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Luden [163]
2 years ago
15

the volume, v of a sphere is given by its radius, r, in the formula v=4/3(pi)(r^3) what is the formula for the radius given the

volume?

Mathematics
1 answer:
MrMuchimi2 years ago
4 0
R=(3V4<span>Home: Kyle's ConverterKyle's CalculatorsKyle's Conversion Blog</span>Volume of a Sphere CalculatorReturn to List of Free Calculators<span><span>Sphere VolumeFor Finding Volume of a SphereResult:
523.599</span><span>radius (r)units</span><span>decimals<span>  -3  -2  -1  0  1  2  3  4  5  6  7  8  9  </span></span><span>A sphere with a radius of 5 units has a volume of 523.599 cubed units.This calculator and more easy to use calculators waiting at www.KylesCalculators.com</span></span> Calculating the Volume of a Sphere:

Volume (denoted 'V') of a sphere with a known radius (denoted 'r') can be calculated using the formula below:

V = 4/3(PI*r3)

In plain english the volume of a sphere can be calculated by taking four-thirds of the product of radius (r) cubed and PI.

You can approximated PI using: 3.14159. If the number you are given for the radius does not have a lot of digits you may use a shorter approximation. If the radius you are given has a lot of digits then you may need to use a longer approximation.

Here is a step-by-step case that illustrates how to find the volume of a sphere with a radius of 5 meters. We'll u

π)⅓
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2 years ago
Is the graph increasing, decreasing, or constant where -3 &lt; X &lt;-1?
Romashka [77]

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3 years ago
A -42+(-17)<br> b 8-(-9)<br> c 8(-9)<br> D<br> E<br> F<br> G
Diano4ka-milaya [45]

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3 years ago
The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.
Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using 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.

If Z \leq -2 or Z \geq 2, the outcome X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

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

By the Central Limit Theorem

Z = \frac{19 - 22}{1}

Z = -3

Z = -3 \leq -2, so yes, the sample mean being less than 19 days would be considered an unusual outcome.

7 0
2 years ago
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