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2 years ago

What is the interquartile range of this data? 6 7 8 9

Mathematics

2 answers:

svetlana [45]2 years ago

6 0

Answer:

Step-by-step explanation:

liq [111]2 years ago

5 0

<h2>The Interquartile Range = 20</h2>

Step 1: Find Q2

Note - Q2 is basically the median of the set

We see that the two numbers in the middle are 28 and 32.

To get the median, find the average of 28 and 32.

28 + 32 = 60, 60/2 = 30

Q2 = 30

Step 2: Find Q1

The median splits the set. So anything before 30, is what we should focus on.

In this case, it's: 12, 16, 20, 24, and 28

Find the median of those five numbers. We get: 20

Q1 = 20

Step 3: Find Q3

Similar to before, focus on the numbers after 30.

In this case, it's: 32, 36, 40, 44, 48

Find the median of those five numbers. We get: 40

Q3 = 40

Step 4: Find the IQR

The IQR (Interquartile Range), as shown in its name, is found by subtracting Q3 and Q1.

Q3 - Q1 = 40 - 20 =20

IQR = 20

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2 years ago

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3 years ago

Read 2 more answers

Triangles P Q R and S T U are shown. Angles P R Q and T S U are right angles. The length of P Q is 20, the length of Q R is 16,

dimaraw [331]

Answer:

$\angle P$

Step-by-step explanation:

Given

$\triangle PRQ = \triangle TSU = 90^o$

$PQ = 20$ $QR = 16$ $PR = 12$

$ST = 30$ $TU = 34$ $SU = 16$

<em>See attachment</em>

Required

Which sine of angle is equivalent to $\frac{4}{5}$

Considering $\triangle PQR$

We have:

$\sin(P) = \frac{QR}{PQ}$ --- i.e. opposite/hypotenuse

So, we have:

$\sin(P) = \frac{16}{20}$

Divide by 4

$\sin(P) = \frac{4}{5}$

Hence:

$\angle P$ is correct

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3 years ago

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Step-by-step explanation:

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