(g) If each customer takes 3 minutes to check out, what is the probability that it will take more than 6 minutes for all the customers currently in line to check out? The probability that it will take more than 6 minutes for all the customers currently in line to

Hello dear,

Your  question is not complete, please paste it complete for better understanding.

Anyhow what i understood from the question, the answer is in the explanation.

Step-by-step explanation:

The mean time between arrivals is 3 minutes. Or else, depending on the way the exponential distribution has been introduced to you, the rate is 1/3.

Recall that an exponential distribution with parameter λ has mean 1λ. So 1λ=3 and therefore λ=13.

A customer has just arrived. Let X be the waiting time until the next customer arrives. Then X has exponential distribution with parameter λ=13. For any positive x,

Pr(X≤x)=∫x013e−t/3dt=1−e−x/3.(1)

Now we can answer the questions. Interpretation is needed, since there are some ambiguities in the questions.

Interpret the question as saying: "Customer Alicia has just arrived. What is the probability that there will be a customer who arrives later than Alicia, but no more than 3 minutes later." Then we want  Pr(X≤3) . By (1), this is 1−e−3/3.

Interpret the question as saying that Alicia has just arrived, and we want the probability that there will be a gap of at least 6 minutes until the next customer arrives. Then we want

Pr(X>6) = 1−Pr(X≤6),

which is 1−(1−e−6/3).

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°

i think the amount would be $140.62