# Math Help - Help with story problems!

1. ## Help with story problems!

1. In a vontorversial school vote, only 19 percent of the teachers vote yes, whereas 95 percent of the students voted yes. If all teachers and students voted and 91 percent of the voters voted yes, what is the student to teacher ratio?

2. The density of the wood of a pine tree is 35lbs/ft^3. The tree is 72 feet high and the diameter is 3 feet. Ten percent of the weight of the tree is contained in the branches and foliage. If the trunk of the tree is a right circular cylindar, estimate the weight fo the tree above the ground.

3. In 1991, the separation between college football goalpost uprights was reduced from 223 feet, 4inches to 18 feet, 6 inches. To the nearest thenth of a percent what was the reduction?

4. A squater with side length 6 inches is rvolved about a diagonal as an axis. Find the volume of the solid created by this revolution.

Any help would be great!
Thanks!

2. Originally Posted by epetrik
1. In a vontorversial school vote, only 19 percent of the teachers vote yes, whereas 95 percent of the students voted yes. If all teachers and students voted and 91 percent of the voters voted yes, what is the student to teacher ratio?
Consider all who voted yes,

$0.19t+0.95s = 0.91(t+s)$

$0.19t+0.95s = 0.91t+0.91s$

$0.04s = 0.72t$

$\frac{s}{t} = \dots$

3. Originally Posted by epetrik
2. The density of the wood of a pine tree is 35lbs/ft^3. The tree is 72 feet high and the diameter is 3 feet. Ten percent of the weight of the tree is contained in the branches and foliage. If the trunk of the tree is a right circular cylindar, estimate the weight fo the tree above the ground.
Pretty much just arithmetic. The volume of the tree trunk is [/tex]72(3.14)(3/2)^2= 508.68 cubic feet and so has weight (508.68)(35)= 17803.8 pounds. Ten percent of that is 1780.38 pounds so the total weight of trunk and branches and foliage is 17803.8+ 1780.38= 19584.18 pounds.

3. In 1991, the separation between college football goalpost uprights was reduced from 223 feet, 4inches to 18 feet, 6 inches. To the nearest thenth of a percent what was the reduction?
I don't believe for a minute that football goalposts uprights used to be more that 223 feet apart! You mean 23 feet 4 inches, right?
The amoumnt of reduction is 23 feet 4 inches - 18 feet 6 inches= 22 feet 16 inches - 18 feet 6 inches= 4 feet 10 inches= 4.83 feet while "23 feet 4 inches" is 23.33 feet . The percent is given by $\frac{4.83}{23.33}= 0.207$ or 20.76%.

4. A squater with side length 6 inches is rvolved about a diagonal as an axis. Find the volume of the solid created by this revolution.
If a square has sides 6 inches long, then it has diagonals of length $6\sqrt{2}$ inches long. You can think of the solid as two cones with base radius and height $3\sqrt{2}$ inches, half the length of the diagonals. The formula for volume of a cone of radius r with height h is $\frac{1}{3}\pi r^2 h= \frac{1}{3}\pi (18\sqrt{2})$= 26.6 cubic inches. Since the figure is two such cones, its volume is 53.3 cubic inches.

Any help would be great!
Thanks![/QUOTE]

4. Thanks for the help!!!!!