A restaurant gat an average of 14 calls in a 2 hr time period. What is the probability that at most 2 calls in 45 min period

0.10512

Step-by-step explanation:

Given the following :

Mean number of calls(μ) in 2 hours = 14

2 hours = 60 * 2 = 120 minutes

Average number of calls in 45 minutes :

= (45 / 120) * 14

= 0.375 * 14

= 5.25

Now find P( x ≤ 2) = p(x = 0) + p( x = 1) + p(x = 2)

Using the poisson probability formula:

P(x, μ) = [(e^-μ) * (μ^x)] / x!

Where :

e = euler's constant

μ = 5.25

x = 0, 1, 2

Using the online poisson probability calculator :

P(x, 5.25) = P( x ≤ 2) = p(x = 0) + p(x = 1) + p(x = 2)

P(x, 5.25) = P( x ≤ 2) = 0.00525 + 0.02755 + 0.07232 = 0.10512

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!

the answer would be 44.79

plz mark brainliest

step-by-step explanation: