A factory makes 10% defective items and items are independently defective. If a sample of 10 items is to be selected, find the probability that 9 or more are NOT defective in two ways. (Round to 3 decimal places) g

In simplifying complex fractions, first, put the individual fractions in simplest terms. Solve the equations based on the order of operations, simplify the fractions, find a common denominator, and combine the fractions. Your skill in Algebraic operations, particularly with polynomials and fractions, play a significant role in solving the problem.

When it comes to simplifying complex fractions, you need to first ensure that the individual fractions are in their simplest terms. For the fraction n-4/n², consider factoring out n from the denominator, while for 2n-15/n +1/n +3, the expression can be simplified to (2n-15+n+3)/n or (3n-12)/n.

Following an order of operations, evaluate the expressions and perform the indicated operations for a simple and correct answer. Briefly, your steps should include: simplifying the complex fractions, finding a common denominator, and combining the fractions. Once you have done this, continue to simplify until you are left with one fraction.

Also note that a key aspect to solving this is applying your knowledge of Algebraic Operations, specifically with polynomials and fractions. It's important to take care not to make any mistakes in signs or operations as they can significantly alter your responses.

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The simplified form is (3n² - n + 11)/n².

To simplify the complex fraction (n - 4)/(n²) - (2n - 15)/n + 1/n + 3, follow these steps:

Find a common denominator for all the fractions involved. Here, the common denominator will be n².

1. Rewrite each term with the common denominator:

(n - 4)/n²

(2n - 15)/n becomes (2n - 15)n/n²

1/n becomes n/n²

3 becomes 3n²/n²

2. Combine all terms to have the same denominator:

(n - 4 - 2n + 15 + n + 3n²)/n²

3. Simplify the numerator:

(n - 4 - 2n + 15 + n + 3n²)

Combine like terms: (3n² + n - 2n + n - 4 + 15)

4. Simplifies to: (3n² - n + 11)

5. Rewrite the fraction:

(3n² - n + 11)/n²

So, the simplified fraction is (3n² - n + 11)/n².

Answer:

(b) 0.251

Step-by-step explanation:

1. Standard deviation equation:

[tex]SD=\sqrt{\frac{\sum\limits^N_i {(x_{i}-X)^{2} } }{N} }[/tex]

Where X is the mean of the data, and N the amount of data. Then, N=5

2. Estimate the Mean:

[tex]X=\frac{0.750+0.750+0.200+0.600+0.200}{5}=\frac{2.5}{5}=0.5\\[/tex]

3. Caclulate Standard deviation:

[tex]SD=\sqrt{\frac{{(0.750-0.500)^{2}+(0.750-0.500)^{2}+(0.200-0.500)^{2}+(0.600-0.500)^{2}+(0.200-0.500)^{2} } }{5} }[/tex]

[tex]SD=\sqrt{\frac{0.315}{5} }=\sqrt{0.063}\\SD=0.251[/tex]

Answer: Third answer

Step-by-step explanation:

Answer:

yes, he was effectively a .400 hitter

Step-by-step explanation:

The mean rate of success ( ÿ ) is 0.394.

The total number of hits was 419.

Under table t the  value at 5% confidence interval is 1.96.  

Here se the asymptotic standard deviation is given as follows

se = [tex]\sqrt{\frac{y(1 - y)}{n} }[/tex]

     = [tex]\sqrt{\frac{0.394(1 - 0.394)}{419} }[/tex]

    = 0.02387  

The confidence interval is calculated as follows:  

y ± 1.96 * se

0.394 ± 1.96  * 0.02387  

the answer is  -0.347 or  0.4407)  

meaning that the mean lies between -0.347 to 0.4407.  

There is strong evidence that 0.4 lies between the above confidence interval.

This means that  Gwynn was effectively a 0.4 hitter

Tony Gwynn's batting average was .394 with 165 hits in 419 at bats. He would have needed approximately 3 more hits to officially reach a .400 average. Despite his impressive performance, it is factually incorrect to label him a .400 hitter for that season.

To test the hypothesis that Tony Gwynn was a .400 hitter in 1994 before the strike ended the season, we need to see if his actual batting average could be considered close to .400 even though it was .394.

Gwynn had 165 hits in 419 at bats, which gives a batting average of 165/419, or approximately .394 when rounded to three decimal places. To have a .400 average, he would need to have 419 * .400 hits, which equals 167.6. Therefore, he needed approximately 3 more hits (rounding up from 2.6) to achieve a .400 average with the same number of at bats.

Given that batting averages are not rounded up after the season ends, stating that Gwynn was a .400 hitter would be incorrect. However, his achievement of .394 is remarkably high, and some may argue that rounding up for discussion purposes could informally suggest he was 'effectively' a .400 hitter.

Answer:

Margin of error = 0.01344

Step-by-step explanation:

Margin of error = critical value × standard deviation.

critical value for 95% confidence interval = 1.96

Standard deviation = √[(p)(q)/n]

p = 0.4, q = 1 - p = 1 - 0.4 = 0.6, n = 5100

Standard deviation = √[(0.6×0.4)/5100] = 0.00686

Margin of error = 1.96 × 0.00686 = 0.0134

Answer:

(a) The probability of the event (X > 84) is 0.007.

(b) The probability of the event (X < 64) is 0.483.

