each number of 3 sided die is same = 1 / 3.

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2 years ago