Arrivals at a fast-food restaurant follow a poisson distribution with a mean arrival rate of 16 customers per hour. what is the

Answer:

0.40

Step-by-step explanation:

The percentage of members who engage only in long-distance running is 8%

Therefore, the probability that a member focuses solely on long-distance running is P(A) = 0.08

The percentage of members who participate exclusively in field events is 32%

Thus, the probability of a member competing only in field events is P(B) = 0.32

The percentage of members acting as sprinters is 12%

So, the probability that a member is a sprinter is P(C) = 0.12

We need to determine the probability that a team member is either an exclusive long-distance runner or an only field event competitor, which equates to finding P(A or B). Since these two events cannot occur simultaneously, we can express this as:

P(A or B) = P(A) + P(B)

Substituting the known values results in:

P(A or B) = 0.08 + 0.32 = 0.40

Thus, the likelihood that a randomly selected team member runs exclusively long-distance or participates solely in field events stands at 0.40

Response:

Detailed explanation:

Greetings!

You have the variable

X: Area eligible for painting with a can of spray paint (feet²)

This variable is normally distributed with a mean of μ= 25 feet² and a standard deviation of δ= 3 feet²

As this variable has a normal distribution, it needs to be converted into the standard normal form to utilize tabulated cumulative probabilities.

a.

P(X>27)

The first step involves standardizing the X value using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Having determined the Z value, you can find it in the table, but since the table includes probabilities for , the following conversion must be applied:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

b.

A sample of 20 cans was taken, and you need to ascertain the probability of averaging a coverage area of 540 feet².

The sample mean maintains the same distribution as its source variable, but its variance is influenced by sample size, thus it is normally distributed with parameters:

X[bar]~N(μ;δ²/n)

To cover 540 feet² with 20 cans, the average coverage must be approximately 540/20= 27 feet² per can.

c.

No, if the distribution is not normal and skewed, the normal distribution should not be applied for calculating probabilities. While the central limit theorem might approximate the sampling distribution to normal when the sample size is 30 or larger, that isn’t applicable here.

I trust this information is helpful!

Answer:

y

y = StartFraction 3 + 6 StartRoot 2 EndRoot Over 4 EndFraction y = StartFraction 3 menos 6 StartRoot 2 EndRoot Over 4 EndFraction

Explicación paso a paso:

La ecuación cuadrática que tenemos es (4y - 3)² = 72

Debemos encontrar el valor de y.

Ahora, 4y - 3 = ± 6√2

⇒ 4y = 3 ± 6√2

⇒ y

Por lo tanto, las soluciones son y = StartFraction 3 + 6 StartRoot 2 EndRoot Over 4 EndFraction y y = StartFraction 3 menos 6 StartRoot 2 EndRoot Over 4 EndFraction (Respuesta)