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Zigmanuir [339]

2 years ago

13

The mapping of DEFG to D'E'F'G' is shown. 2 parallelograms have identical side lengths and angle measures. The second parallelog

ram is a reflection of the first. Which statements are true regarding the transformation? Check all that apply. EF corresponds to E'F'. FG corresponds to G'D'. ∠EDG Is-congruent-to ∠E'D'G' ∠DEF Is-congruent-to ∠D'E'F' The transformation is not isometric. The transformation is a rigid transformation.

2 answers:

Eduardwww [97]2 years ago

6 1

Answer:

1346

Step-by-step explanation:

vodka [1.7K]2 years ago

4 0

Answer:

(A)EF corresponds to E'F'

(C)∠EDG Is-congruent-to ∠E'D'G'

(D)∠DEF Is-congruent-to ∠D'E'F'

(F)The transformation is a rigid transformation.

Step-by-step explanation:

Given:

Parallelogram DEFG is mapped to D'E'F'G'
DEFG and D'E'F'G' have identical side lengths and angle measures.

The following applies:

EF corresponds to E'F'
∠EDG Is-congruent-to ∠E'D'G'
∠DEF Is-congruent-to ∠D'E'F'

Now, a rigid transformation is a transformation of the plane that preserves length. Since the two parallelograms have identical side lengths:

The transformation is a rigid transformation.

Note that a reflection is an isometric transformation. Therefore the statement "The transformation is not isometric" is INCORRECT.

FG and GD are adjacent sides, therefore they may not necessarily be congruent. Thus FG does not corresponds to G'D'

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Options

A. UV = 14 ft and m∠TUV = 45°

B. TU = 26 ft

C. m∠STU = 37° and m∠VTU = 37°

D. ST = 20 ft, UV = 14 ft, and m∠UST = 98°

E. m∠UST = 98° and m ∠TUV = 45°

Answer:

A. UV = 14 ft and m∠TUV = 45°

D. ST = 20 ft, UV = 14 ft, and m∠UST = 98°

Step-by-step explanation:

Given

See attachment for triangle

Required

What proves that: ΔSTU ≅ ΔVTU using SAS

To prove their similarity, we must check the corresponding sides and angles of both triangles

First:

$\angle UST$ must equal $\angle UVT$

So:

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Next:

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2 years ago

larisa86 [58]

Answer:

The correct option is (C).

Step-by-step explanation:

According to the Central Limit Theorem if we have n unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the distribution of sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The information provided is:

<em>μ</em> = 3.5

<em>σ</em> = 1.7078

<em>n</em> = 400 = number of times the experiment is repeated.

As the sample size is quite large, i.e. <em>n</em> = 400 > 30 the central limit theorem can be used to approximate the sampling distribution of the sample mean.

The mean of the distribution of sample means is:

$\mu_{\bar x}=\mu=3.5$

The standard deviation of the distribution of sample mean is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{1.7078}{\sqrt{400}}=0.0854$

The distribution of the sample mean is:

$\bar X\sim N(\mu = 3.5,\ \sigma=0.0854)$ .

Thus, the correct option is (C).

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