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77julia77 [94]
2 years ago
11

a cone with radius of 3 feet and a height of 5 feet is placed on top of a cylinder as shown find the volume of the total figure

in terms
Mathematics
1 answer:
Darina [25.2K]2 years ago
5 0

Answer: Volume of the cone = 47.12 cubic feet.

Step-by-step explanation:

You didn't finish the question so Idid the best with what I had, you also didn't include the image that I think was supposed to be included.

To solve for the full volume including the cylinder, use the cylinder's volume formula.

Volume of a cylinder = height (pi) (r^2) and then add it to the 47.12.

Volume of a cone is pi(r^2)(h/3)

plug and chug!

h stands for height and r stands for radius.

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What is the slope of the line shown below?<br><br> (9, 1) and (-3, -7
Dima020 [189]

Answer:

The slope is 4

Step-by-step explanation:

I think it's right

3 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
dexar [7]

Answer:

The value of the constant C is 0.01 .

Step-by-step explanation:

Given:

Suppose X, Y, and Z are random variables with the joint density function,

f(x,y,z) = \left \{ {{Ce^{-(0.5x + 0.2y + 0.1z)}; x,y,z\geq0  } \atop {0}; Otherwise} \right.

The value of constant C can be obtained as:

\int_x( {\int_y( {\int_z {f(x,y,z)} \, dz }) \, dy }) \, dx = 1

\int\limits^\infty_0 ({\int\limits^\infty_0 ({\int\limits^\infty_0 {Ce^{-(0.5x + 0.2y + 0.1z)} } \, dz }) \, dy } )\, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y }(\int\limits^\infty_0 {e^{-0.1z} } \, dz  }) \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0{e^{-0.2y}([\frac{-e^{-0.1z} }{0.1} ]\limits^\infty__0 }) \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}([\frac{-e^{-0.1(\infty)} }{0.1}+\frac{e^{-0.1(0)} }{0.1} ])  } \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}[0+\frac{1}{0.1}]  } \, dy  }) \, dx =1

10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2y} }{0.2}]^\infty__0  }) \, dx = 1

10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2(\infty)} }{0.2}+\frac{e^{-0.2(0)} }{0.2}]   } \, dx = 1

10C\int\limits^\infty_0 {e^{-0.5x}[0+\frac{1}{0.2}]  } \, dx = 1

50C([\frac{-e^{-0.5x} }{0.5}]^\infty__0}) = 1

50C[\frac{-e^{-0.5(\infty)} }{0.5} + \frac{-0.5(0)}{0.5}] =1

50C[0+\frac{1}{0.5} ] =1

100C = 1 ⇒ C = \frac{1}{100}

C = 0.01

3 0
2 years ago
Draw a histogram for the intervals 16-18,19-21, 22-24, and 25-27 using the following data: 26, 16, 27
Kitty [74]

Answer with explanation:

Class Interval             Variate         Frequency

16-18                              16                     1

19-21                               0                     0

22-24                               0                    0

25-27                              26,27              2

8 0
3 years ago
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Svetach [21]
(4,5)(2,1)
slope = (1 - 5) / (2 - 4) = -4/-2 = 2

(-2,3)(-4,4)
slope = (4 - 3) / (-4 - (-2) = 1/(-4 + 2) = -1/2

2 and -1/2 are negative reciprocals of each other....therefore, ur lines are perpendicular <==


5 0
3 years ago
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2 years ago
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