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faust18 [17]
3 years ago
7

Complete the following pattern: 12, ___, 18, 21 A. 15 B. 16

Mathematics
2 answers:
crimeas [40]3 years ago
8 0

Answer:

15

Step-by-step explanation:

kolezko [41]3 years ago
6 0
A.15 because you are adding 3
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What additional information would allow you to prove the quadrilateral is a parallelogram?
slavikrds [6]

The additional information that would allow us to prove that the image is a parallelogram is that; Line EJ ≅ Line GJ

<h3>How to prove a Parallelogram?</h3>

The six basic properties of parallelograms are primarily;

  • Both pairs of opposite sides are parallel
  • Both pairs of opposite sides are congruent
  • Both pairs of opposite angles are congruent
  • Diagonals bisect each other
  • One angle is supplementary to both consecutive angles (same-side interior)
  • One pair of opposite sides are congruent AND parallel.

Now, looking at the parallelogram properties above and comparing with the given image of the quadrilateral attached, we can say that the additional information that would allow us to prove that the image is a parallelogram is that; Line EJ ≅ Line GJ

Read more about Parallelogram Proof at; brainly.com/question/24056495

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2 years ago
General terms of a sequence
MatroZZZ [7]
(2n-1) /(n+9)
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5 0
3 years ago
What is 0.5 divided by 2.675
eduard

.186915888

that is the answer

if I meant 2.675 divided by .5 it is 5.35

7 0
2 years ago
Read 2 more answers
The number of violent crimes committed in a day possesses a distribution with a mean of 2.2 crimes per day and a standard deviat
VARVARA [1.3K]

Answer:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And replacing we got:

\mu_{\bar X}= 2.2

\sigma_{\bar X}= \frac{6}{\sqrt{100}}= 0.6

Step-by-step explanation:

For this case we have the following info given:

\mu = 2.2 represent the mean

\sigma = 6 represent the deviation

We select a sample size of n=100. This sample is >30 so then we can use the central limit theorem. And we want to find the distribution for the sample mean and we know that the distribution is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And replacing we got:

\mu_{\bar X}= 2.2

\sigma_{\bar X}= \frac{6}{\sqrt{100}}= 0.6

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3 years ago
I need trigonometry help!
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Which question is it


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