For the answer to the question above, I'll show the computation below.
<span>V(cube) = s^3
SA(cube with no top) = 5s^2
s^3 = 5s^2
s = 5
V = 125 sq ft = 216,000 sq in
V(cannonball) = (4/3)πr^3
SA(cannonball) = 4πr^3
(4/3)πr^3 = 4πr^2
r = 3
V = 36π sq in
Packing of spheres in a cube is yielding up to 74% with remainder as space between the spheres. It can be achieved by placing the 2nd row of spheres in the crevices between adjoining first row spheres.
216,000 / 36π = 6,000/π
(6000/π)*.74 = <u><em>1414 spheres can be fit "in" each box</em></u>.
</span>
Answer:
Step-by-step explanation:
find the slope 1/2
then write it as y-4=1/2(x+6)
then u get y=1/2x+7
Estimation can help u find the area of a rectangle or square because if you get an answer and a bar above the last number u can just estimate the answer to make stuff simple
Answer:
from the t-distribution table, at df = 7 and t = 2.23
Lies p-values [ 0.05 and 0.025 ]
Hence;
0.025 < p-value < 0.05
Step-by-step explanation:
Given that;
= 6.5 gpm
μ = 5 gpm
n = eight runs = 8
standard deviation σ = 1.9 gpm
Test statistics;
t = (
- μ) / ![\frac{s}{\sqrt{n} }](https://tex.z-dn.net/?f=%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%20%7D)
we substitute
t = (6.5 - 5) / ![\frac{1.9}{\sqrt{8} }](https://tex.z-dn.net/?f=%5Cfrac%7B1.9%7D%7B%5Csqrt%7B8%7D%20%7D)
t = 1.5 / 0.67175
t = 2.23
the degree of freedom df = n-1 = 8 - 1
df = 7
Now, from the t-distribution table, at df = 7 and t = 2.23
Lies p-values [ 0.05 and 0.025 ]
Hence;
0.025 < p-value < 0.05
Answer:
1/12.
Step-by-step explanation:
6 sides and 2 sides. 2 possible outcomes for each side of the dice