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 0.4% every 100 feet, by what percentage is air pressure reduced by moving from sea level to Mexico City?

By moving from sea level to Mexico City  air pressure is reduced by 25%.

Step-by-step explanation:

Hi, to answer this we have to apply a decrease rate formula

P0 (1-r)^x

Where:  

p0= initial pressure

r = decrease rate. (0.4 = 0.4/100=0.004 decimal form)

x = feet (hundredths, since the air pressure decays every 100 feet we have to divide the input value by 100)

Replacing with the values given

P0 (1- 0.4/100)^(7340/100) =

P0 (0.996)^73.4 =

P0 ( 0.75)=

The air pressure at Mexico City compared to the sea level is 75% (0.75 x 100 =75%)

To obtain the percent decrease  

100% -75% =25%

step-by-step explanation:

we are given

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

add both sides by 9

divide both sides by 4

we get

answer: ur v dumb

step-by-step explanation: