Answer:
The 90% confidence interval to estimate the average difference in scores between the two courses is (-1.088, 20.252).
Step-by-step explanation:
We have the standard deviation for the differences, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 8 - 1 = 7
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 7 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 1.8946
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 9.582 - 10.67 = -1.088
The upper end of the interval is the sample mean added to M. So it is 9.582 + 10.67 = 20.252
The 90% confidence interval to estimate the average difference in scores between the two courses is (-1.088, 20.252).
The graph of g(x) = -x^2 is a reflection in the x-axis of the graph of f(x) = x^2. Both graphs have one x-intercept as both graphs have their vertices at the origin, (0,0).
Answer:
y = 3x + 2
There slope is 3 and intercept is 2.
<u>First find what are the 2 pair that given by slope =3</u>
First consider A. (3, 11) and B. (3, 9)
Tan α = \fraction(11-9)(3-3)
It is undefine.
Next consider A. (3, 11) and BC (5,15)
Tan α = \fraction(15-11)(5-3)
=2
Next consider A. (3, 11) and D. (2,4)
Tan α = \fraction(11-4)(3-2)
=7
Next consider B (3, 9) and C. (5,15)
Tan α = \fraction(15-9)(5-3)
=3
So this is ordered pair is generated from slope =3
Step-by-step explanation:
You should have drawn1 - x-axis and y-axis in light pencil.2 - graphed a down-facing parabola with the top of the frown on the y-axis at y = 2. It should be crossing the x-axis at ±√2. This should be in dark pencil or another color.3 - In dark pencil or a completely new color, draw a rectangle with one of the horizontal sides sitting on top of the x-axis and the other horizontal side touching the parabola at each of the top corners of the rectangle. The rectangle will have half of its base in the positive x-axis and the other half on the negative x-axis. It should be split right down the middle by the y-axis. So each half of the base we will say is "x" units long. So the whole base is 2x units long (the x units to the right of the y-axis, and the x units to the left of the y-axis) I so wish I could draw you this picture... In the vertical direction, both vertical edges are the same length and we will call that y. The area that we want to maximize has a width 2x long, and a height of y tall. So A = 2xy This is the equation we want to maximize (take derivative and set it = 0), we call it the "primary equation", but we need it in one variable. This is where the "secondary equation" comes in. We need to find a way to change the area formula to all x's or all y's. Since it is constrained to having its height limited by the parabola, we could use the fact that y=2 - x2 to make the area formula in only x's. Substitute in place of the "y", "2 - x2" into the area formula. A = 2xy = 2x(2 - x2) then simplify A = 4x - 2x3 NOW you are ready to take the deriv and set it = 0 dA/dx = 4 - 6x2 0 = 4 - 6x2 6x2 = 4 x2 = 4/6 or 2/3 So x = ±√(2/3) Width remember was 2x. So the width is 2[√(2/3)]Height is y which is 2 - x2 = 2 - 2/3 =4/3
