The school cafeteria is holding an election for students to vote on which items they would like to see on the lunch menu. The ch
1 answer:
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Answer:
C. 666
Step-by-step explanation:
USING PEMDAS we can solve this problem.
Answer: Choice D
b greater-than 3 and StartFraction 2 over 15 EndFraction
In other words,
b > 3 & 2/15
or
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Explanation:
Let's convert the mixed number 2 & 3/5 into an improper fraction.
We'll use the rule
a & b/c = (a*c + b)/c
In this case, a = 2, b = 3, c = 5
So,
a & b/c = (a*c + b)/c
2 & 3/5 = (2*5 + 3)/5
2 & 3/5 = (10 + 3)/5
2 & 3/5 = 13/5
The inequality is the same as
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Let's multiply both sides by 15 to clear out the fractions
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Now isolate the variable b
Side note: Another way to go from 47/15 to 3 & 2/15 is to notice how
47/15 = 3 remainder 2
The 3 is the whole part while 2 helps form the fractional part. The denominator stays at 15 the whole time.