Here's an example of solving for dy/dx using implicit differentiation


y^2 + 2xy^3 - 3x^2 = 14

2y(dy/dx) + 2x(3y^2)(dy/dx) + 2y^3 - 6x = 0

2y(dy/dx) + 6xy^2(dy/dx) = 6x - 2y^3

dy/dx(2y +6xy^2) = 6x - 2y^3

dy/dx = (6x-2y^3)/(6xy^2 + 2y)

dy/dx = (3x-y^3)/(3xy^2 + y)

For Normal distribution with mean μ = 400
and standard deviation σ = 100 find probability
that x is between 340 and 714: P( 340 < x < 714)

0.3427

0.5462

0.7249

0.9140

We have mean = 400 and standard deviation of 100, now we want P(X > 619), Z = (619 - 400)/100 = 2.19

Z(2.19) = .9857, so we take 1- .9857 since the probability is greater than, so we get .0143
For the next question, mean is 400 and standard deviation is 100, want: P( 340 < x < 714)
Z = (340 - 400)/100 = -0.6 and Z = (714 - 400)/100 = 3.14

So now we find Z(3.14) and Z(-0.6) and subtract them to get the probability in between. .9992 - .2743 = .7249

A quick note:

For the test just comparing two means when population standard deviation is not known and sample size is small, you can do a t-test . The formula for the test statistic is t = (x1-bar - x2-bar)/sqrt(s1^2/n1 +s2^2/n2), then compare to the critical value t , found using any t-distribution chart.

we calculate Chi-square by getting the sum(Observed - Expected)^2/(Expected)



You get the expected in each cell by taking row total times column total divided by overall total
Expected Low Drunk Driving and Low Under Age is (22)(20)/50 = 8.8
Expected High Drunk Driving and Low Under Age is (28)(20)/50 = 11.2
Expected Low Drunk Driving and High Under Age is (22)(30)/50 = 13.2
Expected High Drunk Driving and High Under Age is (28)(30)/50 = 16.8

so Chi-square statistic is (8-8.8)^2/8.8 + (14-13.2)^2/13.2 + (12 -11.2)^2/11.2 + (16- 16.8)^2/16.8 = 0.216

Now we get the p-value, with df of (row-1)(column - 1). There are 2 rows and 2 columns in the contingency table, so df = 1. I use this site.... http://www.socscistatistics.com/pvalues/chidistribution.aspx
The p-value is 0.642105. Since the p-value > .05

Suppose you want lim x-> a   [f(x) - f(a)](x-a) where a = 2


get [f(x) - f(a)](x-a) with a = 2, then take the limit as x -> 2
f(x) - f(a)/(x-a) =( 4.5x^2 -3x + 2 - (4.5(2)^2 - 3(2) + 2)]/(x- 2)

= (4.5x^2 - 3x + 2 - 14)/(x-2)
= (4.5x^2 -3x -12)/(x-2)
= (4.5x + 6)(x-2)/(x-2)
= 4.5x + 6

Now we take lim x-> 2 (4.5x + 6) = 4.5(2) + 6 = 15

Suppose you have an angle and you want to know where its terminal side lies. How do you go about figuring this out?  It's simply seeing where the angle ends. For example, if you have the 4 quadrants, first ending at 90 degrees, 2nd at 180, 3rd at 270 and 4th at 360, then a 225 degree angle would have its terminal side in the third quadrant since 225 falls between 180 and 270.

If the angle is negative in value, simply add 360 each time until you get a number between 0 and 360 and follow the same guidelines as above.

For example, an angle of -556 degrees would have it's terminal side in quadrant 2. Add 360 to -556 and you get -196, then add  360 again to get 164, which is a second quadrant angle

Here's an easy way to simplify higher powers of i.

First know the first 3 powers of i

i^1 = sqrt(-1)
i^2 = -1
i^3 = -i


To get any power of i greater than 4, simply divide by 4 and the integer value remainder gives the answer. It will be one of the above.

Example:

i^45....   45 divided by 4 is 11 with a remainder of 1, so the answer is sqrt(-1)

i^202..... 202 divided by 4 is 50 with a remainder of 2, so the answer is -1

i^79.... 79 divide by 4 is 19 with a remainder of 3, so the answer is -i