Given that there are three spotted Pacific banana slugs for every four non spotted Pacific banana slug(s) in Arcata's community forest, let xx represent the probability of drawing a single spotted Pacific banana slug from a single draw of both varieties of spotted and non spotted slugs.

The probability of selecting a single spotted Pacific banana slug will be .

Step-by-step explanation:

There are three spotted Pacific banana slugs for every four non spotted Pacific banana slugs.

There are three  spotted Pacific banana slugs in every group of 7 Pacific banana slugs.

There are four non spotted Pacific banana slugs in every group of 7 Pacific banana slugs.

Therefore the probability of selecting a single spotted Pacific banana slug will be

= .

The probability of selecting a single non spotted Pacific banana slug will be = .

81 seats

step-by-step explanation:

if you multiply 0.65 times 125 then you get 81.5 seats. well since you can't have half of a seat you have to round down and get 81 seats in the no children zones. that leave 44 seats outside of the no children zone.

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]