Answer:
The probability that the intersection will come under the emergency program is 0.1587.
Step-by-step explanation:
Lets divide the problem in months rather than in years, because it is more suitable to divide the period to make a better approximation. If there were 36 accidents in average per year, then there should be 3 accidents per month in average. We can give for the amount of accidents each month a Possion distribution with mean 3 and variance 3.
Since we want to observe what happen in a period of one year, we will use a sample of 12 months and we will take its mean. We need, in average, more than 45/12 = 3.75 accidents per month to confirm that the intersection will come under the emergency program.
For the central Limit theorem, the sample mean will have a distribution Normal with mean 3 and variance 3/12 = 0.25; thus its standard deviation is √0.25 = 1/2.
Lets call the sample mean distribution X. We can standarize X obtaining a standard Normal random variable W with distribution N(0,1).
The values of 
, the cummulative distribution function of W, can be found in the attached file. We are now ready to compute the probability of X being greater than 3.75, or equivalently, the probability than in a given year the amount of accidents is greater than 45, leading the intersection into an emergency program
Answer:
D[3, 3], C[1, 4], B[0, 1], A[4, -1]
Step-by-step explanation:
Whenever you are doing a 90° clockwise rotation ABOUT THE ORIGIN, it is in the form of [<em>y</em><em>,</em><em> </em><em>-</em><em>x</em>], meaning you take the <em>y</em><em> </em>and make it your <em>x</em><em>,</em><em> </em>then take your original <em>x</em><em> </em>and put its OPPOSITE.
90° counterclockwise rotation → [<em>-y, x</em>]
90° clockwise rotation → [<em>y, -x</em>]
I hope this helps, and as always, I am joyous to assist anyone at any time.
The greatest common factor is 6
36: 1, 2, 3, 4, 6, 9, 12, 18, 36
42: 1, 2, 3, 6, 7, 14, 21, 42
Hope this helps : )
Given:
The inequality is:
To find:
The graph of the given inequality.
Solution:
We have,
Subtract both sides by 1.
Divide both sides by 3.
The value of t is less than or equal to 
.
Since , it means is included in the solution, therefore there is a closed circle at 
and an arrow approaches to left from .
Therefore, the correct option is A.
The fourth option is the answer. When graphed, Min is 2, Q1 is 3, Median is 7, Q3 is 11, and Max is 24. The fourth option is the one that follow all of those!
