To calculate the IQR, we need the radian, which is the middle number of each column, namely (86,82) for each exam.
Above the middle number, we have 5 values. The middle value (of the upper part) is Q3, which is on the third line, or (92,85).
Below the middle number, we also have 5 values. The middle (of the lower half) is Q1, which is on the 9th line (80,78)
The IQR is the difference between Q3 and Q1, namely
(92-80, 85-78) = (12, 7)
Since 12>7, we conclude that the IQR of the mid-term is higher than the final.
Answer:
approximately 5.6
Step-by-step explanation:
Cross multiply
16a=90
a=5.625
Answer:
(0, ∞)
Step-by-step explanation:
A good place to start is by visualizing what the graph looks like on a number line.
For x > 0, it is an open circle at x=0, and shading to the right extending to infinity.
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So, the left end of the interval is 0, but 0 is not included in the interval.
The right end of the interval is infinity, but there is no such number, so "infinity" is not included in the interval.
"Not included" means you use round brackets ( ) for the corresponding end of the interval. ("Included" would mean you use square brackets [ ].)
So, the interval 0 < x < ∞ is written in interval notation as ...
(0, ∞)
The value of f(-3) is 5.
In order to find this answer, we are going to input -3 in for x (this is what the notation f(-3) means).
f(x)=| x-2 |
f(-3)=| -3 -2 |
f(-3)=| -5 |
Now we have to take the absolute value.
f(-3) = 5
No, it is not a true proportion.
