Find the interquartile range for a data set having the five-number summary: 3.5, 12, 19.1, 25.8, 31.8 A. 28.3 B. 13.8 C. 15.6 D.
2 answers:
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Answer:
Step-by-step explanation:
- Re-write the expression:
÷
- 31/6 ÷ 7/10 = 31/6 × 10/7
- 310/42 = 155/21
I hope this helps!
These are indeed quite a lot of exercises, but a lot of them are almost identical - only small computations will change. So, I'm glad to help you with one exercise from every cathegory, but I encourage you to solve the others on your own.
<h2>Exercise 1</h2>You can call four consecutive integers as
So, the sum of the first and third is
We want this quantity to be six less than the largest, i.e.
So, the equality is
Subtract x from both sides:
Subtract 2 from both sides:
So, the consecutive integers are
In fact, the sum of the first and third is , which is indeed six less than the largest:
If you call the first odd number , the next consecutive odd numbers will be
In fact, we have to count skipping two's, because we only want odd integers. From here, you go on like exercise 1: you write the largest (which is x+6), and set it to be two more than the sum of the other three (x, x+2 and x+4)
<h2>Exercise 3</h2>By the same logic of exercise 2, two consecutive even integers are , assuming that x is even.
So, you set the equation as usual: the smaller (which is x) is 26 less than three times the larger (which means 3(x+2)-26)
<h2>Exercise 4 to 8</h2>These are all pretty identical to exercise 1: you start by listing three or four consecutive integers:
and then you translate the request of each exercise accordingly. Remember that expressions like "three times the second number" means that you have to multiply: , while expression like "six more than the first" or "thirteen less than the first" imply adding/subtracting:
or
.
A multiple of 5 can be written as , for some integer k.
So, three consecutive multiples of 5 are
We want these three numbers to have a sum of 75. So, we have
So, the three numbers are