Try this:
1) note that weight of pure antifreeze before mixing and after mixing is the same. So, if 'x' is weight of pure antifreeze in 50% solution, it is possible to make up equation before mixing: 0.5x+0.2*90.
2) there are 0.2*90=18 gal. of pure antifreeze in the 20% solution. If 'x' gal. is the weight of pure antifreeze in 50% sol. and 18 gal. is the weight of pure antifreeze in 20% sol., it is possible to make up an equation after mixing: 0.4(x+18).
3) using the both parts: 0.5x+0.2*90=0.4(x+18) ⇒ x=54 gal. of <u>pure</u> weight.
4) to find 50% solution of 54 gal. pure weight just 54:0.5=108 gal.
Answer: 108 gal.
Its the first one: Letter A
Okay, so 75 counts for 30% (exams), 80 counts for 20% (term paper), and 85 counts for 50% (final exam)
So you would multiply 30% by 75, 20% by 80, and 50% by 85. That would be 22.5, 16, and 42.5. Add these numbers up, and put them over the total percentage, 100%. So you would have 81/100. The student's final average is 81%.
Answer:
its not factor able i suggest try using math_way
Step-by-step explanation:
