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MaRussiya [10]
3 years ago
7

As shown in the image below, triangle ABC is reflected onto its image. А B с С" A Which of the following lines represents the li

ne of reflection that maps triangle ABC onto its image?​

Mathematics
1 answer:
dybincka [34]3 years ago
6 0

Answer:

Line of reflection will be y = -2

Step-by-step explanation:

From the picture attached,

Triangle ABC has been reflected over an unknown line so that the triangle ABC will overlap triangle A'B'C'.

To find the line of reflection,

Coordinates of point C → (7, 0)

Coordinates of point C' → (7, -4)

Line of reflection = line passing through midpoint of the line joining C and C'

Midpoint of the line joining points C and C' = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})

                                                                        = (\frac{7+7}{2},\frac{0-4}{2})

                                                                        = (7, -2)

Therefore, line of reflection is y = -2.

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Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

       = 3 - 1 = 2

d.f.D = N - k

        = 35 - 3 = 32

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The mean and the variance of each sample :

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$\overline X_1 =50.5$           $\overline X_2 =50.07143$        $\overline X_3 =55.71429$

$s_1^2=19.96154$      $s_2^2=14.68681$         $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

         $=\frac{51+51+...+49+49}{35}$

        = 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

     $=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

   $=\frac{127.1143}{2}$

   = 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

    $=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

   $=\frac{669.8571}{32}$

  = 20.93304

The F-test  statistics value is :

$F=\frac{s_B^2}{s_W^2}$

  $=\frac{63.55714}{20.93304}$

  = 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source       Sum of squares    d.f    Mean square    F

Between    127.1143                 2      63.55714          3.036212

Within        669.8571             32      20.93304

Total           796.9714            34

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Wich will be x=1 so 1 mile
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