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iren [92.7K]
3 years ago
9

P is the set of natural numbers less than 6.

Mathematics
1 answer:
Wewaii [24]3 years ago
6 0

Answer:

Therefore,

P = { 1 , 2 , 3 , 4 , 5 }

Step-by-step explanation:

Natural numbers:

Natural numbers are those numbers starting from 1 , 2 , 3 , 4 ,......... and so on.

Also it is denoted by 'N'.

Zero does not include a natural number.

Therefore the set of natural numbers less than 6 is 1 , 2 , 3 , 4 , and 5.

Here in the question it is given that

P is the set of natural numbers

less than 6.

∴ P = { 1 , 2 , 3 , 4 , 5 }

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Factorise completely:<br> 1. 3xy - x<br>2. 4a² - 9​
Amanda [17]

Answer:

x(3y - 1) and (2a - 3)(2a + 3)

Step-by-step explanation:

(1)

3xy - x ← factor out x from each term

= x(3y - 1)

(2)

4a² - 9 ← is a difference of squares and factors in general as

a² - b² = (a - b)(a + b) , then

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Mathematically verify the outlier(s) in the data set using the 1.5 rule.
statuscvo [17]

Given:

The data values are:

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

To find:

The outliers of the given data set.

Solution:

We have,

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

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(7, 8, 11, 13, 14, 14), (14, 15, 16, 16, 18, 19)

Divide each parenthesis in two equal parts.

(7, 8, 11), (13, 14, 14), (14, 15, 16), (16, 18, 19)

Now,

Q_1=\dfrac{11+13}{2}

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And

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Q_3=\dfrac{32}{2}

Q_3=16

The interquartile range is:

IQR=Q_3-Q_1

IQR=16-12

IQR=4

The data values lies outside the interval [Q_1-1.5IQR,Q_3+1.5IQR] are known as outliers.

[Q_1-1.5IQR,Q_3+1.5IQR]=[12-1.5(4),16+1.5(4)]

[Q_1-1.5IQR,Q_3+1.5IQR]=[12-6,16+6]

[Q_1-1.5IQR,Q_3+1.5IQR]=[6,22]

All the data values lie in the interval [6,22]. So, there are no outliers.

Hence, the correct option is 4.

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Answer:

8n^2+10n-25

Step-by-step explanation:

it's probably wrong that's what I got tough xp

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