SOLUTION: In a typical group of 1000 workers from each of the boatbuilding, iron foundry, and amusement park/arcade industries, there were 30 more injuries in iron foundries (I) than in amus
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Question 723929: In a typical group of 1000 workers from each of the boatbuilding, iron foundry, and amusement park/arcade industries, there were 30 more injuries in iron foundries (I) than in amusement parks/arcades (A). There were 12 more injuries in amusement parks/arcades than in boatbuildings (B). Among these workers, there were 387 nonfatal occupational injuries. How many injuries took place in each industry?
I'm guessing (keyword being guessing) that:
Data:1000 workers, I had 30 more than A and A had 12 more than B and the total is 387.
Variable: A= amusement parks/arcades, B= boatbuildings, I= iron foundry
Plan: B+(B+A)+(B+I)= 387
Equation: B+(B+12)+(B+30)=387
3B+42 = 387
-42 -42
3B = 345
3B/3 345/3
B = 115
Solution: A= 127, B= 115, and I= 145
Am I right? Answer by MathLover1(20850) (Show Source):
You can put this solution on YOUR website!
there were nonfatal occupational injuries
let the boat-building injuries be ,amusement park/arcade injuries be , and iron foundry injuries be
given:
...substitute ...=> ...=> ...plug in all values
...solve for
now find and
check their sum:
check your solution: A= 127, B= 115, and I= 145
Am I right?
your equation: ...you put , but it is given that had more than , or
so, you need to put
since also given that had more than , we know that and we go back to and substitute ...=>...
now, plug it in equation
