Using the median concept, it is found that the third quartile is of is of $1,052.
<h3>What is the median of a data-set?</h3>
The median of the data-set separates the bottom half from the upper half, that is, it is the 50th percentile.
In this problem, the ordered data-set is:
$379, $498, $619, $777, $895, $1,052, $1,256.
Hence, the median of $777 divides the data-set into two halfs:
- The first half is: $379, $498, $619.
- The second half is: $895, $1,052, $1,256.
Hence, the third quartile, which is the median of the second half, is of $1,052.
More can be learned about the median concept at brainly.com/question/25215461
Answer:
Step-by-step explanation:
its a because you go up 200
If the length of one side of the cube 4 units then the area of one its face is 4 × 4 = 16 square units. From the net, we can see that there are six equal faces and so we get the total surface area is 6 × 16 = 96 square units.
F: R → R is given by, f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1
So, f(1.2) = f(1.9), but 1.2 ≠ 1.9
f is not one-one
Now, consider 0.7 ε R
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ε R such that f(x) = 0.7
So, f is not onto
Hence, the greatest integer function is neither one-one nor onto.
The answer was quite complicated but I hope it will help you.
