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: Company C
Step-by-step explanation:
A: $150
B: $160
C: $140
D: $170
So Company C has the best rate.
IDK so so so sorry i am not fameluar with this
Simplifying
15 + -5(4x + -7) = 50
Reorder the terms:
15 + -5(-7 + 4x) = 50
15 + (-7 * -5 + 4x * -5) = 50
15 + (35 + -20x) = 50
Combine like terms: 15 + 35 = 50
50 + -20x = 50
Add '-50' to each side of the equation.
50 + -50 + -20x = 50 + -50
Combine like terms: 50 + -50 = 0
0 + -20x = 50 + -50
-20x = 50 + -50
Combine like terms: 50 + -50 = 0
-20x = 0
Solving
-20x = 0
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Divide each side by '-20'.
x = 0.0
Simplifying
x = 0.0
Answer:
3
Step-by-step explanation:
500 - 10 = 490
long way:
490 -135= 355
355 -135=220
220 -135=85
85 candy left