On average, there are 10 customers arrives a local bank per hour. Bob is the only bank teller on duty at the bank currently and needs to run out for 6 minutes. What is the probability that no customer will arrive in the next 6 minutes?

the probability that no customer will arrive in the next 6 minutes = 0.36788 = 0.368

Step-by-step explanation:

If there are 10 customers per hour, this translates to 1 customer per 6 minutes

So, if there's a mean of 1 customer per 6 minutes, to obtain the probability that no customer will come in a 6 minute interval, this becomes a Poisson distribution problem.

The Poisson distribution formula is given by

P(X = x) = (e^-λ)(λˣ)/x!

where λ = mean = 1 customer per 6 minutes

x = 0 customer per 6 minutes

P(X=0) = (e⁻¹)(1⁰)/0! = 0.36788 = 0.368

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

where are the images of the functions i can only see the question you typed.

step-by-step explanation: