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Varvara68 [4.7K]
2 years ago
12

Trapezoid G H J K is rotated about G 90 degrees counterclockwise to form trapezoid G prime H prime J prime K prime. Trapezoid G

prime H prime J prime K prime is reflected across the line of reflection m to form trapezoid G double-prime H double-prime J double-prime K double-prime.
What is the final transformation in the composition of transformations that maps pre-image GHJK to image G"H"J"K"?

a translation to the right
a reflection across line m
a 90° rotation about point G
a 180° rotation about point G



helppp timed !!!

Mathematics
2 answers:
11Alexandr11 [23.1K]2 years ago
6 0

Answer:

The answer is c i just took the test

Step-by-step explanation:

o-na [289]2 years ago
4 0

Answer:

Option C.

Step-by-step explanation:

It is given that Trapezoid GHJK is rotated about G 90 degrees counterclockwise to form trapezoid G'H'J'K'.

First rule of transformation = Rotation 90 degrees counterclockwise about G

                                             = R(G,90^{\circ})

Trapezoid G'H'J'K' is reflected across the line of reflection m to form trapezoid G''H''J''K''.

Second rule of transformation = Reflection across m = r_m(x,y)

Combination of transformation is known composition of transformation  transformation. We apply the transformation from left to right.

The figure GHJK is Rotated 90 degrees counterclockwise about G and then  reflected across the line m. So, the composition of transformation is

r_m(x,y) \circ R(G,90^{\circ})

The final transformation in the composition of transformations is 90° rotation about point G.

Therefore the correct option is C.

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A study is being conducted in which the health of two independent groups of ten policyholders is being monitored over a one-year
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Answer:

46.91% probability that at least nine participants complete the study in one of the two groups, but not in both groups

Step-by-step explanation:

We use two binomial trials to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Probability of at least nine participants finishing the study in a group.

0.2 probability of a students dropping out. So 1 - 0.2 = 0.8 probability of a student finishing the study. This means that p = 0.8.

10 students, so n = 10

We have to find:

P(X \geq 9) = P(X = 9) + P(X = 10)

Then

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 9) = C_{10,9}.(0.8)^{9}.(0.2)^{1} = 0.2684

P(X = 10) = C_{10,10}.(0.8)^{10}.(0.2)^{0} = 0.1074

P(X \geq 9) = P(X = 9) + P(X = 10) = 0.2684 + 0.1074 = 0.3758

0.3758 probability that at least nine participants complete the study in a group.

Calculate the probability that at least nine participants complete the study in one of the two groups, but not in both groups?

0.3758 probability that at least nine participants complete the study in a group. This means that p = 0.3758

Two groups, so n = 2

We have to find P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{2,1}.(0.3758)^{1}.(0.6242)^{1} = 0.4691

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Math 1314 lab module 2
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A brewery produces cans of beer that are supposed to contain exactly 12 ounces. But owing to the inevitable variation in the fil
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Answer:

T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)

P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solutio to the problem

Let X the random variable that represent the amount of beer in each can of a population, and for this case we know the distribution for X is given by:

X \sim N(12,0.3)  

Where \mu=12 and \sigma=0.3

For this case we select 6 cans and we are interested in the probability that the total would be less or equal than 72 ounces. So we need to find a distribution for the total.

The definition of sample mean is given by:

\bar X = \frac{\sum_{i=1}^n X_i}{n} = \frac{T}{n}

If we solve for the total T we got:

T= n \bar X

For this case then the expected value and variance are given by:

E(T) = n E(\bar X) =n \mu

Var(T) = n^2 Var(\bar X)= n^2 \frac{\sigma^2}{n}= n \sigma^2

And the deviation is just:

Sd(T) = \sqrt{n} \sigma

So then the distribution for the total would be also normal and given by:

T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)

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P(T\leq 72)

And we can use the z score formula given by:

z = \frac{x-\mu}{\sigma}

P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z

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