The answer is: " [tex]42\frac{2}{3}[/tex] miles per hour " .
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Step-by-step explanation:
This is a "unit rate" problem.
"Find the rate in miles per "single [unit] hour."
{GIven: "5 1/3 mi per 7 1/2 min."}.
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Using a technique unknown as "dimension analysis" ;
let us convert the given value; in "mi / min" ;
to its value in "mi/hr" (our answer):
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→ " 5 1/3 mi 60 min ,
________ * ______ = ? mi / hr ? "
7 1/2 min 1 hr
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Note: The units of "min" ["minute(s)"] ; cancel out to "1" ;
→ {since: "min/min = 1 " ;
and we have: " [ (5 1/3 * 60) / (7 1/2) ] mi / hr. " ;
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→ " 5 1/3 = [ (3*5) + 1 ] / 3 " ;
= " (15 + 1) /3 " ;
= " 16 / 3 " ; → (as an "improper fraction") ;
→ " 7 1/2 = [ (2*7) + 1 ] / 2 " ;
= " (14 + 1) / 2) " ;
= " 15 / 2 " ; → (as an "improper fraction") ;
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So we have:
→ " [tex](\frac{16}{3}*\frac{60}{1})[/tex] ÷ [tex]\frac{15}{2}[/tex] " ;
Note: Dividing by a value is the same value as multiplying by the reciprocal of that value.
In this case: The reciprocal of: " [tex](\frac{15}{2}[/tex][tex])[/tex] " ; is: " [tex](\frac{2}{15})[/tex] " ;
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So, we can rewrite the expression:
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→ " [tex](\frac{16}{3}*\frac{60}{1}) * \frac{2}{15}[/tex] " ;
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= " [tex]\frac{(16*60*2)}{(3*1*15)}[/tex] " ;
→ " ( 16 * 60 * 2) "
(3 * 15)
→ Take note of the "60" in the numerator; and the "15" in the denominator ;
" (60 ÷ 15 = 4) " ;
so we can change the "60" to a "15" and cancel out the "15" ; since:
" (15 ÷ 15 = 1 ) " ;
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And rewrite:
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→ " ( 16 * 4 * 2) "
3 ;
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[tex]= \frac{(16 * 4 * 2)}{3} = \frac{(64*2)}{3} = \frac{128}{3} =[/tex] (128 ÷ 3) = " [tex]42\frac{2}{3}[/tex] " .
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The answer is: " [tex]42\frac{2}{3}[/tex] miles per hour " .
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Hope this is helpful!
Best of luck to you!
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