(trying to isolate/get x by itself in the equation)
5(x + 6) = 50 Distributive property [distribute 5 into (x + 6)]
5x + 30 = 50 Subtraction [subtract 30 on both sides of the equation]
5x = 20 Division [divide 5 on both sides]
x = 4
Subtraction then division, the 2nd option
Let
x = first integer
y = second integer
z = third integer
First equation: x + y + z = 194
Second equation: x + y = z + 80
Third equation: z = x - 45
Let's find the values of x, y and z.
Substitute 3rd eq to 1st eq:
x + y + x - 45 = 194
2x + y = 45 + 194
y = -2x + 239
Plug in both we have solved for y and the 3rd eq to the 2nd eq to find x
x + (-2x + 239) = (x - 45) + 80
x - 2x - x = -45 + 80 - 239
-2x = -204
x = -204/-2
x = 102
Solving for y,
y = -2(102) + 239
y = 35
Solving for z,
z = 102 - 45
z = 57
Rewrite the boundary lines <em>y</em> = -1 - <em>x</em> and <em>y</em> = <em>x</em> - 1 as functions of <em>y </em>:
<em>y</em> = -1 - <em>x</em> ==> <em>x</em> = -1 - <em>y</em>
<em>y</em> = <em>x</em> - 1 ==> <em>x</em> = 1 + <em>y</em>
So if we let <em>x</em> range between these two lines, we need to let <em>y</em> vary between the point where these lines intersect, and the line <em>y</em> = 1.
This means the area is given by the integral,
The integral with respect to <em>x</em> is trivial:
For the remaining integral, integrate term-by-term to get
Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line <em>y</em> = 1 and (0, -1)), so its area is 1/2*4*2 = 4.
The interquartile range (IQR) is 15.
To find the IQR, you must:
1) Find the median
2) Split the data set in two at the median.
3) Find the medians from both the groups you made in Step 2
4) Label the smaller one Quartile 1 and the larger one Quartile 3 (Hint: The median is Quartile 2)
5) Subtract: Quartile 3 - Quartile 1