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vladimir2022 [97]
3 years ago
10

Triangle ABC is translated on the coordinate plane below to create triangle A'B'C':

Mathematics
2 answers:
Elena L [17]3 years ago
8 0
<span>the answer for this case is
 since the translation is
 b. (4, 9)
 (x + 3, y + 6)
 for each ordered pair
</span>
sashaice [31]3 years ago
8 0

Answer:  The correct option is (b) (4, 9).

Step-by-step explanation:  Given that the triangle ABC is translated on the coordinate plane shown to create triangle A'B'C'.

And the parallelogram EFGH is translated according to the same rule that translated triangle ABC.

We are to find the ordered pair of the point H'.

From the figure, we note that

the co-ordinates of the vertices of triangle ABC are A(-7, -2), B(-5, -5) and C(-1, -5).

And, the co-ordinates of the vertices of triangle A'B'C' are A'(-4, 4), B'(-2, 1) and C(2, 1).

So, the required translations from the vertices of triangle ABC to A'B'C are

A(-7, -2)   ⇒   A'(-4, 4) = A'(-7+3, -2+6),

B(-5, 5)    ⇒   B'(-2, 1)  = B'(-5+3, -5+6),

C(-1, -5)   ⇒   C'(2, 1)   =  C'(-1+3, -5+6).

So, the required translation rule from triangle ABC to triangle A'B'C' is given by

(x, y)   ⇒   (x+3, y+6).

Now, the co-ordinates of the point H are (1, 3).

So, if the parallelogram EFGH is translated according to the rule (x, y) ⇒ (x+3, y+6), then the co-ordinates of the point H' will be

H(1, 3)  ⇒   H'(1+3, 3+6) = H'(4, 9).

Thus, the required co-ordinates of the point H' are (4, 9).

Option (b) is CORRECT.

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Monica [59]

Answer:

Interest earned = $32.835

Step-by-step explanation:

Given the following data;

Principal = $275

Number of times = 0.5

Interest rate = 2.9% = 0.029

Time = 4 years

To find the interest earned, we would use the compound interest formula;

A = P(1 + \frac{r}{n})^{nt}

Where;

A is the future value.

P is the principal or starting amount.

r is annual interest rate.

n is the number of times the interest is compounded in a year.

t is the number of years for the compound interest.

Substituting into the equation, we have;

A = 275(1 + \frac{0.029}{0.5})^{0.5*4}

A = 275(1 + 0.058)^{2}

A = 275(1.058)^{2}

A = 275(1.1194)

A = $307.835

Interest earned = 307.835 - 275

Interest earned = $32.835

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Tomtit [17]
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3 years ago
An urn contains three white balls and two red balls. The balls are drawn from the urn, oneat a time without replacement, until a
Tpy6a [65]

Answer:

p ( X = 1 ) = 0.6 , p ( X = 2 ) = 0.3 , p ( X = 3 ) = 0.1

Verified

E ( X ) = 1.5

Step-by-step explanation:

Solution:-

- An urn contains the following colored balls:

                       Color                  Number of balls

                       White                            3

                       Red                               2

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

- We will construct cases to determine the distribution of the random-variable X: The number of trials it takes to get the first white ball.

- We have three following case:

1) White ball is drawn on the first attempt ( X =  1 ). The probability of drawing a white ball in the first trial would be:

              p ( X = 1 ) = ( Number of white balls ) / ( Total number of ball )

              p ( X = 1 ) = ( 3 ) / ( 5 )

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 )

      p ( Red on first trial ) = ( 2 ) / ( 5 )

- Then draw a white ball from a total of 4 balls left in the urn ( remember without replacement ).

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

   p ( White on second trial ) = ( 3 ) / ( 4 )

- 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 )

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

                p ( X = 2 ) =  ( 3 / 10 )  

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 )

   p ( Red on second trial ) = ( 1 ) / ( 4 )    

 

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

   p ( White on 3rd trial ) = ( Number of white balls ) / ( number of balls left )

   p ( White on 3rd trial ) = ( 3 ) / ( 3 ) = 1

- Then to draw red on first two trials and white ball on third trial we can express:

                p ( X = 3 ) =  p ( Red on 1st trial )*p ( Red on 2nd trial )*p ( White on 3rd trial )

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

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

- The probability distribution of X is as follows:

    X          1                  2                      3

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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