What is the interquartile range of this set of data? 15,19,20,25,31,38,41

Mathematics

1 answer:

cupoosta [38]3 years ago

7 0

Answer: 19.

Step-by-step explanation:

1: Place your numbers in numerical order from smallest to largest; 15,19,20,25,31,38,41.

2. Find your median in the set of numbers; 15,19,20,25,31,38,41.

3. Find Q1 and Q3 in your set of numbers by placing parentheses around the median you have just found. (15,19,20),25,(31,38,41.)

4. In this instance, you need to find the medians of the two number sets in parentheses; (15,19,20), 19 being Q1, and (31,38,41), 38 being Q3.

5. Using the interquartile range formula; Q3-Q1= IQR, plug in your numbers.

38-19=19, making the answer 19.

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This trapezium is drawn on a centimetre grid. Find the area of the trapezium.

aivan3 [116]

Answer:

20 unit²

Step-by-step explanation:

A trapezium is given to us on the grid and we need to find out the area of the trapezium . In order to find the area , we need to find the measure of the parallel sides and the distance between the parallel sides.

From the grid :-

$\rm\implies Side_1 = 7 \ units$

$\rm\implies Side_2 = 3 \ units$

$\rm\implies \perp \ Distance =4 \ units$

Now here we got the two parallel sides of the trapezium and the distance between the two parallel sides. Now we can find the area as ,

$\rm\implies Area_{Trapezium}= \dfrac{1}{2}\times ( s_1 + s_2) \times \perp \ Distance \\\\\rm\implies Area = \dfrac{1}{2} \times ( 7 + 3 ) \times 4 \ unit^2 \\\\\rm\implies Area = \dfrac{1}{2} \times ( 10) \times 4 \ unit^2 \\\\\rm\implies\boxed{\rm Area = 20 \ unit^2}$