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Mamont248 [21]
4 years ago
13

Point P is reflected across two lines. Point E is its reflection image. Which two lines is point P reflected across?

Mathematics
2 answers:
prisoha [69]4 years ago
7 0

Answer:

The point P reflected across the lines x=3 and y=2.

Step-by-step explanation:

From the given graph it is clear that the coordinates of P are (0,0) and the coordinates of E are (6,4).

P(0,0)\rightarrow E(6,4)                 .... (1)

If the origin reflected across the line x=a then

(0,0)\rightarrow (2a,0)

If the origin reflected across the line y=b, then

(0,0)\rightarrow (0,2b)

If the origin reflected across the line x=a and y=b, then

(0,0)\rightarrow (2a,2b)                .... (2)

From (1) and (2) we get

2a=6\Rightarrow a=3

2b=4\Rightarrow b=2

The value of a is 3 and the value of b is 2. Therefore the point P reflected across the lines x=3 and y=2.

Virty [35]4 years ago
4 0
Line DJ: x=3. The reflection of point P across Line DJ is the point (6,0)
Line y=2. The reflection of point (6,0) acroos line y=2 is the point E=(6,4)

Answer: The point P is reflected across lines x=3 and y=2 
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p ( X = 1 ) = 0.6 , p ( X = 2 ) = 0.3 , p ( X = 3 ) = 0.1

Verified

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Step-by-step explanation:

Solution:-

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                       White                            3

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- A ball is drawn from urn without replacement until a white ball is drawn for the first time.

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2) A red ball is drawn on the first draw and a white ball is drawn on the second trial ( X = 2 ). The probability of drawing a red ball first would be:

      p ( Red on first trial ) = ( Number of red balls ) / ( Total number of balls )

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   p ( White on second trial ) = ( Number of white balls ) / ( number of balls left )

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- Then to draw red on first trial and white ball on second trial we can express:

                p ( X = 2 ) =  p ( Red on first trial ) *  p ( White on second trial )

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3) A red ball is drawn on the first draw and second draw and then a white ball is drawn on the third trial ( X = 3 ). The probability of drawing a red ball first would be ( 2 / 5 ). Then we are left with 4 balls in the urn, we again draw a red ball:

   p ( Red on second trial ) = ( Number of red balls ) / ( number of balls left )

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- Then draw a white ball from a total of 3 balls left in the urn ( remember without replacement ).                  

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                p ( X = 3 ) =  ( 2 / 5 ) * ( 1 / 4 ) * 1

                p ( X = 3 ) =  ( 1 / 10 )  

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p ( X )      0.6               0.3                  0.1

- To verify the above the distribution. We will sum all the probabilities for all outcomes ( X = 1 , 2 , 3 ) must be equal to 1.

          ∑ p ( Xi ) = 0.6 + 0.3 + 0.1

                         = 1 ( proven it is indeed a pmf )

- The expected value E ( X ) of the distribution i.e the expected number of trials until we draw a white ball for the first time:

               E ( X ) = ∑ [ p ( Xi ) * Xi  ]

               E ( X ) = ( 1 ) * ( 0.6 ) + ( 2 ) * ( 0.3 ) + ( 3 ) * ( 0.1 )

               E ( X ) = 0.6 + 0.6 + 0.3

               E ( X ) = 1.5 trials until first white ball is drawn.

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