Kiloliters are bigger so if they are bigger they have to be kiloliters right>:)
10kL<span> = 100hL, so </span>10kL<span> > </span><span>50hL</span>
Answer:
750
Step-by-step explanation:
(a) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.11 + 0.52 + 0.19 = 0.82
(b) P(X ≥ 1) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.52 + 0.19 + 0.12 + 0.06 = 0.89
(c) µ = 0×0.11 + 1×0.52 + 2×0.19 + 3×0.12 + 4×0.06 = 1.5
(d) σ² = (0²×0.11 + 1²×0.52 + 2²×0.19 + 3²×0.12 + 4²×0.06) - µ² = 1.07
σ = √(σ²) ≈ 1.03
we know that
the volume of a solid oblique pyramid is equal to

where
B is the area of the base
h is the height of the pyramid
in this problem we have that
B is a square

where
<u>
</u>
so


substitute in the formula of volume
![V=\frac{1}{3}*x^{2}*(x+2)\\ \\V=\frac{1}{3}*[x^{3} +2x^{2}]\ cm^{3}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B3%7D%2Ax%5E%7B2%7D%2A%28x%2B2%29%5C%5C%20%5C%5CV%3D%5Cfrac%7B1%7D%7B3%7D%2A%5Bx%5E%7B3%7D%20%2B2x%5E%7B2%7D%5D%5C%20cm%5E%7B3%7D)
therefore
<u>the answer is</u>
![V=\frac{1}{3}*[x^{3} +2x^{2}]\ cm^{3}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B3%7D%2A%5Bx%5E%7B3%7D%20%2B2x%5E%7B2%7D%5D%5C%20cm%5E%7B3%7D)