Answer:
The proportion of trees greater than 5 inches is expected to be 0.25 of the total amount of trees.
Step-by-step explanation:
In this problem we have a normal ditribution with mean of 4.0 in and standard deviation of 1.5 in.
The proportion of the trees that are expected to have diameters greater than 5 inches is equal to the probability of having a tree greater than 5 inches.
We can calculate the z value for x=5 in and then look up in a standard normal distribution table the probability of z.

The proportion of trees greater than 5 inches is expected to be 0.25 of the total amount of trees.
Let the number of chemistry books be x and the number of calculus books be y
total number of books in each case is:
x+y=24
x=24-y.....i
Weight of each case:
2707.2/20
=135.36
thus
4.2x+6.0y=135.36..ii substituting i in ii
4.2(24-y)+6y=135.36
100.8-4.2y+6y=135.36
hence
1.8y=34.56
y=19.2
hence
x=24-19.2
x=4.8
thus the number of chemistry books was 4.8*20=96 number of calculus was
19.2*20
=384 books
Step-by-step explanation:
the equation is
c =>
4.50 + 1.70 w = 16