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2 years ago

Homework Jim Thompson! PART 2

Mathematics

1 answer:

BlackZzzverrR [31]2 years ago

6 0

Problem 2

Part (a)

The 3D shape formed when rotating around the y axis forms a pencil tip

The shape formed when rotating around the x axis is a truncated cone turned on its side.

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Part (b)

Check out the two diagrams below.

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Problem 3

Answer: Choice A and Choice C

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Explanation:

Think of stacks of coins. Let's say we had 2 stacks of 10 quarters each. The quarters are identical, so they must produce identical volumes. Those sub-volumes then add up to the same volume for each stack. Now imagine one stack is perfectly aligned and the other stack is a bit crooked. Has the volume changed for the crooked stack? No, it hasn't. We're still dealing with the same amount of coins and they yield the same volume.

For more information, check out Cavalieri's Principle.

With all that in mind, this leads us to choice C. If the bases are the same, and so are the heights, then we must be dealing with the same volumes.

On the other hand, if one base is wider (while the heights are still equal) then the wider based block is going to have more volume. This leads us to choice A.

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I’m plz help fast thanks 16

nataly862011 [7]

Answer:

1/2 <3/4

Step-by-step explanation:

1/2 vs 3/4

We need a common denominator of 4

1/2*2/2 vs 3/4

2/4 vs 3/4

Since the denominators are the same, we compare the numerators

2 <3 so

2/4 < 3/4

1/2 <3/4

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4 years ago

During a winter cold spell the temperature change was -1.2 f per hour for a period of 4.5 hours which expressions can be used to

Gala2k [10]

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3 years ago

Hi! please help me out its due very sooon :((

melomori [17]

Answer:

y= 1/3x -6

Step-by-step explanation:

you just have to plug the given values into the slope intercept formula:

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where M is the slope and B is the y-intercept

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3 years ago

Read 2 more answers

Consider a figure in a coordinate plane. For each of the transformations below, first transform the figure as stated. Then rever

prohojiy [21]

Answer:

transformation is a function that takes points on the plane and maps them to other points on

the plane. Transformations can be applied one after the other in a sequence where you use the

image of the first transformation as the preimage for the next transformation.

Find the image for each sequence of transformations.

 Using geometry software, draw a triangle and label the

vertices A, B, and C. Then draw a point outside the

triangle and label it P.

Rotate △ABC 30° around point P and label the image as

△A′B′C ′. Then rotate △A′B′C ′ 45° around point P and

label the image as △A″B″C ″. Sketch your result.

 Make a conjecture regarding a single rotation that will map △ABC to △A″B″C″.

Check your conjecture, and describe what you did.

 Using geometry software, draw a triangle and label the

vertices D, E, and F. Then draw t

Step-by-step explanation:

7 0

3 years ago

The following observations were made on fracture toughness of a base plate of 18% nickel maraging steel (in ksi √in, given in in

Grace [21]

Answer:

A 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].

Step-by-step explanation:

We are given the following observations that were made on fracture toughness of a base plate of 18% nickel maraging steel below;

68.6, 71.9, 72.6, 73.1, 73.3, 73.5, 75.5, 75.7, 75.8, 76.1, 76.2, 76.2, 77.0, 77.9, 78.1, 79.6, 79.8, 79.9, 80.1, 82.2, 83.7, 93.4.

Firstly, the pivotal quantity for finding the confidence interval for the standard deviation is given by;

P.Q. = $\frac{(n-1) \times s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation = $\sqrt{\frac{\sum (X - \bar X^{2}) }{n-1} }$ = 5.063

$\sigma$ = population standard deviation

n = sample of observations = 22

Here for constructing a 90% confidence interval we have used One-sample chi-square test statistics.

So, 90% confidence interval for the population standard deviation, $\sigma$ is ;

P(11.59 < $\chi^{2}__2_1$ < 32.67) = 0.90 {As the critical value of chi at 21 degrees

of freedom are 11.59 & 32.67}

P(11.59 < $\frac{(n-1) \times s^{2} }{\sigma^{2} }$ < 32.67) = 0.90

P( $\frac{ 11.59}{(n-1) \times s^{2}}$ < $\frac{1}{\sigma^{2} }$ < $\frac{ 32.67}{(n-1) \times s^{2}}$ ) = 0.90

P( $\frac{(n-1) \times s^{2} }{32.67 }$ < $\sigma^{2}$ < $\frac{(n-1) \times s^{2} }{11.59 }$ ) = 0.90

90% confidence interval for $\sigma^{2}$ = [ $\frac{(n-1) \times s^{2} }{32.67 }$ , $\frac{(n-1) \times s^{2} }{11.59 }$ ]

= [ $\frac{21 \times 5.063^{2} }{32.67 }$ , $\frac{21 \times 5.063^{2} }{11.59 }$ ]

= [16.48 , 46.45]

90% confidence interval for $\sigma$ = [ $\sqrt{16.48}$ , $\sqrt{46.45}$ ]

= [4.06 , 6.82]

Therefore, a 90% confidence interval for the standard deviation of the fracture toughness distribution is [4.06, 6.82].

5 0

3 years ago