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3 years ago

Find the volume of a cone with a base area 36pi and a height equal to twice the radius

1 answer:

8 0

The volume of a cone is $V= \frac{1}{3} A_{\text{basis}}\cdot H$ . Since $A_{\text{basis}}=36\pi$ and you know that basis is a circle you can conclude that $\pi R^2=36\pi$ and $R=6$ .

If <span>height $H$ equal to twice the radius $R$ , then $H=2R=2\cdot 6=12$ .</span>

The volume will be: $V= \frac{1}{3} \cdot 36\pi \12=144\pi$ cubic units.
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Step-by-step explanation:

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3 years ago

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The numbers of students in the four sixth-grade classes at Northside school are 26,19,34 and 21. Use properties to find the tota

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Basically there are 3 addition properties and they are associative property, identity property, and commutative property.

Commutative property is just changing the order of the addends. Example: 26+19+34+21= 21+34+19+26

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3 years ago

According to a Yale program on climate change communication survey, 71% of Americans think global warming is happening.† (a) For

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Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .