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Nana76 [90]
2 years ago
6

Suppose that the populations of the United States and China both increase by 12 million people in one year. What would be the re

lative change for each country? Answer to the nearest tenth of a percent (1 decimal place). Make sure that you fill in each answer area before checking "How Did I Do?".
Mathematics
1 answer:
Jlenok [28]2 years ago
3 0

Question:

Suppose that the populations of the United States and China both increase by 12 million people in one year. What would be the relative change for each country? Answer to the nearest tenth of a percent (1 decimal place). Make sure that you fill in each answer area before checking "How Did I Do?".

If you assume that the U.S. population is about 300 million, an increase of 12 million would result in a relative change of Number %.

If you assume that China has about 1 billion people, an increase of 12 million would result in a relative change of Number %.

Answer:

The relative change of the United States and China is 4% and 1.2%, respectively.

Step-by-step explanation:

Given

Represent population with p. So, we have:

United States

\triangle p = 12\ million

p_1 =300\ million

China

\triangle p = 12\ million

p_1 =1\ billion

Required

Determine the relative change for each country

Relative change, C is calculated as:

C = \frac{\triangle p}{p} * 100\%

For United States

C = \frac{12\ million}{300\ million} * 100\%

C = \frac{12}{300} * 100\%

C = \frac{12 * 100}{300}\%

C = \frac{1200}{300}\%

C = 4\%

For China

C = \frac{12\ million}{1\ billion} * 100\%

C = \frac{12}{1000} * 100\%

C = \frac{12* 100}{1000} \%

C = \frac{1200}{1000} \%

C = 1.2\%

The relative change of the United States and China is 4% and 1.2%, respectively.

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UkoKoshka [18]

Answer:

Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

X \sim N(14.7,3.7)  

Where \mu=14.7 and \sigma=3.7

Since the distribution of X is normal then we know that the distribution for the sample mean \bar X is given by:

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

And we have;

\mu_{\bar X}= 14.70

\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59

Step-by-step explanation:

Assuming this question: The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of 14.7 minutes and a standard deviation of 3.7 minutes. Let R be the mean delivery time for a random sample of 40 orders at this restaurant. Calculate the mean and standard deviation of \bar X Round your answers to two decimal places.

Previous concepts

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".

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Solution to the problem

Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

X \sim N(14.7,3.7)  

Where \mu=14.7 and \sigma=3.7

Since the distribution of X is normal then we know that the distribution for the sample mean \bar X is given by:

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

And we have;

\mu_{\bar X}= 14.70

\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59

4 0
3 years ago
The probability distribution for the random variable x follows. x f(x) 20 0.20 25 0.15 30 0.30 35 0.35 (a) Is this probability d
Damm [24]

Answer:

(a) The probability distribution is valid.

(b) The probability that x = 30 is 0.30.

Step-by-step explanation:

The probability distribution of the random variable <em>X</em> is:

<em>    x</em>:  20   |  25  |  30   |   35

f (<em>x</em>): 0.20 | 0.15 | 0.30 | 0.35

(a)

The properties of a probability distribution are:

  1. 0 ≤ P (X) ≤ 1
  2. ∑ P (X) = 1

All the probability value are more than 0 and less than 1.

Compute the sum of all the probabilities as follows:

\sum P(X)=0.20+0.15+0.30+0.35=1

The sum of all probabilities is 1.

Thus, the probability distribution is valid.

(b)

Consider the probability distribution table.

The probability of <em>X</em> = 30 is,

P (X = 30) = 0.30.

Thus, the probability that x = 30 is 0.30.

6 0
3 years ago
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