Answer:
Step-by-step explanation:
Given that
Group Group One Group Two
Mean 26.00 23.00
SD 2.00 4.00
SEM 0.40 1.21
N 25 11
where group I represents female servers and group II male servers.
We have to calculate confidence interval for 90% for difference in means
The mean of Group One minus Group Two equals 3.00
df = 34
standard error of difference = 0.993
t critical = 2.034 for 90% df 34
Hence confid. interval at 90%
=Mean diff ±2.034 * std error of diff
= (0.98, 5.02)
1200 ... i think this you mean
This graph the answer to your problem.
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.
Answer:
Step-by-step explanation:
Vertex A of the triangle ABC when rotated by 90° counterclockwise about the origin,
Rule to be followed,
A(x, y) → P(-y, x)
Therefore, A(1, 1) → P(-1, 1)
Similarly, B(3, 2) → Q(-2, 3)
C(2, 5) → R(-5, 2)
Triangle given in second quadrant will be the triangle PQR.
If the point P of triangle PQR is reflected across a line y = x,
Rule to be followed,
P(x, y) → X(y, x)
P(-1, 1) → X(1, -1)
Similarly, Q(-2, 3) → Y(3, -2)
R(-5, 2) → Z(2, -5)
Therefore, triangle given in fourth quadrant is triangle XYZ.
