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garik1379 [7]
2 years ago
9

Help pleasee i’ll mark you as the brainliest if you answer less than 10 mins

Mathematics
1 answer:
aksik [14]2 years ago
3 0

Answer:

Answer  B: 2

Step-by-step explanation:

You can multiply the numbers of the smaller triangle and determine the answer

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1/15

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6,000 seconds is greater or less than 2 hours?
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is less it 1.67 hours

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How do the slopes of the lines created by each table compare?
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6 0
3 years ago
Graph a triangle (LMN) and rotate it 180° around the origin to create triangle L′M′N′.
andreev551 [17]

When a triangle is rotated, it must be rotated through a center of rotation.

Assume the coordinate of LMN is

\mathbf{L = (1,1)}

\mathbf{M = (1,7)}

\mathbf{N = (3,6)}

The rule of 180-degree rotation is:

\mathbf{(x,y) \to (-x,-y)}

So, we have:

\mathbf{L' = (-1,-1)}

\mathbf{M' = (-1,-7)}

\mathbf{N' = (-3,-6)}

See attachment for the graphs of LMN and L'M'N

<u>(a) Describe the transformation</u>

When triangle LMN is rotated 180 degrees across the origin, the x and y coordinates of LMN are negated to form triangle L'M'N'

<u>(b) Lines LL' and MM'</u>

The distance between point L and the center of origin is the same as the distance between point M' and the center of origin

Similarly, the distance between point L and the center of origin is the same as the distance between point M' and the center of origin

<u>(c) Line NN'</u>

Line NN' will assume the same characteristics as lines LL' and MM'

Read more about transformations at:

brainly.com/question/11707700

7 0
2 years ago
Suppose a simple random sample of size nequals81 is obtained from a population with mu equals 79 and sigma equals 18. ​(a) Descr
Daniel [21]

Answer:

(a) The sampling distribution of\overline{X} = Population mean = 79

(b)  P ( \overline{X} greater than 81.2 ) =  0.1357

(c) P (\overline{X} less than or equals 74.4 ) = .0107

(d) P (77.6 less than \overline{X} less than 83.2 ) = .7401

Step-by-step explanation:

Given -

Sample size ( n ) = 81

Population mean (\nu) = 79

Standard deviation (\sigma ) = 18

​(a) Describe the sampling distribution of \overline{X}

For large sample using central limit theorem

the sampling distribution of\overline{X} = Population mean = 79

​(b) What is Upper P ( \overline{X} greater than 81.2 )​ =

P(\overline{X}> 81.2)  = P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{81.2 - 79}{\frac{18}{\sqrt{81}}})

                    =  P(Z> 1.1)

                    = 1 - P(Z<   1.1)

                    = 1 - .8643 =

                    = 0.1357

(c) What is Upper P (\overline{X} less than or equals 74.4 ) =

P(\overline{X}\leq  74.4) = P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq  \frac{74.4- 79}{\frac{18}{\sqrt{81}}})

                    = P(Z\leq  -2.3)

                    = .0107

​(d) What is Upper P (77.6 less than \overline{X} less than 83.2 ) =

P(77.6< \overline{X}<   83.2) = P(\frac{77.6- 79}{\frac{18}{\sqrt{81}}})< P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq  \frac{83.2- 79}{\frac{18}{\sqrt{81}}})

                                = P(- 0.7< Z<   2.1)

                                 = (Z<   2.1) - (Z<   -0.7)

                                  = 0.9821 - .2420

                                   = 0.7401

3 0
3 years ago
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