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Anton [14]
3 years ago
9

When the Olympic Games were held outside Mexico City in 1968, there was much discussion about the effect the high altitude ( fee

t) would have on the athletes. Assuming air pressure decays exponentially by every feet, by what percentage is air pressure reduced by moving from sea level to Mexico City?
Mathematics
1 answer:
Dmitrij [34]3 years ago
3 0

Complete Question

When the Olympic Games were held outside Mexico City in 1968, there was much discussion about the effect the high altitude (7340 feet) would have on the athletes. Assuming air pressure decays exponentially by 0.5% every 100 feet, by what percentage is air pressure reduced by moving from sea level to Mexico City?

Round your answer to one decimal place.    

....... %

Answer:

The value is  x = 73.4\%

Step-by-step explanation:

From the question we are told that

   The  height of Mexico city is  h =  7340 feet

   The rate at which air pressure decays is  0.5% per 100 feet  

From the data given

    0.5 \%  \to 100 \ feet \\x\% \to  7340

=>  x = \frac{7340}{100}

=>    x = 73.4\%

   

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Answer:

At the start of 2014, the population was growing at 8.34 million people per year.

At the start of 2015, the population was growing at 8.39 million people per year.

Step-by-step explanation:

To find how fast was the population growing at the start of 2014 and at the start of 2015 we need to take the derivative of the function with respect to t.

The derivative shows by how much the function (the population, in this case) is changing when the variable you're deriving with respect to (time) increases one unit (one year).

We know that the population, P(t), of China, in billions, can be approximated by P(t)=1.394(1.006)^t

To find the derivative you need to:

\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=\\\\\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\1.394\frac{d}{dt}\left(1.006^t\right)\\\\\mathrm{Apply\:the\:derivative\:exponent\:rule}:\quad \frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)\\\\1.394\cdot \:1.006^t\ln \left(1.006\right)\\\\\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=(1.394\cdot \ln \left(1.006\right))\cdot 1.006^t

To find the population growing at the start of 2014 we say t = 0

P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(0)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^0\\P(0)' = 0.00833901 \:Billion/year

To find the population growing at the start of 2015 we say t = 1

P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(1)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^1\\P(1)' = 0.00838904 \:Billion/year

To convert billion to million you multiple by 1000

P(0)' = 0.00833901 \:Billion/year \cdot 1000 = 8.34 \:Million/year \\P(1)' = 0.00838904 \:Billion/year \cdot 1000 = 8.39 \:Million/year

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