All categories

3 years ago

Which solid of revolution is produced by rotating the shape below 360° about the given axis?

Mathematics

1 answer:

maxonik [38]3 years ago

5 0

<h3>Answer: Choice A</h3>

Explanation:

When we rotate a rectangle around an axis of symmetry, we end up with a cylinder. Think of a revolving door or a turbine. The flat door moves through 3D space to produce or carve out a 3D cylinder.

This figure is composed of two rectangles, so we'll end up with two cylinders. They will both share the same center, so we'll have a sideways two-layer wedding cake structure when everything is said and done. This points to choice A as the final answer.

You might be interested in

Solve the problem, who will solve this will be marked as brainliest

laila [671]

<h2>I hope it's helpful for you</h2>

5 0

3 years ago

Read 2 more answers

What is the percent of change from 74 to 112? rounded to the nearest percent

ruslelena [56]

Answer:

The percent of change from 74 to 112 is 51%.

Step-by-step explanation:

Given:

The percent of change from 74 to 112.

Now, the difference between the numbers $112-74=38$ .

Then, calculating the percentage of the result by $74$ .

$\frac{38}{74}\times100$

$=0.5135\times100$

$=51.35%$ .

Round to the nearest percent of $51.35%$ is $51%$ .

Therefore, the percent of change from 74 to 112 is 51%.

4 0

3 years ago

There are 5 movies that Bilal wants to see over the break. He only has time to see 3 of the movies. How many different groups of

mihalych1998 [28]

Answer:

Bilal has no friends

Step-by-step explanation:

And so do I

pain

5 0

3 years ago

Read 2 more answers

Are these angles Supplementary or Complementary

gregori [183]

Answer:

they are suplementary

Step-by-step explanation:

because no right angle and less then 90 degrees

7 0

3 years ago

Read 2 more answers

7.2 Given a test that is normally distributed with a mean of 100 and a standard deviation of 10, find: (a) the probability that

kompoz [17]

Answer:

The probability that a single score drawn at random will be greater than 110

P( X > 110) = 0.1587

b)

The probability that a sample of 25 scores will have a mean greater than 105

P( x> 105) = 0.0062

c)

The probability that a sample of 64 scores will have a mean greater than 105

P( x⁻> 105) = 0.002

d)

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = 0.9544

Step-by-step explanation:

a)

Given mean of the Normal distribution 'μ' = 100

Given standard deviation of the Normal distribution 'σ' = 10

Let 'X' be the random variable of the Normal distribution

let 'X' = 110

$Z = \frac{x-mean}{S.D} = \frac{110-100}{10} =1$

The probability that a single score drawn at random will be greater than 110

P( X > 110) = P( Z >1)

= 1 - P( Z < 1)

= 1 - ( 0.5 +A(1))

= 0.5 - A(1)

= 0.5 -0.3413

= 0.1587

let 'X' = 105

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{25} } } = 2.5$

The probability that a single score drawn at random will be greater than 110

P( x> 105) = P( z > 2.5)

= 1 - P( Z< 2.5)

= 1 - ( 0.5 + A( 2.5))

= 0.5 - A ( 2.5)

= 0.5 - 0.4938

= 0.0062

The probability that a single score drawn at random will be greater than 105

P( x> 105) = 0.0062

c)

let 'X' = 105

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{64} } } = 4$

The probability that a single score drawn at random will have a mean greater than 105

P( x> 105) = P( z > 4)

= 1 - P( Z< 4)

= 1 - ( 0.5 + A( 4))

= 0.5 - A ( 4)

= 0.5 - 0.498

= 0.002

The probability that a sample of 64 scores will have a mean greater than 105

P( x⁻> 105) = 0.002

d)

Let x₁ = 95

$Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{95-100}{\frac{10}{\sqrt{16} } } = -2$

Let x₂ = 105

$Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{16} } } = 2$

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = P( -2≤z≤2)

= P(z≤2) - P(z≤-2)

= 0.5 + A( 2) - ( 0.5 - A( -2))

= A( 2) + A(-2) (∵A(-2) =A(2)

= A( 2) + A(2)

= 2 × A(2)

= 2×0.4772

= 0.9544

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = 0.9544

7 0

3 years ago