abbycross167
  • abbycross167
Will someone please help me figure out how to answer this question? At Generic High School, in the year 2001, there were 20 students per every computer. In 2008, there were 4.5 students per computer. During which year were there 10 students per computer?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathstudent55
  • mathstudent55
We will assume this is a linear variation. Use (2001, 20) and (2008, 4.5) as two ordered pairs in a line. Find the equation of the line. Then let y = 10, and solve for x.
abbycross167
  • abbycross167
@mathstudent55 I'm sooo sorry for not replying sooner, I tried but my WiFi is really slow. Could you show me how to find the equation? please?
abbycross167
  • abbycross167
@DERRICKCURRY

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anonymous
  • anonymous
habg on gtv for like 30 min
abbycross167
  • abbycross167
ok
abbycross167
  • abbycross167
@DERRICKCURRY are you going to finish helping me?
abbycross167
  • abbycross167
@RCCB Do you think you could help me please?
RCCB
  • RCCB
well im not 100 percent sure but i think it might be 2003, sorry it took me so long, is it multiple choice?
abbycross167
  • abbycross167
no sir it's not multiple choice, I have to work the answer out
RCCB
  • RCCB
the work has to be written out?
abbycross167
  • abbycross167
somewhat yes sir, I just need to show the steps that I took to get the Answer
RCCB
  • RCCB
haha thank you but you dont have to call me sir of you dont want to, im not sure exactly how to explain it maybe @Whitemonsterbunny17
abbycross167
  • abbycross167
Ok umm I just don't know how to do this and would love it if someone could explain it to me?
RCCB
  • RCCB
@jigglypuff314
RCCB
  • RCCB
im not sure im sorry i could guess but no for sure answer
RCCB
  • RCCB
@TheSmartOne
abbycross167
  • abbycross167
It' s ok @RCCB I try to get someone else to help me anser it... Thank you soo much for trying!
abbycross167
  • abbycross167
@TheSmartOne do you think you'll be able to help me?
mathstudent55
  • mathstudent55
We can use a trick to make the numbers easy to deal with. We will let y be the number of students per computer, but for x, we will let x = the number of years after 2000. That means when x = 1, the year is 2001. When x = 2, the year is 2002, etc. We are told that in 2001, there were 20 computers per student. That gives us the ordered pair (1, 20). Then in 2008, there were 4.5 computers per student. That gives us the ordered pair (8, 4.5). Now we find the equation of the line that contains the points (1, 20) and (8, 4.5). The equation of the line through points \((x_1, y_1)\) and \((x,_2, y_2) \) is: \(y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)\) Now we plug in the coordinates of our two points: \(y - 20 = \dfrac{4.5 - 20}{8 - 1}(x - 1)\) \(y - 20 = \dfrac{-15.5}{7}(x - 1)\) Multiply both sides by 7 to get rid of the denominator 7: \(7y - 140 = -15.5x + 15.5\) \(15.5x + 7y = 155.5\) Multiply both sides by 2 to get integer coefficients: \(31x + 14y = 311\) We now have the equation of the line. Now we want to know the year that corresponds to 10 students per computer. We need to find the value of x that corresponds to y = 10. \(31x + 14(10) = 311\) \(31x + 140 = 311\) \(31x = 171\) \(x = \dfrac{171}{31}\) \(x \approx 5.52\) Since x is approximately equal to 5.5 years, the year is 2005.5, or the middle of 2005. That means the school will have 10 computers per student approximately in the middle of 2005.
mathstudent55
  • mathstudent55
Here is a different method. It is called linear interpolation. It is a way of finding a value in between two other values using a straight line. |dw:1449676964493:dw|
mathstudent55
  • mathstudent55
\(\dfrac{15.5}{7}= \dfrac{10}{x} \) \(x = \dfrac{10 \times 7}{15.5} \) \(x \approx 4.52\) 2001 + 4.52 = 2005.52 The answer is the middle of 2005.
abbycross167
  • abbycross167
thank you sooo much @mathstudent55 you're amazing!!
mathstudent55
  • mathstudent55
You're welcome. Thanks.

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