21 fence posts. i dont know how to draw a picture on this site, but you covert 16in to ft and then divide 32 by 1.5
Answer:
a) 16%
b) 2.5%
Step-by-step explanation:
a)
The mean is 70 with standard deviation(SD) of 3 and you are asked to find out the percentage of staff that have <67(70-3 inch= mean - 1 SD) inch size, which means 1 SD below the mean (<-1 SD). Using 68-95-99.7 rule, you can know that 68% of the population is inside 1 SD range from the mean ( -1 SD to + 1 SD).
To put it on another perspective, there are 32% of the population that have < -1 SD and > +1 SD value. Assuming the distribution is symmetrical, then the value of < - 1 SD alone is 32%/2= 16%
b)
The question asks how many populations have size >76 inches, or mean + 2 SD (70+3*2 inch).
You can also solve this using 68-95-99.7 rule, but you take 95% value as the question asking for 2 SD instead. Since 95% of population is inside 2 SD range from the mean ( -2 SD to + 2 SD), so there are 5% of population that have < -2 SD and > +2 SD value. Assuming the distribution is symmetrical, then the value of > +2 SD alone is 5%/2= 2.5%
Answer:
x = 62
Step-by-step explanation:
62 and x are alternate exterior angles and alternate exterior angles are equal when the lines are parallel
Answer:
It is 91% more likely that the tree was atmost 500 yards from the river.
<h3>Step-by-step explanation:</h3>
We are given with distance and height of 100 young trees near a river.
From that table, in total there are 55 trees which grow more than 3 ft during the year.
And among those 55 trees, 50 trees are atmost 50 yards from river.
Hence it is ≈91% more likely that the tree was atmost 50 yards from the river.
Answer: A. In 2015 there will be more seagulls than chickadees.
Step-by-step explanation: if you look at the chart you can see the seagull population keeps growing and the chickadee population keeps decreasing.
