If sinα=1/4 and it is in q2 it means that the x coordinate is negative.
The triangle has a height of one and a hypotenuse of 4 so:
x=-√(4^2-1)
x=-√15
so tanα=-1/√15
tanα=-√(1/15)
tanα≈ -0.258
10% of the 50 were red.
10% * 50
(10/100) * 50
(1/10) * 50
= 5
5 were red.
<span>Standard deviation of first data set = 5879.1
Standard deviation of second data set = 14768.78
The second data set is more variable.
The basic definition of standard deviation is the square root of the mean of the squares of the difference from the mean. It's a bit of a mouthful, but easy enough to do. For the first data set, first calculate the mean.
(28995 + 37534 + 31361 + 27087 + 20966 + 37741) / 6 = 30614
Now calculate the square of the differences from the mean
(28995 - 30614)^2 = 2621161
(37534 - 30614)^2 = 47886400
(31361 - 30614)^2 = 558009
(27087 - 30614)^2 = 12439729
(20966 - 30614)^2 = 93083904
(37741 - 30614)^2 = 50794129
And now the average of the squares
(2621161 + 47886400 + 558009 + 12439729 + 93083904 +50794129) / 6 = 34563888.67
And finally, take the square root to get the standard deviation.
sqrt(34563888.67) = 5879.1
Now for the second data set of western states. First, the mean
(72964 + 70763 + 101510 + 62161 + 66625 + 54339) / 6 = 71393.67
Now the squares of the differences
(72964 - 71393.67)^2 = 2465946.778
(70763 - 71393.67)^2 = 397740.4444
(101510 - 71393.67)^2 = 906993533.4
(62161 - 71393.67)^2 = 85242133.78
(66625 - 71393.67)^2 = 22740181.78
(54339 - 71393.67)^2 = 290861655.1
And the average of the squares is 218116865.2
Finally, the square root of the average is 14768.78
So the standard deviation of the 2nd data set is 14768.78
And since the standard deviation of the 2nd data set is larger than the standard deviation of the 1st data set, that means that the 2nd data set is more variable.</span>
A) 6x -5y = 5
B) 3x + 5y = 4
Adding the equations:
9x = 9
x = 1
A) 6*1 -5y = 5
A) 6 - 5 = 5y
A) 5y = 1
y = 1/5 = .2
The right answer is -8/3
please see the attached picture for full solution
hope it helps
