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2 years ago

Which pair of transformations moves quadrilateral 1 to quadrilateral 2?

Mathematics

2 answers:

LenaWriter [7]2 years ago

8 0

Answer:

<u>Answer choices:</u>

a. Reflection takes it to Q2, further rotation takes it back to Q3

Incorrect

b. Reflection takes it to Q2, rotation takes to Q4 with right shape.

Correct

c. Rotation takes is to Q4, translation keeps in same Q but with wrong shape

Incorrect

d. Translation takes it to Q4, reflection keeps it in same Q but with wrong shape

Incorrect

malfutka [58]2 years ago

6 0

Translate it 8units to the right then reflect it over the line y=-3

Why?

We can see the Quadrilateral is in Quadrant 3.
If we translate it by 8units right it come to Quadrant 4.
Now reflect it over line y=-3
we will get Quadrilateral 2

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Gelneren [198K]

It is 2.47 dollars in the united states

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2 years ago

The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 ci

Olenka [21]

Answer:

The correct answer is

(0.0128, 0.0532)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$

For this problem, we have that:

In a random sample of 300 circuits, 10 are defective. This means that $n = 300$ and $\pi = \frac{10}{300} = 0.033$

Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool.

So $\alpha$ = 0.05, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 - 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0128$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 + 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0532$

The correct answer is

(0.0128, 0.0532)

4 0

2 years ago

Factor and solve the problems in the photo ……….. pleaseeeee help i haven’t done algebra in 2 years

vladimir1956 [14]

Answer:

30. 2x²-x-1=0

or, 2x²-2x+x-1=0

or, 2x(x-1)+1(x-1)=0

or, (x-1)(2x+1)=0

so, x-1 = 0 or, x = 1

and 2x+1 = 0 or, 2x = -1 or, x = -1/2

so, x = 1 and 1/2

31. 3x²-14x=5

or, 3x²-14x-5=0

or, 3x²-15x+x-5=0

or, 3x(x-5)+1(x-5)=0

or, (x-5)(3x+1)=0

so, x-5=0 or, x=5

and 3x+1=0 or, 3x=-1 or, x=-1/3

so, x = 5 and -1/3

Answered by GAUTHMATH

5 0

2 years ago

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denpristay [2]

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2 years ago

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