Multiply each hourly rate by x ( time to complete the work) add it to the service call for each and then set the equations equal:
50 + 40x = 30 +45x
Subtract 30 from both sides:
20 + 40x = 45x
Subtract 40x from both sides:
20 = 5x
Divide both sides by 5
x = 4
The length is 4 hours.
Times a number by a percent. First turn 30% Ino .3 by putting the decimal point 2 times to the l
.
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:
0.6988
Step-by-step explanation:
Given that the number of people who use the ATM at night outside your local bank branch can be modeled as a Poisson distribution.
Let X be the number of customers arriving between 10 and 11 am.
X is Poisson with mean= 1.2
Required probability
= the probability that in the hour between 10 and 11 PM at least one customer arrives
= P(X≥1)
=1-P(X=0)
=1-0.3012
= 0.6988
Answer:
(a). 15
(b). 78
Step-by-step explanation:
Growth of the population of a fruit fly is modeled by
N(t) = 
where t = number of days from the beginning of the experiment.
(a). For t = 0 [Initial population]
N(0) = 
= 
= 
= 15
Initial population of the fruit flies were 15.
(b).Population of the fruit fly colony on 11th day.
N(11) = 
= 
= 
= 
= 
= 77.82
≈ 78
On 11th day number of fruit flies colony were 78.
