Answer:
0.6844 is the required probability.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = $1,250
Standard Deviation, σ = $125
We are given that the distribution of daily sales is a bell like shaped distribution that is a normal distribution.
Formula:
We have to find
P(sales less than $1,310)
Calculation the value from standard normal z table, we have,
0.6844 is the probability that sales on a given day at this store are less than $1,310.
Answer:
You can use ur ruler the ruler that's in half
I would be interested, they help because they allow you to see how much you can use for whatever it is you need. An example would be how much money you can spend per month to still have enough for your bills
Answer:
0.4
Step-by-step explanation:
Let X be the random variable that represents the number of consecutive days in which the parking lot is occupied before it is unoccupied. Then the variable X is a geometric random variable with probability of success p = 2/3, with probability function f (x) = [(2/3)^x] (1/3)
Then the probability of finding him unoccupied after the nine days he has been found unoccupied is:
P (X> = 10 | X> = 9) = P (X> = 10) / P (X> = 9). For a geometric aeatory variable:
P (X> = 10) = 1 - P (X <10) = 0.00002
P (X> = 9) = 1 - P (X <9) = 0.00005
Thus, P (X> = 10 | X> = 9) = P (X> = 10) / P (X> = 9) = 0.00002 / 0.00005 = 0.4.
The math club must sell 50 pies in order to reach the goal of 200 dollars
