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Archy [21]
3 years ago
7

the diameter of a cylindrical water tank is 8ft and its height is 11 ft what is the volume of the tank?

Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
6 0

Using the formula for the volume of a right cylinder, we can determine the volume of the water tank.

Formula for the volume of a right cylinder: V = π (d /2)^2

h = π * (8/2)^2  ·11 ≈ 552.92

Therefore, the volume of the tank is approximately 552.92 ft.

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How many times do 12 go into 209?
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209/12 is 17.41, so 12 goes into 209 17 times.
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If P(AnB)=2/3 and P(B)=3/4, what is P(A I B)? A.35/36 B.15/16 C.24/25 D.8/9
astraxan [27]

Answer:

\large\boxed{D.\ \dfrac{8}{9}}

Step-by-step explanation:

\text{We know:}\\\\P(A|B)=\dfrac{P(A\ \cap\ B)}{P(B)}\\\\\text{We have:}\\\\P(A\ \cap\ B)=\dfrac{2}{3},\ P(B)=\dfrac{3}{4}.\\\\\text{Substitute:}\\\\P(A|B)=\dfrac{\frac{2}{3}}{\frac{3}{4}}=\dfrac{2}{3}:\dfrac{3}{4}=\dfrac{2}{3}\cdot\dfrac{4}{3}=\dfrac{8}{9}

6 0
3 years ago
Lamar sells another broker's $800,000 listing for $785,000 at a commission rate of 6.5%. Lamar has a 40% commission split with h
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6 0
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Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
3 years ago
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