Um A :)
hope this helps :)))))))))))))
The temperature changed a total of 14° over the course of 7 hours with an average change of 2° per hour drop.
The mean is usually the best measure of central tendency to use when your data distribution is continuous and symmetrical, such as when your data is normally distributed. However, it all depends on what you are trying to show from your data.
1. Let a and b be coefficients such that

Combining the fractions on the right gives



so that

2. a. The given ODE is separable as

Using the result of part (1), integrating both sides gives

Given that y = 1 when x = 1, we find

so the particular solution to the ODE is

We can solve this explicitly for y :


![\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|](https://tex.z-dn.net/?f=%5Cln%7Cy%7C%20%3D%20%5Cln%5Cleft%7C%5Csqrt%5B3%5D%7B%5Cdfrac%7B5x%7D%7B2x%2B3%7D%7D%5Cright%7C)
![\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}](https://tex.z-dn.net/?f=%5Cboxed%7By%20%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B5x%7D%7B2x%2B3%7D%7D%7D)
2. b. When x = 9, we get
![y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}](https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B45%7D%7B21%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B15%7D7%7D%20%5Capprox%20%5Cboxed%7B1.29%7D)
Answer:
b 129%
Step-by-step explanation: