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lapo4ka [179]
3 years ago
15

Help me on this please

Mathematics
1 answer:
zalisa [80]3 years ago
8 0

Answer:

1. (x, y) → (x + 3, y - 2)

Vertices of the image

a) (-2, - 3)

b) (-2, 3)

c) (2, 2)

2. (x, y) → (x - 3, y + 5)

Vertices of the image

a) (-3, 2)

b) (0, 2)

c) (0, 4)

d) (2, 4)

3. (x, y) → (x + 4, y)

Vertices of the image

a) (-1, -2)

b) (1, -2)

c) (3, -2)

4. (x, y) → (x + 6, y + 1)

Vertices of the image

a) (1, -1)

b) (1, -2)

c) (2, -2)

d) (2, -4)

e) (3, -1)

f) (3, -3)

g) (4, -3)

h) (1, -4)

5. (x, y) → (x, y - 4)

Vertices of the image

a) (0, -2)

b) (0, -3)

c) (2, -2)

d) (2, -4)

6. (x, y) → (x - 1, y + 4)

Vertices of the image

a) (-5, 3)

b) (-5, -1)

c) (-3, 0)

d) (-3, -1)

Explanation:

To identify each <u><em>IMAGE</em></u> you should perform the following steps:

  • List the vertex points of the preimage (the original figure) as ordered pairs.
  • Apply the transformation rule to every point of the preimage
  • List the image of each vertex after applying each transformation, also as ordered pairs.

<u>1. (x, y) → (x + 3, y - 2)</u>

The rule means that every point of the preimage is translated three units to the right and 2 units down.

Vertices of the preimage      Vertices of the image

a) (-5,2)                                   (-5 + 3, -1 - 2) = (-2, - 3)

b) (-5, 5)                                  (-5 + 3, 5 - 2) = (-2, 3)

c) (-1, 4)                                   (-1 + 3, 4 - 2) = (2, 2)

<u>2. (x,y) → (x - 3, y + 5)</u>

The rule means that every point of the preimage is translated three units to the left and five units down.

Vertices of the preimage      Vertices of the image

a) (0, -3)                                   (0 - 3, -3 + 5) = (-3, 2)

b) (3, -3)                                   (3 - 3, -3  + 5) = (0, 2)

c) (3, -1)                                    (3 - 3, -1 + 5) = (0, 4)

d) (5, -1)                                    (5 - 3, -1 + 5) = (2, 4)

<u>3. (x, y) → (x + 4, y)</u>

The rule represents a translation 4 units to the right.

Vertices of the preimage   Vertices of the image

a) (-5, -2)                               (-5 + 4, -2) = (-1, -2)

b) (-3, -5)                               (-3 + 4, -2) = (1, -2)

c) (-1, -2)                                (-1 + 4, -2) = (3, -2)

<u>4. (x, y) → (x + 6, y + 1)</u>

Vertices of the preimage      Vertices of the image

a) (-5, -2)                                  (-5 + 6, -2 + 1) = (1, -1)

b) (-5, -3)                                  (-5 + 6, -3 + 1) = (1, -2)

c) (-4, -3)                                   (-4 + 6, -3 + 1) = (2, -2)

d) (-4, -5)                                  (-4 + 6, -5 + 1) = (2, -4)

e) (-3, -2)                                  (-3 + 6, -2 + 1) = (3, -1)

f) (-3, -4)                                   (-3 + 6, -4 + 1) = (3, -3)

g) (-2, -4)                                  (-2 + 6, -4 + 1) = (4, -3)

h) (-2, -5)                                  (-2 + 3, -5 + 1) = (1, -4)

<u>5. (x, y) → (x, y - 4)</u>

This is a translation four units down

Vertices of the preimage      Vertices of the image

a) (0, 2)                                    (0, 2 - 4) = (0, -2)

b) (0,1)                                      (0, 1 - 4) = (0, -3)

c) (2, 2)                                     (2, 2 - 4) = (2, -2)

d) (2,0)                                     (2, 0 - 4) = (2, -4)

<u>6. (x, y) → (x - 1, y + 4)</u>

This is a translation one unit to the left and four units up.

Vertices of the pre-image     Vertices of the image

a) (-4, -1)                                   (-4 - 1, -1 + 4) = (-5, 3)

b) (-4 - 5)                                  (-4 - 1, -5 + 4) = (-5, -1)

c) (-2, -4)                                  (- 2 - 1, -4 + 4) = (-3, 0)

d) (-2, -5)                                 (-2 - 1, -5 + 4) = (-3, -1)

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Step-by-step explanation:

Data given and notation  

X_{1}=253 represent the number with no defects in sample 1

X_{2}=196 represent the number with no defects in sample 1

n_{1}=300 sample 1

n_{2}=300 sample 2

p_{1}=\frac{253}{300}=0.843 represent the proportion of number with no defects in sample 1

p_{2}=\frac{196}{300}=0.653 represent the proportion of number with no defects in sample 2

z would represent the statistic (variable of interest)  

p_v represent the value for the test (variable of interest)  

\alpha=0.05 significance level given

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if is there is a difference in the the two proportions, the system of hypothesis would be:  

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Alternative hypothesis:p_{1} - p_{2} \neq 0  

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Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

z=\frac{0.843-0.653}{\sqrt{0.748(1-0.748)(\frac{1}{300}+\frac{1}{300})}}=5.358    

Statistical decision

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Comparing the p value with the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that we have singificantly differences between the two proportions.  

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