It is defined as the difference between the largest and smallest values in the middle 50% of a set of data<span>. To compute an </span>interquartile range<span> using this definition, first remove observations from the lower quartile. Then, remove observations from the upper quartile.</span>
Answer:
y= -x-2
Step-by-step explanation:
y=ax+b
x= -2; y=0 => 0=-2a+b; b=2a
x= -1; y= -1 => -1=-a+b => b=a-1
2a=a-1=> a= -1; b= -2
=> y= -x-2
Answer:
y = -2x^2(x - 3)
Step-by-step explanation:
<em><u>Preliminary Remark</u></em>
If a cubic is tangent to the x axis at 0,0
Then the equation must be related to y = a*x^2(x - h)
<em><u>(3,0)</u></em>
If the cubic goes through the point (3,0), then the equation will become
0 = a*3^2(3 - h)
0 = 9a (3 - h)
0 = 27a - 9ah
from which h = 3
<em><u>From the second point, we get</u></em>
4 = ax^2(x - 3)
4 = a(1)^2(1 - 3)
4 = a(-2)
a = 4 / - 2
a = -2
<em><u>Answer</u></em>
y = -2x^2(x - 3)
Answer: Yes
Step-by-step explanation: The answer of 1400 is correct.
What I did is I found the area of the figure assuming it was a rectangle, then I subtracted the corners that were removed.
1800 - 150 - 100 - 150 = 1400
