Ask your own question, for FREE!
Mathematics 12 Online
OpenStudy (jv0408):

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

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.

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

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

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

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

OpenStudy (jv0408):

but I will try it

OpenStudy (welshfella):

y =100 because thats the ppm we want to know about

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!