Answer:
41 degrees
Step-by-step explanation:
If the pole is perpendicular to the ground, the angle it creates is 90 degrees. Barry has already lifted it 49 degrees, so
90 - 49 = 41
It must be lifted another 41 degrees.
If you begin with the graph for f(x) and then substitutde x+3 for x, the effect on the graph will be to move the whole graph 3 units to the left.
Given that (-9,-1) is on the graph of f(x), and that we are to move the entire graph 3 units to the left, then the coordinate -9 becomes -9-3, or -12: (-12,-1).
Vertex form is y=a(x-h)^2+k, so we can rearrange to that form...
y=3x^2-6x+2 subtract 2 from both sides
y-2=3x^2-6x divide both sides by 3
(y-2)/3=x^2-2x, halve the linear coefficient, square it, add it to both sides...in this case: (-2/2)^2=1 so
(y-2)/3+1=x^2-2x+1 now the right side is a perfect square
(y-2+3)/3=(x-1)^2
(y+1)/3=(x-1)^2 multiply both sides by 3
y+1=3(x-1)^2 subtract 1 from both sides
y=3(x-1)^2-1 so the vertex is:
(1, -1)
...
Now if you'd like you can commit to memory the vertex point for any parabola so you don't have to do the calculations like what we did above. The vertex of any quadratic (parabola), ax^2+bx+c is:
x= -b/(2a), y= (4ac-b^2)/(4a)
Then you will always be able to do a quick calculation of the vertex :)
Answer:
mean = 1 power failure
variance = 1 (power failure)²
Step-by-step explanation:
Since the mean is computed as
mean = E(x) = ∑ x * p(x) for all x
then for the random variable x=power failures , we have
mean = ∑ x * p(x) = 0 * 0.4 + 1* 0.3 + 2*0.2 + 3* 0.1 = 1 power failure
since the variance can be calculated through
variance = ∑[x-E(x)]² * p(x) for all x
but easily in this way
variance = E(x²) - [E(x)]² , then
E(x²) = ∑ x² * p(x) = 0² * 0.4 + 1²* 0.3 + 2²*0.2 + 3²* 0.1 = 2 power failure²
then
variance = 2 power failure² - (1 power failure)² = 1 power failure²
therefore
mean = 1 power failure
variance = 1 power failure²
B don’t quote me if I’m wrong
