It’s B) -0.8 because it’s going down and it’s pretty strong.
X is a variable, a variable you have to divide to cancel out
if the sphere has a diameter of 5, then its radius is half that, or 2.5.
![\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=2.5 \end{cases}\implies V=\cfrac{4\pi (2.5)^3}{3}\implies V=\cfrac{62.5\pi }{3} \\\\\\ V\approx 65.44984694978736\implies V=\stackrel{\textit{rounded up}}{65.45}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20sphere%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D2.5%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B4%5Cpi%20%282.5%29%5E3%7D%7B3%7D%5Cimplies%20V%3D%5Ccfrac%7B62.5%5Cpi%20%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20V%5Capprox%2065.44984694978736%5Cimplies%20V%3D%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7B65.45%7D)
Answer:


So then we can conclude that we expect the middle 95% of the values within 18 and 30 minutes for this case
Step-by-step explanation:
For this case we can define the random variable X as the amount of time it takes her to arrive to work and we know that the distribution for X is given by:

And we want to use the empirical rule to estimate the middle 95% of her commute times. And the empirical rule states that we have 68% of the values within one deviation from the mean, 95% of the values within two deviations from the mean and 99.7 % of the values within 3 deviations from the mean. And we can find the limits on this way:


So then we can conclude that we expect the middle 95% of the values within 18 and 30 minutes for this case
|Ω| = 6 - number of all results
A = {1, 3} → |A| = 2 - number of (a) results
P(A) = 2/6 = 1/3