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The volume of a right circular cone is 300 cubic inches. What is the volume, in cubic inches, of a right cylinder that

Anna71 [15]

Answer:

The volume of a right cylinder is equal to the 3 times of the volume of a right circular cone.

Step-by-step explanation:

Given that,

The volume of a right circular cone is 300 cubic inches

We need to find the volume of a right cylinder that has the same base and height as the cone.

Volume of a cone is given by :

$V=\dfrac{1}{3}\pi r^2h$

And volume of a cylinder is given by :

$V'=\pi r'^2 h'$

If the base and height of cone and cylinder are same, r=r' and h=h'

$\dfrac{V}{V'}=\dfrac{1/3\pi r^2 h}{\pi r'^2h'}\\\\=\dfrac{1/3\pi r^2 h}{\pi r^2h}\\\\\dfrac{V}{V'}=\dfrac{1}{3}\\\\V'=3V$

So, the volume of a right cylinder is equal to the 3 times of the volume of a right circular cone.

7 0

3 years ago

Need help with this problem

sdas [7]

3 is the correct answer give me brainliest have a nice day

3 0

3 years ago

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The Acculturation Rating Scale for Mexican Americans (ARSMA) is a psychological test that measures the degree to which Mexican A

aleksandr82 [10.1K]

Answer:

The proportion of the population used to develop the test that has scores below 1.7 is 0.063.

The proportion of the population used to develop the test that has scores between 1.7 and 2.1 is 0.0662.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The distribution of ARSMA scores in a population used to develop the test is approximately Normal with mean 3.0 and standard deviation 0.8.

This means that $\mu = 3, \sigma = 0.8$

What proportion of the population used to develop the test has scores below 1.7?

This is the pvalue of Z when X = 1.7. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1.7 - 3}{0.8}$

$Z = -1.63$

$Z = -1.63$ has a pvalue of 0.063

The proportion of the population used to develop the test that has scores below 1.7 is 0.063.

Between 1.7 and 2.1?

This is the pvalue of Z when X = 2.1 subtracted by the pvalue of Z when X = 1.7.

X = 2.1

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{2.1 - 3}{0.8}$

$Z = -1.13$

$Z = -1.13$ has a pvalue of 0.1292

X = 1.7

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1.7 - 3}{0.8}$

$Z = -1.63$

$Z = -1.63$ has a pvalue of 0.063

0.1292 - 0.063 = 0.0662

The proportion of the population used to develop the test that has scores between 1.7 and 2.1 is 0.0662.

3 0

3 years ago

19. If the scale is 1:100 on a plan, and a fence is 7,500 mm,

marshall27 [118]

Answer:

It will measure 75 millimetres on the plan.

Step-by-step explanation:

Given:

Scale on the plan = 1:100

Length of the fence = 7,500 mm

To find:

how many millimetres will it measure on the plan

Solution:

As the ratio on the plan is 1: 100, the for 7500 the ratio will

x : 7500

Now by cross multiplication we get

$7500 \times 1 = x \times 100$

$7500 = 100x$

$x =\frac{ 7500}{100}$

x = 75

4 0

3 years ago

Prove the identity<br> (cot x sin x)(sec x – cos x) = sin^2x

FrozenT [24]

Step-by-step explanation:

$(cot x \: sinx)(secx - cosx) = {sin}^{2} x \\ \\ lhs = (cot x \: sinx)(secx - cosx) \\ = (cot x \: sinx) \times secx \\- (cot x \: sinx) \times cosx \\ \\ = \frac{cosx}{sinx} \times sinx \times secx\\ - \frac{cosx}{sinx} \times sinx \times cosx \\ = cosx \times secx - cosx \times cosx \\ = 1 - {cos}^{2} x \\ = {sin}^{2} x \\ = rhs \\ hence \: proved$

7 0

3 years ago

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