Answer:
B
Step-by-step explanation:
Expressing the ratios as fractions, that is
[tex]\frac{a}{b}[/tex] = [tex]\frac{3}{1}[/tex] ( cross- multiply )
a = 3b
[tex]\frac{b}{c}[/tex] = [tex]\frac{1}{5}[/tex] ( cross- multiply )
c = 5b
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Given
[tex]\frac{2a+3b}{4b+3c}[/tex] , substitute values from above for a and c
= [tex]\frac{2(3b)+3b}{4b+3(5b)}[/tex]
= [tex]\frac{6b+3b}{4b+15b}[/tex]
= [tex]\frac{9b}{19b}[/tex] ← cancel b on numerator/ denominator
= [tex]\frac{9}{19}[/tex] → B
