By the divergence theorem, the surface integral given by
(where the integral is computed over the entire boundary of the surface) is equivalent to the triple integral
where
is the volume of the region
bounded by
.
You have
![\implies \nabla\cdot\mathbf F=\dfrac{\partial}{\partial x}[x^2y]+\dfrac{\partial}{\partial y}[xy^2]+\dfrac{\partial}{\partial z}[4xyz]=8xy](https://tex.z-dn.net/?f=%5Cimplies%20%5Cnabla%5Ccdot%5Cmathbf%20F%3D%5Cdfrac%7B%5Cpartial%7D%7B%5Cpartial%20x%7D%5Bx%5E2y%5D%2B%5Cdfrac%7B%5Cpartial%7D%7B%5Cpartial%20y%7D%5Bxy%5E2%5D%2B%5Cdfrac%7B%5Cpartial%7D%7B%5Cpartial%20z%7D%5B4xyz%5D%3D8xy)
and so the integral reduces to
Answer: did u get an answer?
Step-by-step explanation:
Answer:
easy so first 190-15=175 175 lbs. is the light heavyweight class weight limit
Step-by-step explanation:
Answer: 0.8238
Step-by-step explanation:
Given : Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with 
and 
.
Let x denotes the scores on a certain intelligence test for children between ages 13 and 15 years.
Then, the proportion of children aged 13 to 15 years old have scores on this test above 92 will be :-
![P(x>92)=1-P(x\leq92)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{92-106}{15})\\\\=1-P(z\leq })\\\\=1-P(z\leq-0.93)=1-(1-P(z\leq0.93))\ \ [\because\ P(Z\leq -z)=1-P(Z\leq z)]\\\\=P(z\leq0.93)=0.8238\ \ [\text{By using z-value table.}]](https://tex.z-dn.net/?f=P%28x%3E92%29%3D1-P%28x%5Cleq92%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cleq%5Cdfrac%7B92-106%7D%7B15%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq%20%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq-0.93%29%3D1-%281-P%28z%5Cleq0.93%29%29%5C%20%5C%20%5B%5Cbecause%5C%20P%28Z%5Cleq%20-z%29%3D1-P%28Z%5Cleq%20z%29%5D%5C%5C%5C%5C%3DP%28z%5Cleq0.93%29%3D0.8238%5C%20%5C%20%5B%5Ctext%7BBy%20using%20z-value%20table.%7D%5D)
Hence, the proportion of children aged 13 to 15 years old have scores on this test above 92 = 0.8238
