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fgiga [73]
3 years ago
7

How many 2-inch by 3-inch by 8-inch bricks will you need to build a (uniformly wide) brick walk with the shape shown in the figu

re below? (Lay the bricks so that the largest face is up. Assume that a = 10 ft, b = 6 ft, and c = 22 ft.
Mathematics
1 answer:
GalinKa [24]3 years ago
5 0
Here we find the volume of the bricks, will be given by:
volume=length*width*height
v=2×3×8=48 in³

volume of the bricks is given by:
1 ft= 12 inches
thus our dimensions will be:
width=10*12=120 in
height=6*12=72 in
length=22*12=264 in
V=120×72×264
V=2,280,960 in³
Thus the number of bricks required will be:
2280960/48
=47520 in³



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Answer:

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Step-by-step explanation:

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To do this, we simply calculate the probability of each container.

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P(W) = \frac{n(W)}{n(S)} = \frac{1}{5} = 0.20

P(X) = \frac{n(X)}{n(S)} = \frac{1}{5} = 0.20

P(Y) = \frac{n(Y)}{n(S)} = \frac{1}{5} = 0.20

P(Z) = \frac{n(Z)}{n(S)} = \frac{1}{5} = 0.20

So, the probability model is:

\begin{array}{cccccc}{x} & {V} & {W} & {X} & {Y} & {Z} & P(x) & {0.20} & {0.20} & {0.20} & {0.20} & {0.20} \ \end{array}

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2 years ago
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To prove:

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