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Anestetic [448]

3 years ago

8

Brennan has been playing a game where he can create towns and help his empire expand. Each town he has allows him to create 1.15

times as many villagers as he had in the one before. The game gave Brennan five villages to start with. Explain to Brennan how to create an equation to predict the number of villagers in any specific town. Then show how to use your equation to solve for the number of villagers he can create to live in the 15th town. PLEASE HELP, I WILL GIVE BRAINLIEST!!

1 answer:

PIT_PIT [208]3 years ago

8 0

Answer with Step-by-step explanation:

We are given that:

Brennan has been playing a game where he can create towns and help his empire expand.

Brennan started with 5 villagers.

Each town he has allows him to create 1.15 times as many villagers.

i.e. after creating one town he gets 5×1.15 villagers

after creating 2 towns he gets 5×1.15×1.15 villagers

after creating 3 towns he gets 5×1.15×1.15×1.15 villagers

after creating n towns he gets $5*(1.15)^n villagers$

Hence, after creating 15 towns he gets $5*(1.15)^1 villagers$

≈ 41 villagers

An equation to predict the number of villagers for any number of towns is: Let number of towns be n

Then number of villagers = $5*(1.15)^n$

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Leakage from underground gasoline tanks at service stations can damage the environment. It is estimated that 25% of these tanks

Gnesinka [82]

Answer:

a) 3.75

b) 23.61% probability that fewer than 3 tanks will be found to be leaking

c) 0% the probability that at least 600 of these tanks are leaking

Step-by-step explanation:

For each tank there are only two possible outcomes. EIther they leak, or they do not. The probability of a tank leaking is independent of other tanks. So we use the binomial probability distribution to solve this question.

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:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

To solve question c), i am going to approximate the binomial distribution to the normal.

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

It is estimated that 25% of these tanks leak.

This means that $p = 0.25$

15 tanks chosen at random

This means that $n = 15$

a.What is the expected number of leaking tanks in such samples of 15?

$E(X) = np = 15*0.25 = 3.75$

b.What is the probability that fewer than 3 tanks will be found to be leaking?

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{15} = 0.0134$

$P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{14} = 0.0668$

$P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{13} = 0.1559$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0134 + 0.0668 + 0.1559 = 0.2361$

23.61% probability that fewer than 3 tanks will be found to be leaking

c.Now you do a larger study, examining a random sample of 2000 tanks nationally. What is the probability that at least 600 of these tanks are leaking?

Now we have n = 2000. So

$\mu = E(X) = np = 2000*0.25 = 500$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2000*0.25*0.75} = 19.36$

This probability is 1 subtracted by the pvalue of Z when X = 600. So

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

$Z = \frac{600 - 500}{19.36}$

$Z = 5.16$

$Z = 5.16$ has a pvalue of 0.

0% the probability that at least 600 of these tanks are leaking

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