In a given city, the permissible limit of CO (carbon monoxide) in the air is 100 parts per million (ppm). The city monitors the steady rise of CO from various sources annually. In which year (rounded off to the nearest integer) will the CO level exceed the permissible limit?
2017
2018
2019
2022
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OpenStudy (welshfella):
not enough info to do this
OpenStudy (jv0408):
its
OpenStudy (jv0408):
there sorry
OpenStudy (welshfella):
we have the way it rises over thirteen years
amount it rises is 110-77 = 33 ppm over 13 years
OpenStudy (welshfella):
so we can write an eqaution relating ppm of CO (y) with the year x.
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OpenStudy (welshfella):
what is the slope of the line in the diagram?
OpenStudy (welshfella):
|dw:1454352855407:dw|
OpenStudy (jv0408):
how do I do solpe again?
OpenStudy (jv0408):
rise over run?
OpenStudy (welshfella):
yes
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OpenStudy (welshfella):
rise = 33 run = 13
OpenStudy (jv0408):
so 33*13?
OpenStudy (jv0408):
or 33/13?
OpenStudy (welshfella):
no 33 / 13
divide not multiply
OpenStudy (jv0408):
ok hold on
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OpenStudy (jv0408):
13 only goes 2 time in 33 well 2 full times
OpenStudy (welshfella):
divide it on ur calculator
2.54 is ok
OpenStudy (jv0408):
I wish I had a calculator I dont
OpenStudy (jv0408):
sorry
OpenStudy (welshfella):
so if we count year 2013 as zero and the ppm in 2013 is 77
we can write the equation as
y = 2.54x + 77
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OpenStudy (welshfella):
where y = ppm CO and x is the number of the year after 2013
OpenStudy (welshfella):
(counting 2013 as 0)
OpenStudy (jv0408):
add 77 to 2013?
OpenStudy (welshfella):
so we get
100 = 2.54x + 77
solve for x and add this value to 2013 and you'll get your answer
OpenStudy (jv0408):
im totally lost sorry
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OpenStudy (jv0408):
but I will try it
OpenStudy (welshfella):
y =100 because thats the ppm we want to know about