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)

## Friday, February 26, 2016

## Sunday, February 21, 2016

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

## Wednesday, February 17, 2016

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.

## Tuesday, February 9, 2016

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

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