The least number of buses needed to carry 710 passengers is 5.
Option A)5 is the correct answer.
<h3>What is the least number of buses needed to carry 710 passengers? </h3>
Given that;
- Number of passengers n = 710
- Least number of buses need B = ?
To get the least number of buses, we say;
Number of buses B = 710 ÷ 150
Number of buses B = 4.7 ≈ 5
The least number of buses needed to carry 710 passengers is 5.
Option A)5 is the correct answer.
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14-10=? less than indicates that a minus sign is needed.
Answer:y=5
Step-by-step explanation:
5(y+1)=30
Divide both sides by 5
5(y+1)/5=30/5
We get y+1=6
Subtract 1 from both sides
y+1-1=6-1
We get y=5
Answer:
20%
Step-by-step explanation:
(240/1200)*100
The Least Common Factor of 60 and 72 is 360