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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Answer:
(-3-8)/(-4+4)= -11/0= no slope and undefined
Step-by-step explanation:
From 2000 to 2011, there are 11 years. From the given function, we can calculate the profit for a small business at the end of 2011 by substituting 11 to the x of the given equation.
y = (30,000)(1.06)(11)
y = 349,800
Thus, the profit at the end of 2011 is approximately $349,800.
I'll go with graphing cause when you try graphing a picewise function for example it's much harder to graph because it's to many numbers and you can't figure out what to graph especially for me.) mark me brainliest please
Answer:
57.14
Step-by-step explanation:
12/21 = ?/100
12 time 100=1200
1200 divided by 21 = 57.14
