Constant loss of numbers
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Constant loss of numbers

[From: ] [author: ] [Date: 12-03-20] [Hit: ]
From there we solve it.1=a*50+20 --> -19=a*50 --> a=(-19/50).Plugging this found value into are equation gives us the function y=(-19/50)x+20.We can now plug 21-years into it for x.This gives us 12.02.......
If accomodative power starts at 20 diopters at birth and reaches 1 diopter at 50 years, how much accomodation is left in a 21-year-old? Assume that the accomodative system loses a constant number of diopters each year.

Any help please? an equation I should be using? This isn't even for a math class so I'm not sure if there's an associated formula I can use.

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The statement "Assume that the accomodative system loses a constant number of diopters each year" alludes to this being a linear function (y=ax+b). With this stated and that fact that the starting number is 20 then we know that the constant for this equation is 20. This gives us y=ax+20. From there we can plug in a one of the other given points (i.e. (1 diopter, 50 years)). Since y represents the diopter and x represents the years, it would look like this 1=a*50+20. From there we solve it. 1=a*50+20 --> -19=a*50 --> a=(-19/50). Plugging this found value into are equation gives us the function y=(-19/50)x+20. We can now plug 21-years into it for x. This gives us 12.02. If it is possible to have two hundredths of a diopter then this is the answer. If, like many things, it must be a whole number, then 12 is the answer.

Hopefully this was helpful. Good luck in your class. I must say, I am curious of what class it is now. This sounds just like some word problems that I have received before.

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And this is for a behavioral psychology class. This problem was in regards to our vision and how the diopters change throughout life. More diopters means closer range of focus

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