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Anuta_ua [19.1K]
3 years ago
8

A classroom is arranged with 8 seats in the front row, 10 seats in the middle row, and 12 seats in the back row. The teacher ran

domly assigns seats to students as they enter the classroom. What is the probability that the first student who enters the classroom will be assigned a seat in the front row?
Mathematics
1 answer:
asambeis [7]3 years ago
4 0

8/30 or 4/15 or 26%

8 of the 30 seats are in the front row

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Assume you purchased a car for $2,300 and sold it one week later for $3,100. How much did your net worth change, if at all?
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Answer:

My net wroth would have changed by $800

Step-by-step explanation:

Because $3,100 - $2,300 = $800

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Step-by-step explanation:

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2 years ago
The mean annual income for people in a certain city is 37 thousand dollars, with a standard deviation of 28 thousand dollars. A
Aloiza [94]

Answer:

P( 31 < \bar X< 41)

And we can ue the z score formula given by:

z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using this formula we got for the limits:

z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515

z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01

So we want to find this probability:

P(-1.515

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the annual income of a population, and for this case we know the following info:

\mu=37 and \sigma=28  and we are omitting the zeros from the thousand to simplify calculations

We select a sample size of n=50>30.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we want to find this probability:

P( 31 < \bar X< 41)

And we can ue the z score formula given by:

z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using this formula we got for the limits:

z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515

z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01

So we want to find this probability:

P(-1.515

4 0
3 years ago
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yuradex [85]

We use the ratio x:360 to find portions of the circle (again, through the ratio)

so we use 60/360 = x/30 in^2

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The answer is 5 in^2

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3 years ago
2.What is the solution of the system?
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The solution of the system is A. (2, 1)
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