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Vikentia [17]
3 years ago
9

Classify the following random variables as discrete or continuous:X: the number of automobile accidents per year in Virginia. Y

: the length of timeto play 18 holes of golf. M: the amount of milk produced yearly by a particular cow.N: the number of eggs laid each month by a hen. P: the number of building permitsissued each month in a certain city. Q: the weight of grain produced per acre.
Mathematics
1 answer:
Elena-2011 [213]3 years ago
3 0

Answer:

X: discrete

Y: continuous

M: continuous

N: discrete

P: discrete

Q: continuous

Step-by-step explanation:

First, we have to know the difference between discrete and continuous variables:

  • Discrete variables are those that represent things that are counted, 3 red cars, 2 chickens, etc.. They take positive integer values, {0, 1, 2, ..., n}, being [0, n] the interval from which the variable takes values, that means, there is a finite number of possible values.
  • Continuous variables are those that represent things that are measured, 3.56 km of railway laid, 5.77 l of paint used. They take positive real values, that means that in the interval used for the variable there are infinite possible values.

Now, we classify each variable:

  1. The number of automobile accidents per year in Virginia (X) is a discrete variable, as there can't be half an accident, one counts how many accidents are per year to know X.
  2. The length of time to play 18 holes of golf (Y) is a continuous variable, as it can take 2 hours, or 2.5 hours, or 2 hours, 30 minutes, 2 seconds, one measures how long it took to play 18 holes to know Y.
  3. The amount of milk produced yearly by a particular cow (M) is a continuous variable, as one measures how much milk was produced to know M.
  4. The number of eggs laid each month by a hen (N) is a discrete variable, as one counts how much eggs were laid to know N.
  5. The number of building permits issued each month in a certain city (P) is a discrete variable, as one counts how many permits were issued to know P.
  6. The weight of grain produced per acre (Q) is a continuous variable, as one measures the weight per acre to know Q.
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Mr. Dean had 4/5 he wanted to give his student 1/6 how many does he give them
Sunny_sXe [5.5K]

Answer:

\frac{2}{15}

Step-by-step explanation:

\frac{4}{5} ÷ 6 × 1 = \frac{2}{15}

=> He gave them \frac{2}{15}

8 0
3 years ago
At a new exhibit in the Museum of Science, people are asked to choose between 94 or 220 random draws from a machine. The machine
Tresset [83]

Answer:

0.0869 = 8.69% probability of getting more than 61% green balls.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

The machine is known to have 99 green balls and 78 red balls.

This means that p = \frac{99}{99+78} = 0.5593

Mean and standard deviation:

\mu = p = 0.5593

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5593*0.4407}{99+78}} = 0.0373

a. Calculate the probability of getting more than 61% green balls.

This is 1 subtracted by the pvalue of Z when X = 0.61. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{0.61 - 0.5593}{0.0373}

Z = 1.36

Z = 1.36 has a pvalue of 0.9131

1 - 0.9131 = 0.0869

0.0869 = 8.69% probability of getting more than 61% green balls.

8 0
2 years ago
Finding area of shaded sectors.
tekilochka [14]

Answer:

69.81 sq. m. (rounded to 2 decimal places)

Step-by-step explanation:

The sector of a circle is "part" or "portion" of a circle. The formula for the area of a sector is:

A=\frac{\theta}{360}*\pi r^2

Where

\theta  is the central angle

r is the radius

Given the figure, the arc is given as 80 degrees, but not the central angle of the shaded sector. But from geometry we know that the central angle and the intercepted arc have the same measure. So we can say:

\theta = 80

Also, the radius of the circle shown is 10 meters, so

r = 10

Now, we substitute in formula and find our answer:

A=\frac{\theta}{360}*\pi r^2\\A=\frac{80}{360}*\pi (10)^2\\A=\frac{2}{9}*100\pi\\A=\frac{200\pi}{9}\\A=69.81

Thus,

The area of the shaded sector is 69.81 sq. meters.

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3 years ago
A movie theater just opened 2 new theaters. One theater has 15 seats in each row. The second theater has 20 seats in each row. B
Lana71 [14]
15, 30, 45, 60, 75, 90, 105, 120, 135, 150
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Smallest number would be 60 because it’s the least common number in both theaters.

Let me know if I’m right,

-Hoodie
7 0
2 years ago
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Answer:

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Step-by-step explanation:

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