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3 years ago

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A trough is 12 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1

ft. If the trough is being filled with water at a rate of 14 ft3/min, how fast is the water level rising when the water is 4 inches deep

1 answer:

V125BC [204]3 years ago

8 0

Answer:

$\frac{dh}{dt}=0.5ft$

Step-by-step explanation:

From the question we are told that:

Length $l=12$

Top length $l_t=3ft$

Height $h=1ft$

Rate $R=14 ft3/min$

Water rise $w=4$

Generally the equation for Velocity is mathematically given by

$V=frac{1}{2}wh'(l)\\\\V=frac{1}{2}wh'(12)$

$V=18h'^2$

Therefore

$R=18(2h)(\frac{dh}{dt})$

Where

$h=\frac{3}{4}$

Therefore

$\frac{dh}{dt}=\frac{R}{18(2h)}$

$\frac{dh}{dt}=\frac{14}{18(2.3/4)}$

$\frac{dh}{dt}=0.5ft$

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Step-by-step explanation:

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Step-by-step explanation:

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The probability mass function $p_{X}(x)$ of X is given by

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The expected value of X is

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$Var(Y)=E(Y^2)-(E(Y))^2\\\\Var(Y)=1377-(\frac{73}{2})^2\\\\Var(Y)=1377-\frac{5329}{4}\\\\Var(Y)=\frac{179}{4} \approx 44.75$

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