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3 years ago

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 IMAGE 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.

1. (x, y) → (x + 3, y - 2)

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)

2. (x,y) → (x - 3, y + 5)

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)

3. (x, y) → (x + 4, y)

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)

4. (x, y) → (x + 6, y + 1)

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)

5. (x, y) → (x, y - 4)

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)

6. (x, y) → (x - 1, y + 4)

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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3 years ago

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Step-by-step explanation:

Data given and notation

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$X_{2}=196$ 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:

Null hypothesis: $p_{1} - p_2}=0$

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

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

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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

Since is a two sided test the p value would be:

$p_v =2*P(Z>5.358) = 4.2x10^{-8}$

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3 years ago