Step-by-step explanation:

The random variable X follows a Poisson distribution with parameter λ = 64.

The probability mass function of a Poisson distribution is:

[tex]P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, ...[/tex]

(a)

Compute the probability of the event (X > 84) as follows:

P (X > 84) = 1 - P (X ≤ 84)

                [tex]=1-\sum _{x=0}^{x=84}\frac{e^{-64}(64)^{x}}{x!}\\=1-[e^{-64}\sum _{x=0}^{x=84}\frac{(64)^{x}}{x!}]\\=1-[e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{84}}{84!}]]\\=1-0.99308\\=0.00692\\\approx0.007[/tex]

Thus, the probability of the event (X > 84) is 0.007.

(b)

Compute the probability of the event (X < 64) as follows:

P (X < 64) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 63)

                [tex]=\sum _{x=0}^{x=63}\frac{e^{-64}(64)^{x}}{x!}\\=e^{-64}\sum _{x=0}^{x=63}\frac{(64)^{x}}{x!}\\=e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{63}}{63!}]\\=0.48338\\\approx0.483[/tex]

Thus, the probability of the event (X < 64) is 0.483.

To approximate probabilities for a Poisson distribution with mean 64, one can use the cumulative distribution function of the Poisson distribution. But for large means like 64, using a Normal approximation to the Poisson might be more accurate.

The question is about the Poisson distribution which is a probability distribution used often in statistical modelling where we are counting the number of times an event happens in a fixed interval of time or space. The mean is given as 64. To compute probabilities for values more than or less than a certain value, we use the cumulative distribution function (cdf) of the Poisson distribution.

The cumulative distribution function gives us the probability that the random variable is less than or equal to a certain value. In other words, if we want to find P(X > 84), we can find it by subtracting P(X <= 83) from 1. Likewise, to find P(X < 64), we can directly use the cdf at 63.

Please note however, because of the large mean (64), it may be more accurate to use a Normal approximation to the Poisson with mean 64 and variance 64. In that case, standardize the variables and use the Standard Normal table for the probabilities.

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Answer:

76 ≤ B ≤ 89

Step-by-step explanation:

Mean score (μ) = 73.3

Standard deviation (σ) = 9.7

If B Scores are below the top 5% and above the bottom 62%, then:

62% ≤ B ≤ 95%

In a normal distribution, the 62nd percentile has a corresponding z-score of z = 0.305, while the 95th percentile has a corresponding z-score of z = 1.650.

The grades X1 and X2 which are the limits for a B grade are given by:

[tex]z= \frac{X-\mu}{\sigma}\\0.305= \frac{X_1-73.3}{9,7}\\X_1=76.2\\1.650= \frac{X_2-73.3}{9,7}\\X_2=89.3[/tex]

Rounding to the nearest whole number, a B grade is given to grades between 76 and 89.

Final answer:

To calculate the numerical limits for a B grade in a normally distributed set of test scores with a mean of 73.3 and a standard deviation of 9.7, we find the z-scores corresponding to the top 5% and bottom 62%. The B grade range is approximately between 70 and 89.

Explanation:

To find the numerical limits for a B grade in a normally distributed set of test scores, we must determine the z-scores that correspond to the top 5% and the bottom 38% of scores since a B grade is assigned to scores below the top 5% and above the bottom 62% (which is equivalent to being below the top 38%). Given the mean score is 73.3 and the standard deviation is 9.7, we can use z-score formulas to convert these percentages to raw scores.

For the top 5% cutoff (z = 1.645), the score is calculated as follows:

Top 5% cutoff score = mean + (z x standard deviation)

Top 5% cutoff score = 73.3 + (1.645 x 9.7)

Top 5% cutoff score = 73.3 + 15.9565

Top 5% cutoff score ≈ 89

For the top 38% cutoff (z = -0.31), which is the bottom 62%, the score is calculated as:

Bottom 62% cutoff score = mean + (z x standard deviation)

Bottom 62% cutoff score = 73.3 + (-0.31 x 9.7)

Bottom 62% cutoff score = 73.3 + (-2.997)

Bottom 62% cutoff score ≈ 70

Therefore, a B grade on this test would be for scores roughly between 70 and 89.

Answer: 0.0935

Step-by-step explanation: p(business major) = 0.17, p(receiving financial aids) = 0.55

The two events are independent that's the occurrence of one does not affect the other.

Hence, the probability of that the student is a business major and receiving financial aid is given below as

p(business major and financial aid) = 0.17×0.55 = 0.0935.

Answer:

Which of the following is a quantitative continuous variable?

b. Distance a family travels on vacation

e. Amount spent on the vacation, to the nearest $100.

Step-by-step explanation:

Quantitative variables use numbers to express attributes, there are 2 types:

1. Quantitave continuous variables use numbers to express the attributes but, could have an infinite value between its maximum and minimum values, and are able to be divided into smaller values, such as in options b and e.

2. Quantitative discrete, also use numbers to express attributes, but they may have some, but not all the values, and must be integers, such as in options:

c. Number of vacations a family takes per year

d. Number of consecutive years that a family goes on a vacation (Let a trip of at least 200 miles from home be defined as "vacation").

e. Amount spent on the vacation, to the nearest $100.

And there are the cualitative variables, which use words to express attributes, such as in:

a. Where a family goes on vacation.