Suppose you want to determine the area bounded by two curves denoted by functions f(x) and g(x).
We find the points of intersection by setting f(x) = g(x). Then we determine which curve is the upper curve in the integral and which is the lower curve in the integral. Suppose the points of intersection are x=1 and x=3.
If f(x) is the upper curve then the integral looks like
Integral (1 to 3) [f(x) - g(x)] dx
If g(x) is the upper curve the the integral looks like
Integral (1 to 3) [g(x) - f(x)] dx
Tuesday, March 31, 2015
Saturday, March 28, 2015
Check out my profiles at Learnivore.com and Fiverr.com. I can help those with their math, for a modest but fair fee. I can offer you my nearly 15 years of experience tutoring students in math subjects from elementary school to college.
https://learnivore.com/instructors/mathematics-help-with-tutor-with-15-years-experience
https://www.fiverr.com/mathtutorkk
https://learnivore.com/instructors/mathematics-help-with-tutor-with-15-years-experience
https://www.fiverr.com/mathtutorkk
Saturday, March 21, 2015
Suppose you wish to obtain the exact value of sin(165). We know exact values of the following angles on the unit circle : 0, 30, 45, 60, 90, 120, 135, 150,. 180, 210, 225, 240, 270, 300, 315, 330, 360.
If we pick two angles from the unit circle that add to 165, we can obtain the exact value for sin(165). The reason is we can substitute those values in for this formula.
sin(a + b) = sin(a)cos(b)+cos(a)sin(b)
Notice that 120 + 45 = 165 and both of those values are on the unit circle
sin(120 + 45)= sin(120)cos(45) + cos(120)sin(45)
= (sqrt(3)/2)(sqrt(2)/2) + (-1/2)(sqrt(2)/2)
= sqrt(6)/4 - sqrt(2)/4
= (sqrt(6) - sqrt(2))/4
If we pick two angles from the unit circle that add to 165, we can obtain the exact value for sin(165). The reason is we can substitute those values in for this formula.
sin(a + b) = sin(a)cos(b)+cos(a)sin(b)
Notice that 120 + 45 = 165 and both of those values are on the unit circle
sin(120 + 45)= sin(120)cos(45) + cos(120)sin(45)
= (sqrt(3)/2)(sqrt(2)/2) + (-1/2)(sqrt(2)/2)
= sqrt(6)/4 - sqrt(2)/4
= (sqrt(6) - sqrt(2))/4
Wednesday, March 18, 2015
Here's a great little video on hypothesis testing. It really explains the concept simply, in easy to understand terminology.
The Most Simple Introduction to Hypothesis Testing
The Most Simple Introduction to Hypothesis Testing
Thursday, March 12, 2015
Remember there is a difference between finding that Z value for a confidence interval and the Z (critical) value for tests and probability.
For example, suppose you want a 95 percent confidence interval for the sample mean. The formula is mean +/- standard error which is Z(alpha/2)standard deviation/square root(n). alpha is 1-.95 = .05. So the Z value is 1.96.
Now if we want an x value that 95% of the data falls below, then we need Z(.05) which is 1.645.
Be careful to understand the difference between the two.
For example, suppose you want a 95 percent confidence interval for the sample mean. The formula is mean +/- standard error which is Z(alpha/2)standard deviation/square root(n). alpha is 1-.95 = .05. So the Z value is 1.96.
Now if we want an x value that 95% of the data falls below, then we need Z(.05) which is 1.645.
Be careful to understand the difference between the two.
Friday, March 6, 2015
Notice the graph of a quadratic function f(x) = x^2 + 3x + 8 is a parabola that opens up. What if you know how to find the vertex but forget how to determine whether the vertex is the maximum or minimum point, therefore not knowing that is opens up or down. You can take the second derivative and set equal to zero to determine whether the function is concave up or concave down.
The first derivative f'(x) = 2x + 3
Second derivative f"(x) = 2
Notice the second derivative is positive for all values of x therefore the function is concave up, therefore opening up.
The first derivative f'(x) = 2x + 3
Second derivative f"(x) = 2
Notice the second derivative is positive for all values of x therefore the function is concave up, therefore opening up.
Tuesday, March 3, 2015
When finding the area of a sector of a circle take the angle of the sector in degrees and divide by 360. Take the result and multiply by Pi times r^2.
For example, suppose the radius of a circle is 6 and the angle of the sector is 45 degrees. Therefore the area of the sector is
(45/360)Pi(6)(6) = (1/9)(36)Pi = 14.14
For example, suppose the radius of a circle is 6 and the angle of the sector is 45 degrees. Therefore the area of the sector is
(45/360)Pi(6)(6) = (1/9)(36)Pi = 14.14
Subscribe to:
Posts (Atom)