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2 years ago

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The population of a town grew from 20,000 to 28,000. The continuous growth rate is 15%. The equation mc024-1.jpg represents the

situation, where t is the number of years the population has been growing. About how many years has the population of the town been growing? Use a calculator and round your answer to the nearest whole number.

2 answers:

Hunter-Best [27]2 years ago

8 0

To calculate the number of years that the population of the town had been growing we proceed as follows;
A=p(1+r/100)^n
where;
A=final population
p=initial population
n=time
r=rate;
thus, substituting the values in the above formula we get:
28000=20000(1+15/100)^n
1.4=(1+0.15)^n
1.4=1.15^n
introducing natural logs we get:
ln1.4=nln1.15
n=(ln1.4)/(ln1.15)
n=2.4
thus;
n=2.4 years

Lera25 [3.4K]2 years ago

3 0

Answer:

a.) 2 years on edge 2020

Step-by-step explanation:

hope this helps!!!

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Answer:

see the explanation

Step-by-step explanation:

we know that

A gross is equal to 120 ones or ten dozen

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A bank gets an average of 12 customers per hour. Assume the variable follows a Poisson distribution. Find the probability that t

Alina [70]

Answer:

75.76% probability that there will be 10 or more customers at this bank in one hour.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

A bank gets an average of 12 customers per hour.

This means that $\mu = 12$

Find the probability that there will be 10 or more customers at this bank in one hour.

Either there are less than 10 customers, or there are 10 or more. The sum of the probabilities of these events is 1. Then

$P(X < 10) + P(X \geq 10) = 1$

We want $P(X \geq 10)$

Then

$P(X \geq 10) = 1 - P(X < 10)$

In which

$P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)$

So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-12}*12^{0}}{(0)!} \approx 0$

$P(X = 1) = \frac{e^{-12}*12^{1}}{(1)!} = 0.0001$

$P(X = 2) = \frac{e^{-12}*12^{2}}{(2)!} = 0.0004$

$P(X = 3) = \frac{e^{-12}*12^{3}}{(3)!} = 0.0018$

$P(X = 4) = \frac{e^{-12}*12^{4}}{(4)!} = 0.0053$

$P(X = 5) = \frac{e^{-12}*12^{5}}{(5)!} = 0.0127$

$P(X = 6) = \frac{e^{-12}*12^{6}}{(6)!} = 0.0255$

$P(X = 7) = \frac{e^{-12}*12^{7}}{(7)!} = 0.0437$

$P(X = 8) = \frac{e^{-12}*12^{8}}{(8)!} = 0.0655$

$P(X = 9) = \frac{e^{-12}*12^{9}}{(9)!} = 0.0874$

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Then

$P(X \geq 10) = 1 - P(X < 10) = 1 - 0.2424 = 0.7576$

75.76% probability that there will be 10 or more customers at this bank in one hour.

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2 years ago

Janet is designing a frame for a client. She wants to prove to her client that m∠AGE ≅ m∠CHE in her sketch. What is the missing

Amiraneli [1.4K]

Answer:

I believe the answer is transative property. I'm currently taking the test so I'm not 100% sure if I'm right yet.

Step-by-step explanation:

In step one, AGE and HGB are verticle angles since they are opposite of each other. then step two is alternate interior angles because they are interior angles opposite of each other, self-explanatory. the reason I choose the transitive property over the corresponding angles theorem is because the transitive property states: among A B and C, if A=B and B=c then A=C and in this case since AGE=HGB andHGB=CHE then AGE=CHE. I hope this helps and if you know the right answer just put it in the comments, if I'm wrong ofc.

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2 years ago