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skelet666 [1.2K]
3 years ago
5

find the set of five items of data that has a range of 9, a mean of 11, a meadian of 12, and a mode of 15.

Mathematics
1 answer:
ValentinkaMS [17]3 years ago
3 0
Here's what I've done:

Represent the set of 5 items by {a,b,c,d,e}.  Assume that these items have already been arranged in ascending order.  

The middle item, c, has the value 12.

Since "mode" means "most frequently appearing item," let's assume that d and e are both 15.

If the range is 9, we could then write 15-9 = 6, meaning that a = 6.

Then we have {6,b,12,15,15}

To find b, sum up {6,b,12,15,15}, equate the result to 11, and solve for b.
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3K= I-14j which then equals K=(I-14J)/3
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An international company has 27,8000 employees in one country. If this represents 32.7% of the company's employees, how many emp
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Answer:

850152.29052

Step-by-step explanation:

278000 rep 32.7%

278000=32.7%, 100% is the total

100%×278000= 278000000÷32.7=850,152.29052

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A pharmacist found at the end of the day she had 5/4 as many prescriptions for antibiotics then tranquilizers. She had 18 prescr
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Answer:

16. 5/3

Step-by-step explanation:

let the no of tranquilizer be x

therefore

5/4 + x =18

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If you are a dog lover, having your dog with you may reduce your stress level. Does having a friend with you reduce stress? To e
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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
URGENT URGENT URGENT!!!<br> HELP PLEASE!!
viva [34]

Answer:

Step-by-step explanation:

I think it’s b correct me if I’m wrong

3 0
3 years ago
Read 2 more answers
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