About 6,00 dollars worth of hamburger was sold last month
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:
1
Step-by-step explanation:
An event that is certain to happen has a probability of 1.
Answer:
- (√30 - 1, 0)
- (-√30 - 1, 0)
Step-by-step explanation:
Making the equation of the parabola :
⇒ y = a(x - h)² + k
⇒ y = (x + 1)² - 17 - 13
⇒ y = (x + 1)² - 30
x-intercepts have y = 0 :
- 0 = (x + 1)² - 30
- (x + 1)² = 30
Taking the square root on each side :
- √(x + 1)² = √30
- x + 1 = ±√30
- x = ±√30 - 1
The x-intercepts are :
- (√30 - 1, 0)
- (-√30 - 1, 0)
Answer:
20
Step-by-step explanation: