Answer:
The amount stored in small silo are [tex]\frac{1}{3}[/tex]
Step-by-step explanation:
It is given that He stored [tex]\frac{5}{9}[/tex] of his corn in one large silo and [tex]\frac{3}{4}[/tex] of remaining in small silo.
Since [tex]\frac{5}{9}[/tex] is removed, remaing is [tex]\frac{4}{9}[/tex] ( 1- [tex]\frac{5}{9}[/tex])
[tex]\frac{3}{4}[/tex] of remaining means,
= ([tex]\frac{3}{4}[/tex])([tex]\frac{4}{9}[/tex])
= [tex]\frac{1}{3}[/tex]
Thus total is [tex]\frac{1}{3}[/tex] + [tex]\frac{5}{9}[/tex] = [tex]\frac{8}{9}[/tex]
taken to market is (1 - [tex]\frac{8}{9}[/tex]) = [tex]\frac{1}{9}[/tex]
