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strojnjashka [21]
4 years ago
10

What is the volume of a rubber ball that has a radius of 1.7 cm

Mathematics
2 answers:
tino4ka555 [31]4 years ago
5 0
I believe the volume of one rubber ball is 20.58780
Genrish500 [490]4 years ago
5 0

Answer:

20.59 cubic centimeters

Step-by-step explanation:

For a rubber ball

radius (r) = 1.7 cm

<u>Volume of a Sphere </u>

<em>= 4/3 * π * r^3</em>

= 4/3 * 22/7 * 1.7 * 1.7 * 1.7 cm

= 20.59 cm^3

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Which number is a factor of 15 but not a multiple of 3 ???
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The answer is 5 because 5 is a factor of 15 but not a multiple of 5
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Tính diện tích S của hình phẳng giới hạn bởi các đường y=x³,y=2-x² và x=-1
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Answer: uhhh

Step-by-step explanation:

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3 years ago
Consider a binomial distribution of 200 trials with expected value 80 and standard deviation of about 6.9. Use the criterion tha
zavuch27 [327]

Answer:

120 has a z-score higher than 2.5. So yes, it would be unusual to have more than 120 successes out of 200 trials.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

\mu = 80, \sigma = 6.9

Use the criterion that it is unusual to have data values more than 2.5 standard deviations above the mean or 2.5 standard deviations below the mean

This means that z-scores higher than 2.5 or lower than -2.5 are considered unusual.

Would it be unusual to have more than 120 successes out of 200 trials

We have to find the Z-score of X = 120.

So

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

Z = \frac{120 - 80}{6.9}

Z = 5.8

120 has a z-score higher than 2.5. So yes, it would be unusual to have more than 120 successes out of 200 trials.

4 0
3 years ago
If √3 = 1.732 find the value of √27 - 3√75+5√48+2√108​
saw5 [17]

Answer:

Exact Form:

3√3−3√155+12√333-3155+123

Decimal Form:

−34.57898412…

4 0
3 years ago
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