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agasfer [191]
3 years ago
7

Kelly has a rectangular fish aquarium that measures 18 inches long, 8 inches wide, and 12 inches tall.

Mathematics
1 answer:
BigorU [14]3 years ago
8 0

Answer:

a) The maximum amount of water the aquarium can hold is 1728 in³.

b) The cover has to be 624 in² big.

Step-by-step explanation:

a) The maximum volume of water that the aquarium can have is the same as the volume of the aquarium:

V=L×W×H=18 in × 8 in × 12 in= 1728 in³

b) The protective covering is used on the four walls of the aquarium. We have to find the areas of the four walls:

One of them is given by the multiplication of the width and the height:

A₁= W×H= 8 in × 12 in=96 in²

The other area is given by the multiplication of the length and the height:

A₂= L×H= 18 in × 12 in= 216 in²

As the fish aquarium is rectangular the area of the walls is given by:

A_T=2A_1+2A_2=2×96 in²+2×216 in²=624 in²

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a spherical storage tank has a diameter of 24 ft and is full of water. how many delivery trucks with a cylindrical tanks measuri
lbvjy [14]

Diameter of the spherical storage tank = 24 ft

So, radius of the tank = \frac{24}{2}=12 ft

Volume of the spherical tank = 4/3 * PI * r³

= \frac{4}{3}\times3.14\times12\times12\times12

= 7234.56 cubic feet

Diameter of cylindrical tank = 4 ft

So, radius = \frac{4}{2}=2 ft

Height of cylindrical tank = 10 ft

Volume of cylindrical tank = PI * r² * h

= 3.14\times2\times2\times10=125.60 cubic feet

So, the number of cylindrical tanks that can be filled from the spherical tank are =

\frac{7234.56}{125.60}=57.6

Hence, 57 cylindrical tanks can be completely filled from the big spherical tank.


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3 years ago
Hayley cut a 10 2/3 foot rope into pieces that are each 8/9 of a foot long. How many pieces of rope did she cut??
Ivenika [448]
The answer is 7.5 or 15/2
4 0
3 years ago
Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z
kompoz [17]

a.

f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}

is a proper joint density function if, over its support, f is non-negative and the integral of f is 1. The first condition is easily met as long as C\ge0. To meet the second condition, we require

\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}

b. Find the marginal joint density of X and Y by integrating the joint density with respect to z:

f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz

\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}

Then

\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy

\approx\boxed{0.12886}

c. This probability can be found by simply integrating the joint density:

\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz

\approx\boxed{0.012262}

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3 years ago
Help me this is due in 10 minutes
Nina [5.8K]

i am pretty sure the answer is 8.

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3 years ago
Read 2 more answers
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