B i think on #2
hope this helps
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:
(x - 7)² + (y - 4)² = 49
Step-by-step explanation:
Given
Equation: x² + y² = 49
Required:
New Equation when translated 7 units right and 4 units up
Taking it one step at a time.
When the equation is translated 7 units right, this implies a negative unit along the x axis.
The equation becomes
(x - 7)² + y² = 49
When the equation is translated 4 units up, this implies a negative unit along the y axis.
(x - 7)² + (y - 4)² = 49
The expression can be further simplified but it's best left in the form of
(x - 7)² + (y - 4)² = 49