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3 years ago

Is she correct ?????? HELP !!

Mathematics

1 answer:

garik1379 [7]3 years ago

8 0

Here, volume of two shapes doesn't relate with each other, so if one has double dimensions than the other, then it's volume won't be exactly double than the other, 'cause they formula's are very different.

Volume of a Cylinder = πr²h
v = π(4)²(6)
v = 96π

Volume of a cone = 1/3 πr²h
v = 1/3 π (8)²(12)
v = 256π

So, Ratio = 256π/96π = 128/48 = 64/24 = 32/12 = 16/6 = 8/3 [ Not 2 ]

In short, She is Incorrect!

Hope this helps!

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Find the volume of the solid by rotating the region bounded by y=x^3, y=8, and x=0 about the y axis.

dimulka [17.4K]

Answer:

Solution : Volume = 96/5π

Step-by-step explanation:

If we slice at an arbitrary height y, we get a circular disk with radius x, where x = y^(1/3). So the area of a cross section through y should be:

A(y) = πx^2 = π(y^(1/3))^2 = πy^(2/3)

And now since the solid lies between y = 0, and y = 8, it's volume should be:

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3 years ago

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Step-by-step explanation:

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3 years ago

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uranmaximum [27]

I would tell her that her prediction is inaccurate because it isn’t guaranteed that the coin will land on both sides an equal amount of times. It’s possible that the coin could land heads up more often than it could land tails up or the other way around.

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3 years ago

An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand

Elena L [17]

Answer:

a)From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

b) We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

Part b

We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

4 0

3 years ago