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il63 [147K]
3 years ago
5

The following appeared in the magazine Financial Times, March 23,1995: "When Elvis Presley died in 1977, there were 48 professio

nal Elvis impersonators. Today there are an estimated 7328. If that growth is projected, by the year 2016 one person in four on the face of the globe will be an Elvis impersonator." This is an example of______________.
Mathematics
1 answer:
IgorLugansk [536]3 years ago
6 0

Answer:

This is an example of Extrapolation

Step-by-step explanation:

Extrapolation is the forecasting of the future based on assumption that present or past trends will continue.

It also assumes the method used for estimating past trends will still be used in future.

For example forecasts of world population.

In this instance, there were 48 impersonators of Elvis Presley, presently it has increased to to 7,328. Estimating that in 2016 based on the present trend, that one of four people will be an Elvis Presley impersonator, is also an example Extrapolation.

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If you flip two coins 76 times, what is the best prediction possible for the number of times both coins will land on tails? ​
ioda

Answer:

38 times

Step-by-step explanation:

A coin has two sides, so there is a 50% chance it will land on either side (assuming perfect conditions and that the coin cannot land vertically). With that in mind, if you flip a coin 76 times, 50% or 1/2 of the times the coin should land on tails. That is 36 times. (Once again, this is under perfect testing conditions. In real life, you probably wouldn't get an even split after only 76 flips.)

Hope this helps :)

3 0
3 years ago
Pls help this is due today I need help ASAP if its right i will give brainlest
makvit [3.9K]

Answer:

20.76 ✨

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6 0
3 years ago
Read 2 more answers
Explain how to tell weather an event is as likely as not
Degger [83]
I think the answer may be that an event is as likely as not to happen when the probability of the event happening is 50%, and the probability of the event not happening is 50%. Like a coin toss, the chance you will flip a "heads" is 1/2 or 50%, and the chance that you WON'T flip a heads is also 1/2 or 50%
5 0
3 years ago
Given:
Colt1911 [192]
Your question is very confusing but x=8 so just plug in 8 where you see X
3 0
3 years ago
According to a report by App Annie, a business intelligence company that produces tools and reports for the apps and digital goo
e-lub [12.9K]

Answer:

a) P(29\leq \bar X >31)=P(\frac{29-30}{1}\leq Z\leq \frac{31-30}{1})=P(-1\leq Z \leq 1)=P(Z

b) P(28\leq \bar X >32)=P(\frac{28-30}{1}\leq Z\leq \frac{32-30}{1})=P(-2\leq Z \leq 2)=P(Z

c) P(29\leq \bar X >31)=P(\frac{29-30}{0.5}\leq Z\leq \frac{31-30}{0.5})=P(-2\leq Z \leq 2)=P(Z

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable number of apps used per month by smartphone users. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =30,\sigma =5)

We take a sample of n=25 . That represent the sample size.

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=30, \frac{5}{\sqrt{25}})

a. what is the probability that the sample mean is between 29 and 31?

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}

And we want to find this probability:

P(29\leq \bar X >31)=P(\frac{29-30}{1}\leq Z\leq \frac{31-30}{1})=P(-1\leq Z \leq 1)=P(Z

b. what is the probability that the sample mean is between 28 and 32?

P(28\leq \bar X >32)=P(\frac{28-30}{1}\leq Z\leq \frac{32-30}{1})=P(-2\leq Z \leq 2)=P(Z

c. If you select a random sample of 100 smartphone owners, what is the probability that the sample mean is between 29 and 31?

On this case we have a new distribution for the sample mean given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=30, \frac{5}{\sqrt{100}})

And we want to find this probability:

P(29\leq \bar X >31)=P(\frac{29-30}{0.5}\leq Z\leq \frac{31-30}{0.5})=P(-2\leq Z \leq 2)=P(Z

On this case the results for parts a and b shows the probability that we have values within one and two deviations from the mean. And the result for part c its equal for part b since the new deviation for the sampel mean with a sample size of 100 its the half of the standard deviation when we use a random sample of 25.

7 0
3 years ago
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