A concrete beam is used as a counterweight. The beam is 6 feet long with an 8-inch by 8-inch square base. If the beam has to be replaced by bags of sand, what is the volume (in ft^3) of the sand that will have the same weight as the beam?

Use the following densities for the materials involved (in pounds per cubic foot):

concrete 150
sand 100

4 cubic feet

Step-by-step explanation:

Step 1:

Length of the concrete beam = 6 feet

Side for the square base = 8 inch = 8/12 feet

So, volume of the concrete beam = 6*(8/12)*(8/12)= 2.67 cu ft

Hence mass of the beam = density of the beam * volume

= 150 * 2.67 = 400 kg

Step 2:

we need the the weight of sand equivalent to the concrete beam.

Mass of the sand bags = den of sand * vol

=> 100 * volume of sand = 400

=> Volume of sand = 400/100 = 4 cu feet.

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11