Here is a note: the sum of the 2 shorter sides have to be greater than the longest side. Sorry that i cant help more.
![\dfrac\partial{\partial y}\left[e^{2y}-y\cos xy\right]=2e^{2y}-\cos xy+xy\sin xy](https://tex.z-dn.net/?f=%5Cdfrac%5Cpartial%7B%5Cpartial%20y%7D%5Cleft%5Be%5E%7B2y%7D-y%5Ccos%20xy%5Cright%5D%3D2e%5E%7B2y%7D-%5Ccos%20xy%2Bxy%5Csin%20xy)
![\dfrac\partial{\partial x}\left[2xe^{2y}-y\cos xy+2y\right]=2e^{2y}+y\sin xy](https://tex.z-dn.net/?f=%5Cdfrac%5Cpartial%7B%5Cpartial%20x%7D%5Cleft%5B2xe%5E%7B2y%7D-y%5Ccos%20xy%2B2y%5Cright%5D%3D2e%5E%7B2y%7D%2By%5Csin%20xy)
The partial derivatives are not equal, so the equation is not exact.
X weight 131
---- = -------------
18.8% 100%
Cross multiply and get 24.6 pounds
Answer:
There is not sufficient evidence to support the claim that the technique performs differently than the traditional method.
Step-by-step explanation:
The null hypothesis is:

The alternate hypotesis is:

The test statistic is:

In which X is the sample mean,
is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
A researcher used the technique with 260 students and observed that they had a mean of 94 hours. Assume the standard deviation is known to be 6.
This means, respectively, that 
The test-statistic is:



The pvalue is:
2(P(Z < -2.69))
P(Z < -2.69) is the pvalue of Z when X = -2.69, which looking at the z-table, is 0.0036
2*(0.0036) = 0.0072
0.0072 < 0.01, which means that the null hypothesis is accepted, that is, there is not sufficient evidence to support the claim that the technique performs differently than the traditional method.