mmend98
  • mmend98
Diana lost 35 points on a project for school. She can make changes and receive three-tenths of the missing points back. She can do that 10 times. Create the formula for the sum of this geometric series and explain your steps in solving for the maximum number of points Diana can earn back.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
triciaal
  • triciaal
|dw:1439420130504:dw|
mmend98
  • mmend98
is there any way you could type that haha its very hard to interpret this. but thank you @triciaal
Nnesha
  • Nnesha
i'll type it :=) \[\huge\rm S_n =a+ar^{n-1}\] \[\large \rm (\frac{ 3 }{ 10 })^9 =a(1+r^{n-1})\] \[\large \rm 35 =\frac{ a(1+(\frac{ 3 }{ 10 })^9) }{ 1 -\frac{3}{10} }\] (formula: \[S_n = \frac{ a(1-r^n) }{ 1-r }\])

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

mmend98
  • mmend98
So how would I plug this in?
mmend98
  • mmend98
@triciaal
mmend98
  • mmend98
Help ! haha please good person @ganeshie8
ganeshie8
  • ganeshie8
triciaal seems to be having the right method - i think you just need to add up the first 10 terms
ganeshie8
  • ganeshie8
remember the "partial sum formula" for geometric series ?
mmend98
  • mmend98
yes i do
ganeshie8
  • ganeshie8
\[\large S_n=a_1*\dfrac{1-r^n}{1-r}\] does it look something like above in ur notes ?
mmend98
  • mmend98
yes. but there is the sigma thing in the front
ganeshie8
  • ganeshie8
\[\large \sum\limits_{i=1}^na_i=a_1*\dfrac{1-r^n}{1-r}\] like that ?
mmend98
  • mmend98
yes
mmend98
  • mmend98
ok one sec
ganeshie8
  • ganeshie8
No wait, there is a mistake
mmend98
  • mmend98
ok
ganeshie8
  • ganeshie8
scratch that, il type it again fixing the mistake
mmend98
  • mmend98
ok thanks
ganeshie8
  • ganeshie8
Okay, good. Lets see whats going on in the given word problem. She can make changes and receive three-tenths of the missing points back, so this becomes the common ratio : \(r=\dfrac{3}{10}\) To start with, the missing points were 35, so the first term must be : \(a_1 = 35*\dfrac{3}{10} = 10.5\) Plug them in the partial sum formula and evaluate
ganeshie8
  • ganeshie8
\[\large \sum\limits_{i=1}^{10}a_i=10.5*\dfrac{1-\left(\frac{3}{10}\right)^{10}}{1-\frac{3}{10}}\] simplify..
mmend98
  • mmend98
yes thank you
ganeshie8
  • ganeshie8
let me know what you get and please medal triciaal, not me
mmend98
  • mmend98
I got 149
ganeshie8
  • ganeshie8
Wow! 149 ? She cannot receive 149 points back! double check ur work
mmend98
  • mmend98
oh sorry put it in way wrong haha
mmend98
  • mmend98
let me do it over
ganeshie8
  • ganeshie8
take your time..
mmend98
  • mmend98
wait yea i got 149 again lol
ganeshie8
  • ganeshie8
lets check with wolfram
mmend98
  • mmend98
ok
ganeshie8
  • ganeshie8
http://www.wolframalpha.com/input/?i=10.5*%281-%283%2F10%29%5E10%29%2F%281-3%2F10%29
mmend98
  • mmend98
ahh ok
ganeshie8
  • ganeshie8
wolf says\(14.9999\), which approximates to \(15\) looks you have just missed one decimal point somewhere..
mmend98
  • mmend98
but doesn't the sigma change it to 149
ganeshie8
  • ganeshie8
why ?
ganeshie8
  • ganeshie8
what do you know about sigma
mmend98
  • mmend98
like when u account for sigma in the problem.
ganeshie8
  • ganeshie8
what do you mean by that, could you elaborate
mmend98
  • mmend98
yea you have to take in to account the numbers above the sigma
mmend98
  • mmend98
idk maybe I'm crazy haha
ganeshie8
  • ganeshie8
\(\sum\) is just a short cut for "sum of terms" don't let that confuse you here
ganeshie8
  • ganeshie8
check this quick http://www.mathsisfun.com/algebra/sigma-notation.html
mmend98
  • mmend98
ohh thanks so much
ganeshie8
  • ganeshie8
so how many points is she gonna get back after 10 attempts ? whats the maximum possible points that she can get back if she is allowed to attempt infinitely many times ?
mmend98
  • mmend98
15 points !
ganeshie8
  • ganeshie8
careful here, \(14.999\) is not exactly same as \(15\) In fact, she can never get back \(15\) points even if she is allowed to attempt 100 or 1000 or any other big finite number of times
mmend98
  • mmend98
ah. so it would just be 14.999
mmend98
  • mmend98
thanks again
ganeshie8
  • ganeshie8
She gets back 14.999 after 10 attempts The maximum points that she can get back is 15 if she is allowed to try "infinite" times (which we know is practically impossible here)
ganeshie8
  • ganeshie8
http://www.wolframalpha.com/input/?i=Table%5B10.5*%281-%283%2F10%29%5En%29%2F%281-3%2F10%29%2C+%7Bn%2C+1%2C15%7D%5D
ganeshie8
  • ganeshie8
Notice that with each attempt, the sum is by increasing a tiny bit and getting closer to 15, but it never quite reaches 15. We say the sum "converges" to 15.

Looking for something else?

Not the answer you are looking for? Search for more explanations.