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Kisachek [45]
3 years ago
5

A parallelogram has coordinates A(1, 1), B(5, 4), C(7, 1), and D(3, -2). What are the coordinates of parallelogram A′B′C′D′ afte

r a 180° rotation about the origin and a translation 5 units to the right and 1 unit down? I need Help
Mathematics
1 answer:
saul85 [17]3 years ago
5 0

Hey there! I'm happy to help!

First, we need to rotate our points 180° about the origin. To find the coordinates after such a rotation, we simply find the negative version of each number in the ordered pair, which can be written as (x,y)⇒(-x,-y).

Let's convert this below

A: (1,1)⇒(-1,-1)

B: (5,4)⇒(-5,-4)

C: (7,1)⇒(-7,-1)

D: (3,-2)⇒(-3,2)

Now, we need to translate these new points five units to the right and one unit down. This means we will add 5 to our x-value and subtract 1 from our y-value. This will look like (x,y)⇒(x+5,y-1). Let's do this below.

A: (-1,-1)⇒(4,-2)

B: (-5,-4)⇒(0,-5)

C: (-7,-1)⇒(-2,-2)

D: (-3,2)⇒(2,1)

Therefore, this new parallelogram has coordinates of A'(4,-2), B'(0,-5), C'(-2,-2), and D'(2,1)

Now you know how to find the coordinates of translated figures! Have a wonderful day! :D

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2 years ago
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A quality analyst of a tennis racquet manufacturing plant investigates if the length of a junior's tennis racquet conforms to th
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Answer:

Confidence interval : 21.506 to 24.493

Step-by-step explanation:

A quality analyst selects twenty racquets and obtains the following lengths:

21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

So, sample size = n =20

Now we are supposed to find Construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

Since n < 30

So we will use t-distribution

Confidence level = 99.9%

Significance level = α = 0.001

Now calculate the sample mean

X=21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

Sample mean = \bar{x}=\frac{\sum x}{n}

Sample mean = \bar{x}=\frac{21+25+23+22+24+21+25+21+23+ 26+ 21+24+22+ 24+23+21+ 21+ 26+23+ 24}{20}

Sample mean = \bar{x}=23

Sample standard deviation = \sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}

Sample standard deviation = \sqrt{\frac{(21-23)^2+(25-23)^2+(23-23)^2+(22-23)^2+(24-23)^2+(21-23)^2+(25-23)^2+(21-23)^2+(23-23)^2+(26-23)^2+(21-23)^2+(24-23)^2+(22-23)^2+(24-23)^2+(23-23)^2+(21-23)^2+(21-23)^2+(26-23)^2+(23-23)^2+(24-23)^2}{20-1}}

Sample standard deviation= s = 1.72

Degree of freedom = n-1 = 20-1 -19

Critical value of t using the t-distribution table t_{\frac{\alpha}{2} = 3.883

Formula of confidence interval : \bar{x} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}

Substitute the values in the formula

Confidence interval : 23 \pm 1.73 \times \frac{1.72}{\sqrt{20}}

Confidence interval : 23 -3.883 \times \frac{1.72}{\sqrt{20}} to 23 + 3.883 \times \frac{1.72}{\sqrt{20}}

Confidence interval : 21.506 to 24.493

Hence Confidence interval : 21.506 to 24.493

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