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3 years ago

8

Mike just hopped on the edge of a merry-go-round. What are his linear and angular speeds if the diameter of the merry-go-round i

s 14 feet and it takes 6 seconds for it
to make a complete revolution? Round the solutions to two decimal places.

1 answer:

dsp733 years ago

8 0

Step-by-step explanation:

Given that,

The diameter of the merry-go-round, d = 14 feet

Time taken, t = 6 seconds

Radius, r = 7 feet

The linear speed of the merry-go-round is given by :

$v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 7}{6}\\\\=7.33\ m/s$

Also,

$v=r\omega$

Where

$\omega$ is the angular speed

So,

$\omega=\dfrac{v}{r}\\\\\omega=\dfrac{7.33}{7}\\\\=1.04\ rad/s$

Hence, his linear and angular speeds are 7.33 m/s and 1.04 rad/s.

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

SpyIntel [72]

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

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

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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