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Ostrovityanka [42]
3 years ago
5

What is the interquartile range (IQR) of this data set?

Mathematics
2 answers:
Andreas93 [3]3 years ago
7 0

Answer:

11 is the median of the entire data set

6 is the median of the lower half and 16 is the median of the upper half data scores...

Subtract 16 from 6 gives 10 so the answer is C) 10

Sever21 [200]3 years ago
6 0
<h3>Answer:  Choice C)  10</h3>

=============================================================

Explanation:

The first step is to sort the data from smallest to largest. Luckily, that's already been done for us.

Next, we find the median. This is the middle most number of the set. In this case, the median is 11 since we have 3 values below it (2,6,7) and 3 values above it (15,16,17).

Let

L = {2,6,7}

U = {15,16,17}

where L and U are the lower and upper sets respectively.

In other words, anything in set L is lower than the median while anything in set U is above the median. The median itself (11) is not part of either set.

Notice how 6 is the midpoint of set L, while 16 is the midpoint of set U.

This means Q1 = 6 and Q3 = 16

Therefore,

IQR = Q3 - Q1

IQR = 16 - 6

IQR = 10

which is why choice C is the final answer.

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Prove that (81/16)^-3/4 ×[(25/9)^-3/2 ÷ (5/2)^-3]=1​
andreev551 [17]

Answer:

First write them in positive exponent form

(16/81)¾ × [ (9/25)^3/2 ÷ (2/5)³ ]

(2⁴×¾)/ (3⁴×¾) × [ (3² × ^3/2) / (5² ×^3/2) ÷ 2³/5³)

Simplify the terms

2³/3³ × ( 3³ / 5³ ÷ 2³/5³)

Solve the terms in the bracket

2³/3³ × (3³/5³×5³/2³)

You will get

2³/3³ × 3³/2³ = 1

They will cancel each other so the answer will be 1

Hope this helps.

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Answer: it’s 135

Step-by-step explanation:

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Don't trust those link my guy
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What are the values of x and y
Gelneren [198K]

Unfortunately, there is no way I can find the value of X and Y because there is no question for it.

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