Answer:
{x| –2 ≤ x < 5}
Step-by-step explanation:
There is a box function plotted on the graph.
The function is g(x) = –⌊x⌋ + 3.
Now, we know that a box function represents a step graph having horizontal segments that are each 1 unit long. The left end of each segment is a closed circle. The right end of each segment is an open circle.
It is given that the left-most segment of the given graph goes from (-2,5) to (-1,5) and the rightmost segment goes from (4,-1) to (5,-1).
So, for the left most segment the domain is -2 ≤ x < -1
And for the right most segment the domain is 4 ≤ x < 5
Therefore, the total domain of g(x) will be {x| –2 ≤ x < 5} (Answer)
Answer: 17.75
Step-by-step explanation:
The interquartile range(IQR) is the 3rd quartile - the 1st quartile.
How to get quartiles:
First get the median:
3.5, 10.4, 16, 21.7, 27.7
10.4, 16, 21.7
16
Then find the median of the first half of data(3.5, 10.4)
(3.5+10.4)/2 = 6.95
Then find the median of the last half of data(21.7, 27.7)
(21.7+27.7)/2 = 24.7
Then to get the IQR subtract 6.95 from 24.7 to get 17.75
Hope it helps <3
Answer:
Creo que la respuesta sería B: 2 horas y 15 minutos.
Step-by-step explanation:
If you mean "factor over the rational numbers", then this cannot be factored.
Here's why:
The given expression is in the form ax^2+bx+c. We have
a = 3
b = 19
c = 15
Computing the discriminant gives us
d = b^2 - 4ac
d = 19^2 - 4*3*15
d = 181
Note how this discriminant d value is not a perfect square
This directly leads to the original expression not factorable
We can say that the quadratic is prime
If you were to use the quadratic formula, then you should find that the equation 3x^2+19x+15 = 0 leads to two different roots such that each root is not a rational number. This is another path to show that the original quadratic cannot be factored over the rational numbers.
Answer:
(2,4) (3,1) (5,-5)
Step-by-step explanation: