First locate the median $Q_{2}$ at the centre of the data arranged in ascending order. Then locate the lower and upper quartiles $Q_{1}$ and $Q_{3}$ located at the centre of the data to the left and right of the median.

Note that if any of the above are not whole values then they are the average of the values either side of the centre.

rearrange data in ascending order

15 16 ↓17 17 18 21 22 ↓23 25

↑

$Q_{2}$ = 18

$Q_{1}$ = $\frac{16+17}{2}$ = 16.5

$Q_{3}$ = $\frac{22+23}{2}$ = 22.5

IQR = $Q_{3}$ - $Q_{1}$ = 22.5 - 16.5 = 6

4 0

2 years ago

There are 336 students attending summer

ra1l [238]

Answer:

126 are boys

Step-by-step explanation:

6 0

2 years ago

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50 POINTS PLEASE HELP

Rainbow [258]

first simplify the numerator (x+3)(x-2)=x^2-2x+3x-6=x^2+x-6

now factor the numerator (x+3)(x-2)

now find the restriction 3x-6=0 3x=6 x=2

now factor the denominator 3(x-2)

((x+3)(x-2))/3(x-2)

now cross off the (x-2)

and you are left with (x+3)/3

ANSWERS:

The restriction is X=2

The simplified fraction is (x+3)/3

7 0

2 years ago

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Tell whether each relationship is a function.

Serhud [2]

Answer:

3 is a function.

1 and 2 are not.

Step-by-step explanation:

1 and 2 are not functions because we can draw a vertical line which passes through more than one point on the graphs.

You cant do this on graph 3 so it is a function.

8 0

2 years ago