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Mathematics 20 Online
OpenStudy (anonymous):

Phillips received 75 points on a project. he can make changes and receive two tenths of the missing points back. He can make corrections as many times as he wants. Create the formula for the sum of the geometric series, and explain your steps in solving for the maximum grade Phillip can receive. Identify this as converging or diverging

OpenStudy (anonymous):

@TheSmartOne

OpenStudy (anonymous):

@perl

OpenStudy (perl):

can we assume that the student got 75 points out of 100 points possible?

OpenStudy (anonymous):

Sure

OpenStudy (anonymous):

That's why I'm asking on here... Sorry :(

OpenStudy (perl):

no problem :)

OpenStudy (perl):

the missing points is 25 to start out

OpenStudy (anonymous):

Okay

OpenStudy (perl):

2/10 = 1/5 , in lowest form

OpenStudy (anonymous):

Yup

OpenStudy (anonymous):

I guess this helps, thanks :)

OpenStudy (perl):

We start with 100 - 75 = 25 points For the first correction he can get 1/5 * 25 points. That leaves you with 4/5 * 25 points For the second correction he can get 1/5 * (4/5 * 25) points That leaves you with 4/5 * (1/5 * (4/5 * 25)) points

OpenStudy (perl):

do you see a pattern emerging ?

OpenStudy (anonymous):

Yeah

OpenStudy (perl):

correction 1 : 1/5 * (4/5)^0 * 25 correction 2: 1/5 * (4/5)^1 * 25 correction 3: 1/5 * (4/5)^2 * 25 correction 4: 1/5 * (4/5)^3 * 25 . . . correction n: 1/5 * (4/5)^(n-1) * 25

OpenStudy (perl):

We start with 100 - 75 = 25 points For the first correction he can get 1/5 * 25 points. That leaves you with 4/5 * 25 points For the second correction he can get 1/5 * (4/5 * 25) points That leaves you with 4/5 * (4/5 * 25)) points For the third correction he can get 1/5 * (4/5)^2 * 25 Leaving you with 4/5* (4/5)^2 * 25

OpenStudy (perl):

2/10 is the same as 20% , so i subtracted from 100% to get 80% 80% = 80/100 = 4/5

OpenStudy (anonymous):

Thanks @perl

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