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const2013 [10]
3 years ago
10

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

Mathematics
1 answer:
beks73 [17]3 years ago
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.
<span />

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Answer:

a) 0.2741 = 27.41% probability that at least 13 believe global warming is occurring

b) 0.7611 = 76.11% probability that at least 110 believe global warming is occurring

Step-by-step explanation:

Binomial probability distribution

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

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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:

p = 0.71

(a) For a sample of 16 Americans, what is the probability that at least 13 believe global warming is occurring?

Here n = 16, we want P(X \geq 13). So

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)

In which

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

P(X = 13) = C_{16,13}.(0.71)^{13}.(0.29)^{3} = 0.1591

P(X = 14) = C_{16,14}.(0.71)^{14}.(0.29)^{2} = 0.0835

P(X = 15) = C_{16,15}.(0.71)^{15}.(0.29)^{1} = 0.0273

P(X = 16) = C_{16,16}.(0.71)^{16}.(0.29)^{0} = 0.0042

P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741

0.2741 = 27.41% probability that at least 13 believe global warming is occurring

(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?

Now n = 160. So

\mu = E(X) = np = 160*0.71 = 113.6

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{160*0.71*0.29} = 5.74

Using continuity correction, this is P(X \geq 110 - 0.5) = P(X \geq 109.5), which is 1 subtracted by the pvalue of Z when X = 109.5. So

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

Z = \frac{109.5 - 113.6}{5.74}

Z = -0.71

Z = -0.71 has a pvalue of 0.2389

1 - 0.2389 = 0.7611

0.7611 = 76.11% probability that at least 110 believe global warming is occurring

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