A research paper recommends using Poisson process to model the number of failures in commercial water pipes The paper also gives estimates the failure rate in units of failures per 100 miles of pipe per day; for four different types of pipe and for many different years_ For example, for cast iron pipe in 2005, the authors' estimate is the failure rate is 0.0864 failures per 100 miles per day: Suppose a town had 2500 miles of cast iron pipe underground in 2005 What is the probability of at least two failures
Answer:
where a>0.
To graph the the polynomial, begin in the left top of quadrant 2. Then draw downwards to the first real zero on the x-axis at -2. Cross the x-axis and then curve back up to 1/2 on the x-axis. Cross through again and curve back down to cross for the last time at 3 on the x-axis. The graph then ends going down towards the right in quadrant 4. It forms an s shape.
Step-by-step explanation:
The real zeros are the result of setting each factor of the polynomial to zero. By reversing this process, we find:
- zero 1/2 is factor (2x-1)
We write them together with an unknown leading coefficient a which is negative so -a.
where a>0
The leading coefficient of a polynomial determines the direction of the graph's end behavior.
- A positive leading coefficient has the end behavior point up when an even degree and point opposite directions when an odd degree with the left down and the right up.
- A negative leading coefficient has the end behavior point down when an even degree and point opposite directions when an odd degree with the left up and the right down.
- This graph has all odd multiplicity. The graph will cross through the x-axis each time at its real zeros.
To graph the the polynomial, begin in the left top of quadrant 2. Then draw downwards to the first real zero on the x-axis at -2. Cross the x-axis and then curve back up to 1/2 on the x-axis. Cross through again and curve back down to cross for the last time at 3 on the x-axis. The graph then ends going down towards the right in quadrant 4. It forms an s shape.
Answer:
6 lengths
Step-by-step explanation:
You essentially want the smallest integer solution to ...
60x ≥ 350
x ≥ 350/60
x ≥ 5 5/6
The smallest integer solution to this is x = 6.
The minimum number of lengths of hose needed is 6.
_____
Informally, you know that dividing the required total length by the length of one hose will tell you the number of required hoses. You also know the ratio 350/60 is equivalent to 35/6 and that this will be between 5 and 6. (5·6 = 30; 6·6 = 36) The next higher integer value will be 6.
Answer:
55.6%
Step-by-step explanation:
1. The difference between 45 and 70 is 70-45=25 (If it is postive is an increase, if negative a decrease)
2.The base of comparison is the original number: 45
3. With both, you divide the difference with the base:
25/45 and you multiply it by 100 to get a %
4. Equation: ((70-45)/45)*100 = 55,55555555555
5. Rounded to the nearest above: 55,6%
Work is shown!! have a good day