649373837 okay good well i hoped that work for u
Answer: c).
Both options are true. You can see that the whiskers of data set 2 (The lines extending on either side of the box plots) represent a much larger range of data than data set 1, and that the median in data set 2 (the line down the middle of the boxes) is greater than data set 1.
Hope this helps!
Sin2x = 2sinxcosx;
cos2x = (cosx)^2 - (sinx)^2;
tan2x = (sin2x)/(cos2x);
cosx = 5/13 from formula (sinx)^2 + (cosx)^2 = 1;
=> sin2x = 120/169;
.................................
Answer:
Solving for x, the answer is x = 1 - y/3
Hope this helps :D
Speed of the plane: 250 mph
Speed of the wind: 50 mph
Explanation:
Let p = the speed of the plane
and w = the speed of the wind
It takes the plane 3 hours to go 600 miles when against the headwind and 2 hours to go 600 miles with the headwind. So we set up a system of equations.
600
m
i
3
h
r
=
p
−
w
600
m
i
2
h
r
=
p
+
w
Solving for the left sides we get:
200mph = p - w
300mph = p + w
Now solve for one variable in either equation. I'll solve for x in the first equation:
200mph = p - w
Add w to both sides:
p = 200mph + w
Now we can substitute the x that we found in the first equation into the second equation so we can solve for w:
300mph = (200mph + w) + w
Combine like terms:
300mph = 200mph + 2w
Subtract 200mph on both sides:
100mph = 2w
Divide by 2:
50mph = w
So the speed of the wind is 50mph.
Now plug the value we just found back in to either equation to find the speed of the plane, I'll plug it into the first equation:
200mph = p - 50mph
Add 50mph on both sides:
250mph = p
So the speed of the plane in still air is 250mph.
