You want to find the value of x for which the area under the curve to the left of x is 0.6. One way to do that is to create the cumulative distribution function (CDF) for the given PDF, then see where it is equal to 0.6.
Doing that, we find a = 5.
1a) f(x) = I x+2 I. This is a piece-wise graph ( V form)
x = 0 →f(x) =2 (intercept y-axis)
x = -2→f(x) = 0 (intercept x-axis)
x = -3→f(x) = 1 (don't forget this is in absolute numbers)
x = -4→f(x) = 2 (don't forget this is in absolute numbers)
Now you can graph the V graph
1b) Translation: x to shift (-3) units and y remains the same, then
f(x-3) = I x - 3 + 2 I = I x-1 I
the V graph will shift one unit to the right, keeping the same y. Proof:
f(x) = I x-1 I . Intercept x-axis when I x-1 I = 0, so x= 1
Answer:
x = 4
Step-by-step explanation:
3 x ( 4x -7 ) = 27
3 x 4x = 12x
3 x -7 = -21
12x -21 = 27
12x -21 + 21 = 27 + 21
12x = 48
12x/12 = 48/12
x = 4
