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Zielflug [23.3K]
3 years ago
14

What is the answer for # 24?A B C or D

Mathematics
1 answer:
Rainbow [258]3 years ago
8 0

The volume of the space not filled by the sphere is the difference between the volume of a cube with edge length 6 inches and the volume of a sphere with radius 3 inches.

<h3><u>Cube</u></h3>

The volume of a cube of edge length s is

... V = s³

When the edge length is 6 in, the volume is

... V = (6 in)³ = 216 in³

<h3><u>Sphere</u></h3>

The volume of a sphere with radius r is

... V = (4/3)π·r³

When the radius is 3 in, the volume is

... V = (4/3)π·(3 in)³ = 36π in³

<h3><u>Space</u></h3>

Then the volume of the space between the cube and the sphere is

... Vcube - Vsphere = 216 in³ - 36π in³ ≈ 102.9 in³ . . . . corresponding to choice C

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What is it and why? having a good explanation would really help me. thx &lt;3
Alexeev081 [22]

Answer:

B. (-4-2)

Step-by-step explanation:

since the numerator is one you can divide -8 by the denominator of each fraction. -8/2 = -4, -8/4 = -2

(-4-2)

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2 years ago
jamie says the value of the expression 1.43 x (-20/39) is close to -0.75. Does jamie's estimate seem reasonable? explain. Jamie'
kherson [118]
Yes. J<span>amie's estimate seem reasonable.
This is because, when we expand the expression </span><span>1.43 x (-20/39) we get;

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Her estimate was -0.75 while the actual answer is </span>-0.73333...
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3 years ago
Wires manufactured for use in a computer system are specified to have resistances between .12 and .14 ohms. The actual measured
White raven [17]

Answer: a) 0.9544996

b) 0.9999366

Step-by-step explanation:

Given : The actual measured resistances of wires produced by company A have a normal probability distribution with mean \mu=0.13 ohm and standard deviation s=0.005 ohm.

Wires manufactured for use in a computer system are specified to have resistances between .12 and .14 ohms.

Let x be the random variable that represents the value of resistance in wires.

Using formula for z-score , z=\dfrac{x-\mu}{s}

The z-value at x= 0.12 will be

z=\dfrac{0.12-0.13}{0.005}=-2

The z-value at x= 0.14 will be

z=\dfrac{0.14-0.13}{0.005}=2

The p-value : P(-2

=0.9772498-(1-P(z

Hence, the probability that a randomly selected wire from company A’s production will meet the specifications = 0.9544996

b) Sample size : n= 4

Using formula for z-score , z=\dfrac{x-\mu}{\dfrac{s}{\sqrt{n}}}

The z-value at x= 0.12 will be

z=\dfrac{0.12-0.13}{\dfrac{.005}{\sqrt{4}}}=-4

The z-value at x= 0.14 will be

z=\dfrac{0.14-0.13}{\dfrac{.005}{\sqrt{4}}}=4

The p-value : P(-4

=0.9999683-(1-P(z

The probability that all four in a randomly selected system will meet the specifications = 0.9999366

3 0
3 years ago
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