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Klio2033 [76]
3 years ago
14

Select the equation in which the graph of the line has a negative slope, and the y-intercept equals 10. Answer 10x + 5y = –25 5x

– 10y = –12 20x + 5y = 50 2x + 6y = 10
Mathematics
1 answer:
adoni [48]3 years ago
4 0
The answer is 20x + 5y = 50
If u move the 20x over u get:
5y=-20x + 50

Solve for y by divided by 5

This gets y=-4x + 10

Now the answer is in slope intercept form: the -4 is the slope and 10 is the y-intercept
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any perpendicular bisector of one of the sides".

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The outlier of a dataset is a data element that is relatively far from the remaining data elements

  • <em>99 is an outlier of pet group</em>
  • <em>See attachment for the parallel box plots</em>

<u>(a) Prove that 99 is an outlier for Pet</u>

We have:

<em>Pets: 58 64 65 68 69 69 69 70 70 72 76 79 85 86 99</em>

n = 15

The quartiles positions are:

Q_1 = \frac{n + 1}{4}

Q_1 = \frac{15 + 1}{4}

Q_1 = \frac{16}{4}

Q_1 = 4th

Q_3 = Q_1 \times 3

Q_3 = 4th \times 3

Q_3 = 12th

So, we have:

Q_1 = 4th

Q_3 = 12th

From the pet group:

The data elements at the 4th and 12th positions are 68 and 79

So, we have:

Q_1 = 68

Q_3= 79

The lower and upper limits of the outlier are:

L = Q_1 - 1.5 \times (Q_3 - Q_1)

U = Q_3 + 1.5 \times (Q_3 - Q_1)

So, we have:

L = 68 - 1.5 \times (79 - 68)

L = 51.5

U = 79+ 1.5 \times (79 - 68)

U = 95.5

This means that data below 51.5 or above 95.5 are outliers.

<em>Hence, 99 is an outlier because 99 is greater than 95.5</em>

<u>(b) The parallel box plot</u>

The three groups are:

<em>Pets: 58 64 65 68 69 69 69 70 70 72 76 79 85 86 99</em>

<em>Erlento: 88 80 80 81 92 87 88 81 82 80 87 92 87 80 82 </em>

<em>Alone: 62 70 73 75 77 80 84 84 84 87 87 87 90 91 99</em>

<em />

See attachment for the parallel box plots

Read more about box plots and outliers at:

brainly.com/question/14940764

5 0
3 years ago
Find solution to logarithm. 
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3 years ago
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4)) Solve for w.<br> 100 = w + 2<br> w=[
Anettt [7]

Answer: w= 98

Step-by-step explanation:

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Need help ASAP please!! Thank you!
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What’s the question
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