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Maggy started a savings account in March of 2003. On March 2007, she had $4,200. On March 2015, she had $10,400. If Maggy's saving is modeled by a linear function, what was…
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Maggy started a savings account in March of 2003. On March 2007, she had $4,200. On March 2015, she had $10,400. If Maggy's saving is modeled by a linear function, what was…
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Maggy started a savings account in March of 2003. On March 2007, she had $4,200. On March 2015, she had $10,400. If Maggy's saving is modeled by a linear function, what was her initial deposit? A) $1,100 B) $1,300 C) $1,500 D) $700
Mathematics
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Denis Koreyba
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Denis Koreyba
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First find the slope of the linear equation...
slope=m=(y2-y1)/(x2-x1)
m=(10400-4200)/(2015-2007)=6200/8=775 so we so far have a line of:
s=775(y-2003)+b, using either original point we can solve for b, the y-intercept which is the initial value as well...I'll use point (2007, 4200) and we get:
4200=775(2007-2003)+b
4200=3100+b
b=$1100
s(y)=775(y-2003)+1100
The initial deposit was $1100.00.
John Rah
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Sep 10, 2018
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