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damaskus [11]
3 years ago
13

Population Growth The population P (in thousands) of Oriando,Florida from 1980 through 2009 can be modeled by p = 130e0.0205t,wh

ere t = 0 corresponds to 1980.
(a) What was the population of Orlando in 2009?
(b) In what year will Orlando have a population of 300,000?
Mathematics
1 answer:
svetoff [14.1K]3 years ago
6 0

Answer: a) 236,000.

b) 2021

Step-by-step explanation:

Given : Population Growth The population P (in thousands) of Oriando,Florida from 1980 through 2009 can be modeled by p = 130e^{0.0205t}    (1)

where t = 0 corresponds to 1980.

Then , for 2009

t= 2009-1980=29

a. ⇒The population of Phoenix in 2009 =  p = 130e^{0.0205(29)}

p = 130e^{0.5945}

p =130(1.81212465608)=235.57620529\approx236

Hence, the population of Orlando in 2009 was about 236,000.

b) Substitute p= 300 in (1) , we get

300 = 130e^{0.0205t}

2.3077= e^{0.0205t} (Divide both sides by 130)

Taking Natural log on both sides,

\ln(2.3077) = 0.0205t\\\\\Rightarrow\ t=\dfrac{\ln(2.3077)}{0.0205}

t=\dfrac{0.836251357528}{0.0205}=40.7927491477\approx41

Hence, The year Orlando will have a population of 300,000 =1980+41 =2021

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A coyote can run up to 43 miles per hour while a rabbit can run up to 35 miles per hour.write two equivalent expression and then
yaroslaw [1]
D=s*t
distaance=speed times time

cd=coyote distance*time=ds*dt
rd=rabbit diatance*time=rs*rt

given
t=6 for all, so dt=rt=6

and ds=43
rs=35


cd=43*6=258miles
rd=35*6=210miles

how much more?
258-210=48

48 more miles
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2 years ago
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vredina [299]

Answer:

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Step-by-step explanation:

5 0
3 years ago
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9
Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that \mu = 8.9, \sigma = 2.9

Sample of 37:

This means that n = 37, s = \frac{2.9}{\sqrt{37}}

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

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

By the Central Limit Theorem

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

Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}

Z = -0.53

Z = -0.53 has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

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

Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}

Z = -3.08

Z = -3.08 has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

7 0
2 years ago
Can someone help me with this please? :)))​
Natali5045456 [20]

Answer:

88

Step-by-step explanation:

86 is 98% of 87.755102040816 which rounded to the nearest whole number is 88.

7 0
2 years ago
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