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umka2103 [35]
3 years ago
14

Square EFGH has vertices: E = (-1,1), F = (-1,2), G = (-2,1), H = (-2, 2). In which quadrant will the image of vertex E' be if t

he square is rotated 270� clockwise about the origin O?
Mathematics
1 answer:
AleksAgata [21]3 years ago
3 0
The square is originally located in the second quadrant. If the square is rotated 270 degrees clockwise from its original position, the square will be in the third quadrant.
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Exterior of scalene triangle sides are 25 and 15 and x. solve x.
Lyrx [107]

-- All three sides of a scalene triangle have different lengths.
So 'x' can't be 15 and it can't be 25.

-- 'x' must be 10 or more in order to reach between the ends
of the 25 and the 15.

-- 'x' must be less than 40 in order for the 25 and the 15 to reach
between its ends. 

So the value of 'x' must satisfy these conditions:

<em>0 < x < 15</em>
<em>15 < x < 25</em>
<em>25 < x < 40</em>

Any number that satisfies these conditions is an acceptable value for 'x'.


5 0
3 years ago
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If the function f(x)=x^3 has the domain {-3, 0, 2, 4}, what is its range?
mash [69]

Answer: {-16,4,92}. 2) {-16,10 ,42}. 3) {0,10,42}. 4) {0,4,92}. 5 The function f(x) = 2x. 2 + 6x - 12 has a domain.

Step-by-step explanation:

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3 years ago
Alguien que me ayude con esto :(
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Answer:

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Step-by-step explanation:

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5 0
2 years ago
An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen
devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                              P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean = \frac{\sum X}{n} = 441

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = 14.34

            n = sample size = 16

            \mu = population mean

<em>Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.</em>

<em />

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.131 < t_1_5 < 2.131) = 0.95  {As the critical value of t at 15 degrees of

                                             freedom are -2.131 & 2.131 with P = 2.5%}  

P(-2.131 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.131) = 0.95

P( -2.131 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.131 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.131 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.131 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X -2.131 \times {\frac{s}{\sqrt{n} } } , \bar X +2.131 \times {\frac{s}{\sqrt{n} } } ]

                                      = [ 441-2.131 \times {\frac{14.34}{\sqrt{16} } } , 441+2.131 \times {\frac{14.34}{\sqrt{16} } } ]

                                      = [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0
3 years ago
Find the perimeter of the<br> polygon if ZB = ZD.
Butoxors [25]

Answer:

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Step-by-step explanation:

4 0
3 years ago
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