Answer:
(C) They have the same coefficient of variation
Step-by-step explanation:
The coefficient of variation (CV) is calculated using the formula:
Where 
represents standard deviation and 
represents the mean.
Bob's average weight is 200 pounds with a standard deviation of 16 pounds
This indicates that 
.
Thus, his coefficient of variation is
Mary's average weight is 125 pounds, with a standard deviation of 10 pounds.
This implies 
Therefore, her coefficient of variation is
Since both have the same coefficient of variation, the accurate response is.
(C) They have the same coefficient of variation
Answer:
Step-by-step explanation:
a. Create a direction field for the specified differential equation
b. By observing the direction field, comment on the behavior of the solutions as t becomes large.
The solutions seem to oscillate
All solutions appear to approach the function y0(t)=4
All solutions seem to converge to the function y0(t)=0
All solutions appear to have negative slopes eventually and thus decrease indefinitely
All solutions seem to have positive slopes eventually and therefore increase without limit
C
As t approaches infinity
All solutions seem to exhibit positive slopes eventually and thus decrease indefinitely
The solutions seem to gradually approach the function y0(t)=0
All solutions appear to eventually have negative slopes leading to decrease without bounds
All solutions seem to converge to the function y0(t)=4
The results are oscillatory
-6, as 96 divided by 16 equals 6. However, because it starts with -16, this results in a negative answer of -6!
Answer:
Step-by-step explanation:
Review the provided matrix
![A=\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%26-2%263%5C%5C2%2617%260%5C%5C3%2622%268%5Cend%7Barray%7D%5Cright%5D)
Let matrix B be defined as
![B=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Db_%7B11%7D%26b_%7B12%7D%26b_%7B13%7D%5C%5Cb_%7B21%7D%26b_%7B22%7D%26b_%7B23%7D%5C%5Cb_%7B31%7D%26b_%7B32%7D%26b_%7B33%7D%5Cend%7Barray%7D%5Cright%5D)
It is stated that
![\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%26-2%263%5C%5C2%2617%260%5C%5C3%2622%268%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Db_%7B11%7D%26b_%7B12%7D%26b_%7B13%7D%5C%5Cb_%7B21%7D%26b_%7B22%7D%26b_%7B23%7D%5C%5Cb_%7B31%7D%26b_%7B32%7D%26b_%7B33%7D%5Cend%7Barray%7D%5Cright%5D)
By comparing the corresponding elements from both matrices, we derive
Consequently, the needed values are .
Only amend the first item. Please refer to the graph in the attached file. 1) a rectangle with A(3,3), B(3,6), C(7,6), D(7,3) having adjacent sides (in green) 2) a parallelogram with A(2,0), B(3,2) having nonperpendicular sides (in blue) C(6,3), D(5,1) 3) a square with A(3,3), B(2,5), C(4,6), D(5,4) (in red) 4) a rhombus with A(2,-2), B(3,0), C(4,-2), D(3,-4) having adjacent sides (in black)
