The interquartile range (IQR) can be found by finding the difference of the upper and lower quartiles
If you go to sc and slide to filter you’ll find one that solves the problem
Answer:
3sqrt(2) ................
Answer:
Diameter 16, radius 7, circumference 37.68, diameter 8
Step-by-step explanation:
Largest area will be in a circle with the largest radius. So let's find all radii.
d = 8, so r = 4;
r = 7;
circumference = 2pi*r = 37. 68, so r = 6
diameter 18, so r = 8
Answer:
Step-by-step explanation:
y = 6x − 4
y = 5x − 3
A: chose a value for x(input) and get the out put (y) because the equation already in the slope intercept mode:y=mx+b
x y=6x-4 y=5x-3
1 y=2 y=2
2 y=8 y=7
0 -4 -3
slope :
y=6x-4 slope is 6 y intercept(0,-4)
y=5x-3 slope is 5 y intercept (0,-3)
solution of the pair of equation is the point of intersection :
6x-4=5x-3
6x-5x=-3+4
x=1 and y=6x-4 ⇒ y=6-4=2
<h2>(1,2)</h2>