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.

6 0

3 years ago

Which of the following is the simplified form of 6x−5−7x≥−12?

Anna35 [415]

-x>-7 is the answer the line
_

5 0

3 years ago

Find the mean and the standard deviation for each set of values. Show your work. 15,17,19,20,14,23,12

vekshin1

1. Mean.

$\overline{x}=\dfrac{15+17+19+20+14+23+12}{7}=\dfrac{120}{7}\approx\boxed{17.14}$

2. <span>Standard deviation.

First we have to square each value and calculate the mean of that squares:

$\overline{x^2}=\dfrac{15^2+17^2+19^2+20^2+14^2+23^2+12^2}{7}=\boxed{\dfrac{2144}{7}}$

so the variance:

$s^2=\overline{x^2}-\overline{x}^2=\dfrac{2144}{7}-\left(\dfrac{120}{7}\right)^2\approx\boxed{12.41}$

and the standard deviation:

$s=\sqrt{s^2}=\sqrt{12.41}\approx\boxed{3.52}$
</span>

3 0

3 years ago