Answer:
The correct option is (b).
Step-by-step explanation:
If X
N (µ, σ²), then
, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z
N (0, 1).
The distribution of these z-variate is known as the standard normal distribution.
The mean and standard deviation of the active minutes of students is:
<em>μ</em> = 60 minutes
<em>σ </em> = 12 minutes
Compute the <em>z</em>-score for the student being active 48 minutes as follows:

Thus, the <em>z</em>-score for the student being active 48 minutes is -1.0.
The correct option is (b).
80/100 x 30 = 24 because you just multiply percents
notice, the circle is missing 1/4, so the area of it is just 3/4 of the whole area of the circle.
![\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=8 \end{cases}\implies A=\pi 8^2\implies A=64\pi \\\\\\ \stackrel{whole}{\cfrac{4}{4}}-\stackrel{one~quarter}{\cfrac{1}{4}}=\cfrac{3}{4}~\hfill \cfrac{3}{4}\cdot 64\pi \implies 48\pi \implies \stackrel{\pi =3.14}{150.72} \\\\\\ ~\hspace{34em}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D8%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%208%5E2%5Cimplies%20A%3D64%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bwhole%7D%7B%5Ccfrac%7B4%7D%7B4%7D%7D-%5Cstackrel%7Bone~quarter%7D%7B%5Ccfrac%7B1%7D%7B4%7D%7D%3D%5Ccfrac%7B3%7D%7B4%7D~%5Chfill%20%5Ccfrac%7B3%7D%7B4%7D%5Ccdot%2064%5Cpi%20%5Cimplies%2048%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Cpi%20%3D3.14%7D%7B150.72%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chspace%7B34em%7D)
Answer:
Unlikely
Step-by-step explanation:
The probability of picking a number divisible by 5 is 2/12 or 1/6. So its pretty unlikely.