We want to simplify this as much as possible. 11/110=1/10 since 110=11*10. Therefore, the answer is 0.1
Answer:
Option A is correct.
Step-by-step explanation:
As we see the graph, we can say that the correct statement is :
A.)All repairs requiring 1 hour or less have the same labor cost. We can see that the coat from 0 hours to 1 hour is $50. So, the number of hours falling in this range has the same repairing cost.
B.) Labor costs the same no matter how many hours are used for a repair. This is wrong as the graph is increasing after 1 hour.
C.) Labor costs for a repair are more expensive as the number of hours increases. This is wrong as the hours are increasing from 0.25 to 0.5 then to 0.75 but they all have the same cost.
D.)There is no cost of labor for a repair requiring less than 1 hour. This is also wrong. The cost is $50.
Answer:
As few as just over 345 minutes (23×15) or as many as just under 375 minutes (25×15).
Imagine a simpler problem: the bell has rung just two times since Ms. Johnson went into her office. How long has Ms. Johnson been in her office? It could be almost as short as just 15 minutes (1×15), if Ms. Johnson went into her office just before the bell rang the first time, and the bell has just rung again for the second time.
Or it could be almost as long as 45 minutes (3×15), if Ms. Johnson went into her office just after the bells rang, and then 15 minutes later the bells rang for the first time, and then 15 minutes after that the bells rang for the second time, and now it’s been 15 minutes after that.
So if the bells have run two times since Ms. Johnson went into her office, she could have been there between 15 minutes and 45 minutes. The same logic applies to the case where the bells have rung 24 times—it could have been any duration between 345 and 375 minutes since the moment we started paying attention to the bells!
Step-by-step explanation:
Answer:
x = 2, y = 1
Step-by-step explanation:
x + 4y = 6 and y = 3 - x have to be rearranged: x + 4y = 6 and x + y = 3
You subtract the equations to eliminate one of the variables (x) so that the other can be found
