Answer:1728
Step-by-step explanation:
The marginal distribution for gender tells you the probability that a randomly selected person taken from this sample is either male or female, regardless of their blood type.
In this case, we have total sample size of 714 people. Of these, 379 are male and 335 are female. Then the marginal probability mass function would be
![\mathrm{Pr}[G = g] = \begin{cases} \dfrac{379}{714} \approx 0.5308 & \text{if }g = \text{male} \\\\ \dfrac{335}{714} \approx 0.4692 & \text{if } g = \text{female} \\\\ 0 & \text{otherwise} \end{cases}](https://tex.z-dn.net/?f=%5Cmathrm%7BPr%7D%5BG%20%3D%20g%5D%20%3D%20%5Cbegin%7Bcases%7D%20%5Cdfrac%7B379%7D%7B714%7D%20%5Capprox%200.5308%20%26%20%5Ctext%7Bif%20%7Dg%20%3D%20%5Ctext%7Bmale%7D%20%5C%5C%5C%5C%20%5Cdfrac%7B335%7D%7B714%7D%20%5Capprox%200.4692%20%26%20%5Ctext%7Bif%20%7D%20g%20%3D%20%5Ctext%7Bfemale%7D%20%5C%5C%5C%5C%200%20%26%20%5Ctext%7Botherwise%7D%20%5Cend%7Bcases%7D)
where G is a random variable taking on one of two values (male or female).
The ABE and DBC are both made of DBE and and ABD or EBC so if the firsts are equal than these are also equal
Answer:
Simple random should be the technique used
<span><span>1.
</span>The music hall has 20 boxes :
=> on thurs 2/5 were occupied
=> on friday 2/4 were occupied
=> and on saturday 8/10 were occupied.
Let’s find out which day has the most occupied space of the music hall.
2/5 + 2/4 + 8/10 = 6/20 + 7/20 + 10/20
Thurs 2/5 were occupied
=> 6/20 = 0.3
=> 20 * 0.3 = 6 boxes
Friday 2/4 were occupied
=> 7/20 = 0.35
=> 20 * 0.35 = 7 boxes
Saturday 8/10 were occupied.
=> 10/20 = 0.5
=> 20 * .5 = 10 boxes
Thus, during Saturday has the most number of occupied boxes.</span>
