304 students went on a field trip. seven buses were filled and 17 students traveled in cars. how many students were in each bus?

Using the Normal distribution, the Empirical rule and the Central limit theorem, it is found that n is 36.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

In this problem, since 68% of the students had sample means between 48.5 and 51.5 mm, we have that:

Using one of them, we can find the standard error. So:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]1 = \frac{51.5 - 50}{s}[/tex]

[tex]s = 1.5[/tex]

Then, since [tex]\sigma = 9[/tex], the sample size is found solving the following equation:

[tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

[tex]1.5 = \frac{9}{\sqrt{n}}[/tex]

[tex]1.5\sqrt{n} = 9[/tex]

[tex]\sqrt{n} = \frac{9}{1.5}[/tex]

[tex](\sqrt{n})^2 = (\frac{9}{1.5})^2[/tex]

[tex]n = 36[/tex]

n is 36.

A similar problem is given at https://brainly.com/question/13002303

Answer:

B

Step-by-step explanation:

In a calendrical system where leap years occur every four years without any exceptions, there would be 37 leap years in a 150-year period.

In a calendrical system where leap years occur every four years without exception, the number of leap years in a 150-year period can be calculated with a simple division operation. This is because every 4 years there is a leap year consistent with the rules of this calendar system.

To calculate the maximum number of leap years in a 150-year period, you would divide 150 by 4 as this represents the frequency of leap years. This results in a total of 37.5. Given that a half year cannot be a leap year, we eliminate the fraction and we have 37 leap years.

While the Gregorian calendar also has a leap year every four years, there are exceptions for century years not divisible by 400. But since the question specified a system without such rule, there would be more leap years in this system compared to the Gregorian calendar for the same period of time.

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Answer:

x = 9 ±√23

Step-by-step explanation:

The probability that worker 1 will outproduce worker 2 in a single day is approximately 0.352 or 35.2%.

To find the probability that worker 1 will outproduce worker 2 in a single day, we need to compare their production rates using their respective means and standard deviations.

Let X be the production of worker 1 and Y be the production of worker 2. We are interested in finding P(X > Y).

Given:

Worker 1: X ~ N(77, 22^2)

Worker 2: Y ~ N(66, 19^2)

First, we standardize the variables:

Z1 = (X - 77) / 22

Z2 = (Y - 66) / 19

Now, we find P(X > Y) = P(X - Y > 0):

= P((X - 77) - (Y - 66) > 0)

= P((X - Y) > (77 - 66))

Now, we find the mean and standard deviation of the difference:

Mean of (X - Y) = 77 - 66 = 11

Standard deviation of (X - Y) = sqrt(22^2 + 19^2) = sqrt(484 + 361) = sqrt(845) ≈ 29.07

Now we standardize the difference:

Z = (11 - 0) / 29.07 ≈ 0.378

Finally, we find the probability using the standardized normal distribution table:

P(Z > 0.378) ≈ 0.352

So, the probability that worker 1 will outproduce worker 2 in a single day is approximately 0.352 or 35.2%.

the probability that Worker 1 will outproduce Worker 2 in a single day is approximately 0.7202 or 72.02%.

To find the probability that Worker 1 will outproduce Worker 2 in a single day, we can use the z-score formula for normal distributions.

The z-score formula is given by:

[tex]\[ z = \frac{{X - \mu}}{{\sigma}} \][/tex]

Where:

- ( X ) is the value we want to find the z-score for (in this case, Worker 1's productivity).

- [tex]\( \mu \)[/tex]is the mean (average) productivity.

- [tex]\( \sigma \)[/tex] is the standard deviation.

For Worker 1:

- [tex]\( X_1 = 77 \)[/tex] units per day (average productivity)

- [tex]\( \mu_1 = 77 \)[/tex]units per day (average productivity)

- [tex]\( \sigma_1 = 22 \)[/tex] units per day (standard deviation)

For Worker 2:

-[tex]\( X_2 = 66 \)[/tex]units per day (average productivity)

- [tex]\( \mu_2 = 66 \)[/tex]units per day (average productivity)

- [tex]\( \sigma_2 = 19 \)[/tex] units per day (standard deviation)

First, we find the z-score for Worker 1's productivity relative to Worker 2's productivity:

[tex]\[ z = \frac{{X_1 - \mu_2}}{{\sigma_2}} = \frac{{77 - 66}}{{19}} = \frac{{11}}{{19}} \approx 0.579 \][/tex]

Next, we look up the z-score in the standard normal distribution table or use a calculator/tool that can calculate probabilities from the standard normal distribution. The z-score of 0.579 corresponds to a probability of approximately 0.7202.

Therefore, the probability that Worker 1 will outproduce Worker 2 in a single day is approximately 0.7202 or 72.02%.

Answer:

The answer "attorney and notary fees" made me get the question wrong on brainly. The correct answer it showed me after I got it wrong was C. Credit check and administrative costs associated with loan processing

Step-by-step explanation:

The answer to the question is y=6x

Answer:

Option D [tex]\frac{b}{15}[/tex]

Step-by-step explanation:

Let

b-----> the number of bagels

we know that

To find the algebraic expression of "the bagels were divided equally into fifteen bags" , divided the number of bagels by fifteen

so

[tex]\frac{b}{15}[/tex]

We have been given that prices increase at a monthly rate of 0.5%.

Number of months in a year = 12

Let us suppose the item cost $1, then the percentage is given by

[tex]P=P_0(1+r)^n\\ \\ P=1(1+0.005)^{12}\\ \\ P=1.062[/tex]\\

Therefore, increase in a year is

[tex]1.062-1\\ =0.062[/tex]

Therefore, the required percentage in a year is 6.2%