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DaniilM [7]
3 years ago
5

PLEASE ANSWER FAST I WILL GIVE BRAINIEST TO WHOEVER GETS IT RIGHT !! a certain forest covers an area of 2000 square kilometers s

uppose that each year this area decreases by 6% what is the function that best represents the aries of the forest each year and how much area remains
a: f(x) = 2,000(0.94)^x, 952 square kilometers
b: f(x) =2,000(1.06)^x,4,204 square kilometers
c: f(x) =2,000(0.94)^x, 1,239 square kilometers
d: f(x)=2,000 (0.94)^x, 1,432 square kilometers
Mathematics
1 answer:
barxatty [35]3 years ago
7 0

Answer:

it's A=    f(x) = 2,000(0.94)^x, 952 square kilometers.

Step-by-step explanation:

2,000 square kilometers

decreases by 6% per year

12 years.


f(x) = P(1+r)^x


hope this helps, sorry if it is wrong.


-lavira :)

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Let X represent the amount of gasoline (gallons) purchased by a randomly selected customer at a gas station. Suppose that the me
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a) 18.94% probability that the sample mean amount purchased is at least 12 gallons

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Step-by-step explanation:

To solve this question, we use the normal probability distribution and the central limit theorem.

Normal probability distribution

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In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

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Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums, we can apply the theorem, with mean \mu and standard deviation s = \sqrt{n}*\sigma

In this problem, we have that:

\mu = 11.5, \sigma = 4

a. In a sample of 50 randomly selected customers, what is the approximate probability that the sample mean amount purchased is at least 12 gallons?

Here we have n = 50, s = \frac{4}{\sqrt{50}} = 0.5657

This probability is 1 subtracted by the pvalue of Z when X = 12.

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

By the Central Limit theorem

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

Z = \frac{12 - 11.5}{0.5657}

Z = 0.88

Z = 0.88 has a pvalue of 0.8106.

1 - 0.8106 = 0.1894

18.94% probability that the sample mean amount purchased is at least 12 gallons

b. In a sample of 50 randomly selected customers, what is the approximate probability that the total amount of gasoline purchased is at most 600 gallons.

For sums, so mu = 50*11.5 = 575, s = \sqrt{50}*4 = 28.28

This probability is the pvalue of Z when X = 600. So

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

Z = \frac{600 - 575}{28.28}

Z = 0.88

Z = 0.88 has a pvalue of 0.8106.

81.06% probability that the total amount of gasoline purchased is at most 600 gallons.

c. What is the approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers.

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

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X - 575 = 28.28*1.645

X = 621.5

The approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers is 621.5 gallons.

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